NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

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# NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

Edited By Ravindra Pindel | Updated on Sep 15, 2022 02:01 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 3 Matrices is one of the most interesting chapters to study. Matrices are a Mathematical tool that helps in finding the answer to linear equations. Matrices are much faster and efficient than the usual direct solving method. Matrices are not only used in Mathematics but also various other subjects like Economics, Genetics, etc. NCERT Exemplar Class 12 Math chapter 3 solutions covers various matrices related topics like the types, the operations, invertible matrices etc.

## More About NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

Matrix is a topic that is interesting and complex for some students. It is a highly scoring chapter of NCERT Class 12 Maths Solutions that a student can utilize to gain more score in their exams as well. But, the aim should not be about gaining more score only. Instead, students should focus on understanding the topic and its applications. Class 12 Math NCERT exemplar solutions chapter 3 is used not only in linear equations but has a widespread real-world application in higher education. It is used in genetics, modern psychology, economics, etc., therefore, building the base from the start is useful for students in Class 12.

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Question:1

If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements?

In mathematics, a matrix is a rectangular array which includes numbers, expressions, symbols and equations which are placed in an arrangement of rows and columns. The number of rows and columns that are arranged in the matrix is called as the order or dimension of the matrix. By rule, the rows are listed first and then the columns.
It is given that the matrix has 28 elements.
So, according to the rule of matrix,
If the given matrix has mn elements then the dimension of the order can be given by m$\ast$ n, where m and n are natural numbers.
So, if a matrix has 28 elements which is mn=28, then the following possible orders can be found:
$\because mn = 28$
Take m and n to be any number, so that, when they are multiplied, we get 28.
So, let m = 1 and n = 28.
Then, $m \times n = 1 \times 28 (=28)$
$\Rightarrow 1 \times 28$ is a possible order of the matrix with 28 elements
Take m = 2 and n = 14.
Then, $m \times n = 2 \times 14 (=28)$
$\Rightarrow 2 \times 14$ is a possible order of the matrix with 28 elements.
Take m = 4 and n = 7.
Then, $m \times n = 4 \times 7 (=28)$
$\Rightarrow 4 \times 7$ is a possible order of the matrix with 28 elements.
Take m = 7 and n = 4.
Then, $m \times n = 7 \times 4 (=28)$
$\Rightarrow 7 \times 4$ is a possible order of the matrix with 28 elements.
Take m = 14 and n = 2.
Then, $m \times n = 14 \times 2 (=28)$
$\Rightarrow 14 \times 2$ is a possible order of the matrix having 28 elements.
Take m = 28 and n = 1.
Then, $m \times n = 28 \times 1 (=28)$
$\Rightarrow 28 \times 1$ is a possible order of the matrix with 28 elements.
The following are the possible orders which a matrix having 28 elements can have:
$1 \times 28, 2 \times 14, 4 \times 7, 7 \times 4, 14 \times 2 and 28 \times 1$
If the given matrix consisted of 13 elements then its possible order can be found out in the similar way as given above:
Here, mn = 13.
Take m and n to be any number so that when multiplied we get 13
Take m = 1 and n = 13.
Then, $m \times n = 1 \times 13 (=13)$
$\Rightarrow 1 \times 13$ is a possible order of the matrix with 13 elements.
Take m = 13 and n = 1.
Then, $m \times n = 13 \times 1 (=13)$
$\Rightarrow 13 \times 1$ is a possible order of the matrix with 13 elements.
Thus, the possible orders of the matrix consisting of 13 elements are as follows:
$1 \times 13 and 13 \times 1$

Question:2

In the matrix $A=\left[\begin{array}{ccc} a & 1 & x \\ 2 & \sqrt{3} & x^{2}-y \\ 0 & 5 & -\frac{2}{5} \end{array}\right]$, write:
(i) The order of the matrix A
(ii) The number of elements
(iii) Write elements $a_{23}, a_{31}, a_{12}$

We have the matrix
A for element in (i=) 1st row and (j=) 2nd column.
$A=\left[\begin{array}{ccc} a & 1 & x \\ 2 & \sqrt{3} & x^{2}-y \\ 0 & 5 & -\frac{2}{5} \end{array}\right]$
A matrix, as we know, is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.
(i). We need to find the order of the matrix A.
And we know that,
The number of rows and columns that a matrix has is called its order or its dimension. By convention, rows are listed first; and columns second.
So,
Here, in matrix A:
There are 3 rows.
Elements in 1st row = a, 1, x
Elements in 2nd row = 2, $\sqrt 3, x^2 - y$
Elements in 3rd row = 0, 5, -2/5
$\Rightarrow M = 3$
And,
There are 3 columns.
Elements in 1st column = a, 2, 0
Elements in 2nd column = 1, $\sqrt 3, 5$
Elements in 3rd column = x, x2 – y, -2/5
$\Rightarrow N = 3$
Since, the order of matrix = $M \times N$
$\Rightarrow$ The order of matrix A = $3 \times 3$
Thus, the order of the matrix A is $3 \times 3$
(ii). We need to find the number of elements in the matrix A.
And we know that,
Each number that makes up a matrix is called an element of the matrix.
So,
If a matrix has M rows and N columns, the number of elements is MN.
Here, in matrix A:
There are 3 rows.
$\Rightarrow$ M = 3
And,
There are 3 columns.
$\Rightarrow$ N = 3
Then, number of elements = MN
$\Rightarrow$ Number of elements = $3 \times 3$
$\Rightarrow$ Number of elements = 9
The elements are namely, a, 2, 0, 1, $\sqrt 3$, 5, x, x2 – y, -2/5.
Thus, the number of elements is 9.
(iii). We need to find the elements a23, a31 and a12.
We know that,
aij is the representation of elements lying in the ith row and jth column.
For a23:
On comparing aij with a23, we get
i = 2
j = 3
Check in matrix A for element in (i=) 2nd row and (j=) 3rd column.
$A=\left[\begin{array}{ccc} a & 1 & x \\ 2 & \sqrt{3} & x^{2}-y \\ 0 & 5 & -\frac{2}{5} \end{array}\right]$
The element which is common in both 2nd row and 3rd column is x2 – y
$\Rightarrow a_{23 }= x^2 -y$
For a31:
On comparing aij with a31, we get
i = 3
j = 1
Check matrix A for element in $(i=) 3\textsuperscript{rd} row and (j=) 1\textsuperscript{st} column.$
$A=\left[\begin{array}{ccc} a & 1 & x \\ 2 & \sqrt{3} & x^{2}-y \\ 0 & 5 & -\frac{2}{5} \end{array}\right]$
The Element which is common in both 3rd row and 1st column is 0
$\Rightarrow$ a31 = 0
For a12:
On comparing aij with a12, we get
i = 1
j = 2
Check in matrix A for element in (i=) 1st row and (j=) 2nd column.
The element that is common between the first and second row is 1
⇒ a12 = 1
Thus, a23 = x2 - y, a31 = 0 and a12 = 1.

Question:3

Construct $a_{2\times2 }$ matrix where
(i) $a_{ij} =\frac{\left ( \hat i - 2 \hat j \right )}{2}$
(ii) $a_{ij} = |- 2i + 3j|$

We know that,
A matrix, us a rectangular formation in which symbols, numbers, alphabets and expressions are arranged in rows and columns.
Also,
We know that, the notation $A = [a_{ij}]_{m \times m}$ indicates that A is a matrix having the order $m \times n, also 1 \leq i \leq m, 1 \leq j \leq n; i, j \in N.$
(i).We need to construct a matrix, $a_{2 \times 2}$, where
$a_{ij} =\frac{\left ( \hat i - 2 \hat j \right )}{2}$
For $a_{2 \times 2},$
$1 \leq i \leq m$
$\Rightarrow 1 \leq i \leq 2 [\because m = 2]$
And,
$\\1 \leq j \leq n\\ \\ \Rightarrow 1 \leq j \leq 2 [\because n = 2]$
Put i = 1 and j = 1.
$\\ \mathrm{a}_{11}=\frac{(1-2(1))^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{(1-2)^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{(-1)^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{1}{2} \\ \text { Put } \mathrm{i}=1 \text { and } \mathrm{j}=2 \\ \mathrm{a}_{12}=\frac{(1-2(2))^{2}}{2} \\ \Rightarrow \mathrm{a}_{12}=\frac{(1-4)^{2}}{2}$
$\\ \Rightarrow a_{12}=\frac{(-3)^{2}}{2} \\ \Rightarrow a_{12}=\frac{9}{2} \\ \text { Put } i=2 \text { and } j=1 \\ a_{21}=\frac{(2-2(1))^{2}}{2} \\ \Rightarrow a_{21}=\frac{(2-2)^{2}}{2} \\ \Rightarrow a_{21}=\frac{0}{2} \\ \Rightarrow a_{21}=0 \\ \text { Put } i=2 \text { and } j=2 \\ a_{22}=\frac{(2-2(2))^{2}}{2}$
$\\ \Rightarrow \mathrm{a}_{22}=\frac{(2-4)^{2}}{2} \\ \Rightarrow \mathrm{a}_{22}=\frac{(-2)^{2}}{2} \\ \Rightarrow \mathrm{a}_{22}=\frac{4}{2} \\ \Rightarrow \mathrm{a}_{22}=2$
Let the matrix formed be named A.
\begin{aligned} &A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]\\ &\text { By substituting the values of } a_{11}, a_{12}, a_{21} \text { and } a_{22}, \end{aligned} the matrix formed is
$A=\left[\begin{array}{ll} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{array}\right]$
(ii). We need to construct a matrix, $a_{2 \times 2}$, where
$a\textsubscript{ij} = \vert -2i + 3j \vert$
For $a_{2 \times 2}$,
$1 \leq i \leq m$
$\Rightarrow 1 \leq i \leq 2 [\because m = 2]$
And,
$\\1 \leq j \leq n \\ \Rightarrow 1 \leq j \leq 2 [\because n = 2]$
Put i = 1 and j = 1.
$\\a\textsubscript{11} = \vert -2(1) + 3(1) \vert \\ \Rightarrow a\textsubscript{11} = \vert -2 + 3 \vert \\ \Rightarrow a\textsubscript{11} = \vert 1 \vert \\ \Rightarrow a\textsubscript{11} = 1$
Put i = 1 and j = 2.
$\\a\textsubscript{12} = \vert -2(1) + 3(2) \vert \\ \Rightarrow a\textsubscript{12} = \vert -2 + 6 \vert \\ \Rightarrow a\textsubscript{12} = \vert 4 \vert \\ \Rightarrow a\textsubscript{12} = 4$
Put i = 2 and j = 1.
$\\a\textsubscript{21} = \vert -2(2) + 3(1) \vert \\ \Rightarrow a\textsubscript{21} = \vert -4 + 3 \vert \\ \Rightarrow a\textsubscript{21} = \vert -1 \vert \\ \Rightarrow a\textsubscript{21} = 1$
Put i = 2 and j = 2. .
$\\a\textsubscript{22} = \vert -2(2) + 3(2) \vert \\ \Rightarrow a\textsubscript{22} = \vert -4 + 6 \vert \\ \Rightarrow a\textsubscript{22} = \vert 2 \vert \\ \Rightarrow a\textsubscript{22} = 2$
Let the matrix formed be A.
\begin{aligned} &A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]\\ &\text { By substituting the values of } a_{11}, a_{12}, a_{21} \text { and } a_{22}, \text { the matrix formed is }\\ &A=\left[\begin{array}{ll} 1 & 4 \\ 1 & 2 \end{array}\right] \end{aligned}

Question:4

Construct a 3 × 2 matrix whose elements are given by $a_{ij} = e^{ix}\sin jx$

A matrix, in mathematics is a rectangular array of numbers, alphabets, symbols, or expressions, arranged in rows and columns.
Also,
We know that, the notation $A = [a_{ij}]_{m \times m}$ indicates that the matrix A has the order of A $m \times n, also 1 \leq i \leq m, 1 \leq j \leq n; i, j \in N.$
We need to construct a $3 \times 2$ matrix whose elements are as follows:
$a\textsubscript{ij} = e\textsuperscript{i.x} sin jx$
For $a_{3 \times 2}:$
$\\1 \leq i \leq m \\ \Rightarrow 1 \leq i \leq 3 [\because m = 3] \\1 \leq j \leq n \\ \Rightarrow 1 \leq j \leq 2 [\because n = 2]$
Put i = 1 and j = 1.
$\\a\textsubscript{11} = e\textsuperscript{(1)x} sin (1)x \\ \Rightarrow a\textsubscript{11} = e\textsuperscript{x} sin x$
Put i = 1 and j = 2.

$\\a\textsubscript{12} = e\textsuperscript{(1)x} sin (2)x \\ \Rightarrow a\textsubscript{12} = e\textsuperscript{x} sin 2x$

Put i = 2 and j = 1.

$\\a\textsubscript{21} = e\textsuperscript{(2)x} sin (1)x \\ \Rightarrow a\textsubscript{21} = e\textsuperscript{2x}sin x$

Put i = 2 and j = 2.
$\\a\textsubscript{22} = e\textsuperscript{(2)x} sin (2)x \\ \Rightarrow a\textsubscript{22} = e\textsuperscript{2x} sin 2x$
For i = 3 and j = 1.
$\\a\textsubscript{31} = e\textsuperscript{(3)x} sin (1)x \\ \Rightarrow a\textsubscript{31} = e\textsuperscript{3x} sin x$
For i = 3 and j = 2.
$\\a\textsubscript{32} = e\textsuperscript{(3)x} sin (2)x \\ \Rightarrow a\textsubscript{32} = e\textsuperscript{3x} sin 2x$
Let the matrix formed be A.
\begin{aligned} &A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right]\\ &\text { By substituting the values of } a_{11}, a_{12}, a_{21}, a_{22}, a_{31} \text { and } a_{32}, \text { we get the following matrix }\\ &A=\left[\begin{array}{ll} e^{x} \sin x & e^{x} \sin 2 x \\ e^{2 x} \sin x & e^{2 x} \sin 2 x \\ e^{3 x} \sin x & e^{3 x} \sin 2 x \end{array}\right] \end{aligned}

Question:5

Find values of a and b if A = B, where
$\begin{array}{l} A=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right] \\ B=\left[\begin{array}{cc} 2 a+2 & b^{2}+2 \\ 8 & b^{2}-5 b \end{array}\right] \end{array}$

We have the matrices A and B, where
$\begin{array}{l} A=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right] \\ B=\left[\begin{array}{cc} 2 a+2 & b^{2}+2 \\ 8 & b^{2}-5 b \end{array}\right] \end{array}$
We need to find the values of a and b.
We know that, if
$\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right]$
Then,
$\\a\textsubscript{11} = b\textsubscript{11} \\a\textsubscript{12} = b\textsubscript{12} \\a\textsubscript{21} = b\textsubscript{21} \\a\textsubscript{22} = b\textsubscript{22}$
Also, A = B.
$\Rightarrow\left[\begin{array}{cc} \mathrm{a}+4 & 3 \mathrm{~b} \\ 8 & -6 \end{array}\right]=\left[\begin{array}{cc} 2 \mathrm{a}+2 & \mathrm{~b}^{2}+2 \\ 8 & \mathrm{~b}^{2}-5 \mathrm{~b} \end{array}\right]$
This means,
$\\a + 4 = 2a + 2 \ldots (i) \\3b = b\textsuperscript{2} + 2 \ldots (ii) \\ 8 = 8 \\ -6 = b^2 -5b \ldots (iii)$
From equation (i), we can find the value of a.
$\\a + 4 = 2a + 2 \\ \Rightarrow 2a - a = 4 - 2 \\ \Rightarrow a = 2$
From equation (ii), we can find the value of $b\textsuperscript{2}$.
$\\3b = b\textsuperscript{2} + 2 \\ \Rightarrow b\textsuperscript{2}= 3b -2$
By substituting the value of $b\textsuperscript{2}$ in equation (iii), we get
$\\-6 = b\textsuperscript{2} - 5b \\ \Rightarrow -6 = (3b - 2) - 5b \\ \Rightarrow -6 = 3b - 2 - 5b \\ \Rightarrow -6 = 3b - 5b - 2 \\ \Rightarrow -6 = -2b - 2 \\ \Rightarrow 2b = 6 - 2 \\ \Rightarrow 2b = 4$
\begin{aligned} &\Rightarrow \mathrm{b}=\frac{4}{2}\\ &\Rightarrow b=2\\ &\text { Hence, } a=2 \text { and } b=2 \end{aligned}

Question:6

If possible, find the sum of the matrices A and B, where

$\begin{array}{l} A=\left[\begin{array}{ll} \sqrt{3} & 1 \\ 2 & 3 \end{array}\right] \\ B=\left[\begin{array}{lll} x & y & z \\ a & b & 6 \end{array}\right] \end{array}$

According to the convention, the number of rows and columns in a matrix is called its order or dimension and the rows of the matrix are listed first and then the columns are listed.
We know that,
For adding or subtracting any two matrices, the need to be of the same order
That is,
If we need to add matrix A and B, then the order of matrix A is m x n then the order of matrix B should be m x n
We have matrices A and B, where
$\begin{array}{l} A=\left[\begin{array}{ll} \sqrt{3} & 1 \\ 2 & 3 \end{array}\right] \\ B=\left[\begin{array}{lll} x & y & z \\ a & b & 6 \end{array}\right] \end{array}$
We know what order of matrix is,
If a matrix has M rows and N columns, then the matrix has the order $M \times N.$
In matrix A:
Number of rows = 2
$\Rightarrow$ M = 2
Number of column = 2
$\Rightarrow$ N = 2
Then, order of matrix A = $M \times N$
$\Rightarrow$ Order of matrix A = $2 \times 2$
In matrix B:
Number of rows = 2
$\Rightarrow$ M = 2
Number of columns = 3
$\Rightarrow$ M = 3
Then, order of matrix B = $M \times N$
$\Rightarrow$ order of matrix B = $2 \times 3$
Since,
Order of matrix A $\neq$ Order of matrix B
$\Rightarrow$ Matrices A and B cannot be added.
Therefore, matrix A and matrix B cannot be added.

Question:7

If

$\begin{array}{l} X=\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right] \\ Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right] \end{array}$ find
(i) X + Y
(ii) 2X - 3Y
(iii) A matrix Z such that X + Y + Z is a zero matrix.

If you want to add or subtract any two matrices, make sure these two matrices have the same order
That is,
If A and B are two matrices and to add them, if we have order of A as m × n, then order of B must be m × n.
We have matrices X and Y, where
$\begin{array}{l} X=\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right] \\ Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right] \end{array}$

According to convention,
If a matrix has M rows and N columns, the order of matrix is $M \times N.$
(i). We need to find the X + Y.
Let us first find the order of X and Y.
Order of X:
Number of rows = 2
$\Rightarrow$ M = 2
Number of columns = 3
$\Rightarrow$ N = 3
Then, order of matrix X = $M \times N.$
$\Rightarrow$ Order of matrix X =$2 \times 3$
Order of Y:
Number of rows = 2
$\Rightarrow$ M = 2
Number of columns = 3
$\Rightarrow$ N = 3
Then, order of matrix Y = $M \times N.$
$\Rightarrow$ Order of matrix Y = $2 \times 3$
Since, order of matrix X = order of matrix Y
$\Rightarrow$ Matrices X and Y can be added.
So,
\begin{aligned} &X+Y=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]+\left[\begin{array}{ccc} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{lll} (3+2) & (1+1) & (-1-1) \\ (5+7) & (-2+2) & (-3+4) \end{array}\right]\\ &\Rightarrow \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right]\\ &\text { Thus, }\\ &X+Y=\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right] \end{aligned}
(ii). We need to find 2X - 3Y.
Let us calculate 2X.
We have,
\begin{aligned} &X=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]\\ &\text { Then, multiplying by } 2 \text { on both sides, we get }\\ &2 \mathrm{X}=2 \times\left[\begin{array}{ccc} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}=\left[\begin{array}{lll} 2 \times 3 & 2 \times 1 & 2 \times-1 \\ 2 \times 5 & 2 \times-2 & 2 \times-3 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}=\left[\begin{array}{ccc} 6 & 2 & -2 \\ 10 & -4 & -6 \end{array}\right] \end{aligned}
Also,
\begin{aligned} &Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]\\ &\text { Multiplying by } 3 \text { on both sides, we get }\\ &3 Y=3 \times\left[\begin{array}{rrr} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]\\ &\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{lll} 3 \times 2 & 3 \times 1 & 3 \times-1 \\ 3 \times 7 & 3 \times 2 & 3 \times 4 \end{array}\right]\\ &\Rightarrow 3 Y=\left[\begin{array}{ccc} 6 & 3 & -3 \\ 21 & 6 & 12 \end{array}\right] \end{aligned}
Now subtract 3Y from 2X.
\begin{aligned} &2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 6 & 2 & -2 \\ 10 & -4 & -6 \end{array}\right]-\left[\begin{array}{ccc} 6 & 3 & -3 \\ 21 & 6 & 12 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -11 & -10 & -18 \end{array}\right]\\ &\text { Thus, }\\ &2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -11 & -10 & -18 \end{array}\right] \end{aligned}
We need to find matrix Z, such that X + Y + Z is a zero matrix.
That is,
X + Y + Z = 0
Or,
Z = -X - Y
Or,
Z = -(X + Y)
We have already found X + Y in part (i).
So, from part (i):
\begin{aligned} &X+Y=\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right]\\ &\text { Then, }\\ &Z=-\left(\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right]\right)\\ &\Rightarrow \mathrm{Z}=\left[\begin{array}{ccc} -5 & -2 & 2 \\ -12 & 0 & -1 \end{array}\right]\\ &\mathrm{Thus},Z=\left[\begin{array}{ccc} -5 & -2 & 2 \\ -12 & 0 & -1 \end{array}\right] \end{aligned}

Question:8

Find non-zero values of x satisfying the matrix equation:
$\mathrm{x}\left[\begin{array}{cc} 2 \mathrm{x} & 2 \\ 3 & \mathrm{x} \end{array}\right]+2\left[\begin{array}{cc} 8 & 5 \mathrm{x} \\ 4 & 4 \mathrm{x} \end{array}\right]=2\left[\begin{array}{cc} \left(\mathrm{x}^{2}+8\right) & 24 \\ (10) & 6 \mathrm{x} \end{array}\right]$

A matrix, as we know, is a rectangular array which includes numbers, symbols, or expressions, arranged in rows and columns.
Also,
Addition or subtraction of any two matrices is possible only if they have the same order.
If a given matrix has m rows and n columns, then the order of the matrix is m x n.
We have matrix equation,
$\\x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]+2\left[\begin{array}{cc}8 & 5 x \\ 4 & 4 x\end{array}\right]=2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right] \\Take matrix \left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right] \\And multiply it with \mathrm{x}\\ x\left[\begin{array}{cc}2 x & 2 \\ 3 & x\end{array}\right]=\left[\begin{array}{cc}x \times 2 x & x \times 2 \\ x \times 3 & x \times x\end{array}\right]$
$\Rightarrow \mathrm{x}\left[\begin{array}{cc}2 \mathrm{x} & 2 \\ 3 & \mathrm{x}\end{array}\right]=\left[\begin{array}{cc}2 \mathrm{x}^{2} & 2 \mathrm{x} \\ 3 \mathrm{x} & \mathrm{x}^{2}\end{array}\right]$
Take matrix $\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]$
Multiply it with 2 ,
$\\ 2\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]=\left[\begin{array}{lll}2 \times 8 & 2 \times 5 x \\ 2 \times 4 & 2 \times 4 x\end{array}\right] \\\Rightarrow 2\left[\begin{array}{ll}8 & 5 x \\ 4 & 4 x\end{array}\right]=\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]_{\ldots .(i i)}$
Take matrix $\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]$
Multiply it with 2,
$\\ 2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]=\left[\begin{array}{cc}2 \times\left(x^{2}+8\right) & 2 \times 24 \\ 2 \times 10 & 2 \times 6 x\end{array}\right] \\\Rightarrow 2\left[\begin{array}{cc}\left(x^{2}+8\right) & 24 \\ 10 & 6 x\end{array}\right]=\left[\begin{array}{cc}2\left(x^{2}+8\right) & 48 \\ 20 & 12 x\end{array}\right] \ldots..(iii)$
By adding equation (i) and (ii) and make it equal to equation (iii), we get
$\\ \mathrm{x}\left[\begin{array}{cc}2 \mathrm{x} & 2 \\ 3 & \mathrm{x}\end{array}\right]+2\left[\begin{array}{ll}8 & 5 \mathrm{x} \\ 4 & 4 \mathrm{x}\end{array}\right]=2\left[\begin{array}{cc}\left(\mathrm{x}^{2}+8\right) & 24 \\ 10 & 6 \mathrm{x}\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}2 x^{2} & 2 x \\ 3 x & x^{2}\end{array}\right]+\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]=\left[\begin{array}{cc}2\left(x^{2}+8\right) & 48 \\ 20 & 12 x\end{array}\right]$
By adding left side of the matrix equation as they have same order, we get
$\Rightarrow\left[\begin{array}{cc}2 x^{2}+16 & 2 x+10 x \\ 3 x+8 & x^{2}+8 x\end{array}\right]=\left[\begin{array}{cc}2\left(x^{2}+8\right) & 48 \\ 20 & 12 x\end{array}\right]$
We need to find the value of x by comparing the elements in the two matrices.
If,
$\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right]$
Then,
$\\a\textsubscript{11} = b\textsubscript{11} \\a\textsubscript{12} = b\textsubscript{12} \\a\textsubscript{21} = b\textsubscript{21} \\a\textsubscript{22} = b\textsubscript{22}$
So,
$\\2x\textsuperscript{2} + 16 = 2(x\textsuperscript{2} + 8) \ldots (i) \\2x + 10x = 48 \ldots (ii) \\3x + 8 = 20 \ldots (iii) \\x\textsuperscript{2} + 8x = 12x \ldots (iv)$
We have got equations (i), (ii), (iii) and (iv) to solve for x.
So, take equation (i).
$\\2x\textsuperscript{2} + 16 = 2x\textsuperscript{2} + 16$
We cannot find the value of x from this equation as they are similar.
Now, take equation (ii).
2x + 10x = 48
$\Rightarrow 12x = 48$
$\Rightarrow x = 4$
From equation (iii),
$3x + 8 = 20$
$\\ \Rightarrow 3x = 20 - 8 \\ \Rightarrow 3x = 12$
$\Rightarrow x = 4$
From equation (iv),
$\\x\textsuperscript{2} + 8x = 12x \\ \Rightarrow x\textsuperscript{2} = 12x -8x \\ \Rightarrow x\textsuperscript{2} = 4x \\ \Rightarrow x\textsuperscript{2} - 4x = 0 \\ \Rightarrow x(x - 4) = 0 \\ \Rightarrow x = 0 or (x - 4) = 0 \\ \Rightarrow x = 0 or x = 4 \\ \Rightarrow x = 4 (\because x = 0 does not satisfy equations (ii) and (iii))$
So, by solving equations (ii), (iii) and (iv), we can conclude that
x = 4
Hence, the value of x is 4.

Question:9

If $A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]_{\text {and }} B=\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]$ , show that $(A + B) (A - B) \neq A^2 - B^2.$

We have the matrices A and B, where
$A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]_{\text {and }} B=\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]$
We need to prove that $(A + B) (A - B) \neq A^2 - B^2.$
Take L.H.S: (A + B) (A - B)
First, let us compute (A + B).
If two matrices have the same order, m x n, then they can be added or subtracted from each other. For example,
\begin{aligned} &\text { If we have matrices }\left[\begin{array}{ll} \mathrm{a}_{11} & \mathrm{a}_{12} \\ \mathrm{a}_{21} & \mathrm{a}_{22} \end{array}\right] \text { and }\left[\begin{array}{ll} \mathrm{b}_{11} & \mathrm{~b}_{12} \\ \mathrm{~b}_{21} & \mathrm{~b}_{22} \end{array}\right]_{.} \text {Then, they can be added as }\\ &\left[\begin{array}{ll} \mathrm{a}_{11} & \mathrm{a}_{12} \\ \mathrm{a}_{21} & \mathrm{a}_{22} \end{array}\right]+\left[\begin{array}{ll} \mathrm{b}_{11} & \mathrm{~b}_{12} \\ \mathrm{~b}_{21} & \mathrm{~b}_{22} \end{array}\right]=\left[\begin{array}{ll} \mathrm{a}_{11}+\mathrm{b}_{11} & \mathrm{a}_{12}+\mathrm{b}_{12} \\ \mathrm{a}_{21}+\mathrm{b}_{21} & \mathrm{a}_{22}+\mathrm{b}_{22} \end{array}\right]\\ &\text { So, }\\ &A+B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]\\ &\Rightarrow A+B=\left[\begin{array}{ll} 0+0 & 1-1 \\ 1+1 & 1+0 \end{array}\right]\\ \end{aligned}
$\Rightarrow A+B=\left[\begin{array}{ll} 0 & 0 \\ 2 & 1 \end{array}\right]$
Now, let us compute (A - B).
In the same manner, two matrices which have the same order can be subtracted.
So,
$\begin{array}{l} A-B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]-\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] \\ \Rightarrow A-B=\left[\begin{array}{ll} 0-0 & 1-(-1) \\ 1-1 & 1-0 \end{array}\right] \end{array}$
$\begin{array}{l} \Rightarrow A-B=\left[\begin{array}{cc} 0 & 1+1 \\ 0 & 1 \end{array}\right] \\ \Rightarrow A-B=\left[\begin{array}{ll} 0 & 2 \\ 0 & 1 \end{array}\right] \end{array}$
Now, let us compute (A + B) (A - B).
For multiplying two given matrices A and B, we must check if the number of columns in A are equal to the number of rows in B.
Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
$(A+B)(A-B)=\left[\begin{array}{ll} 0 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 0 & 2 \\ 0 & 1 \end{array}\right]$
$\left[\begin{array}{ll} 0 & 0 \\ 2 & 1 \end{array}\right]\left[\begin{array}{ll} 0 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\0 & 5 \end{array}\right]$
So, we get
$(A+B)(A-B)=\left[\begin{array}{ll} 0 & 0 \\ 0 & 5\end{array}\right]$
Take R.H.S: $A^2 - B^2$
Let us compute $A^2$ first.
$A^2$ = A.A
So, we need to compute A.A.
$A \cdot A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix A, then finally sum them up.
(0, 1).(0, 1) = (0 × 0) + (1 × 1)
⇒ (0, 1).(0, 1) = 0 + 1
⇒ (0, 1).(0, 1) = 1
Multiply 2nd row of matrix A by matching members of 2nd column of matrix A, and finally sum them up.
(1, 1).(1, 1) = (1 × 1) + (1 × 1)
⇒ (1, 1).(1, 1) = 1 + 1
⇒ (1, 1).(1, 1) = 2
$\Rightarrow(1,1):(1,1)=2 \left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right] \\So, A^{2}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right] \\Now, let us compute \mathrm{B}^{2}. B^{2}=B . B \\We need to compute B.B.\\ B . B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
Multiply 1st row of matrix B by matching members of 1st column of matrix B, and end by summing them up.
(0, -1).(0, 1) = (0 × 0) + (-1 × 1)
⇒ (0, -1).(0, 1) = 0 - 1
⇒ (0, -1).(0, 1) = -1
$\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}-1 & \end{array}\right]$
Multiply 1st row of matrix B by matching members of 2nd column of matrix B, and finally then
sum them up.
$\\(0,-1) :(-1,0)=(0 \times-1)+(-1 \times 0) \\\Rightarrow(0,-1) \cdot(-1,0)=0+0 \\\Rightarrow(0,-1) \cdot(-1,0)=0 \\\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0\end{array}\right]$
Multiply 2nd row of matrix B by matching members of 1st column of matrix B, and end by summing them up.
$\\ (1,0) \cdot(0,1)=(1 \times 0)+(0 \times 1) \\\Rightarrow(1,0) \cdot(0,1)=0+0 \\\Rightarrow(1,0) \cdot(0,1)=0 \\\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0\end{array}\right]$
Multiply 2nd row of matrix B by matching members of 2nd column of matrix B, and then finally sum them up.
$\\ (1,0) \cdot(-1,0)=(1 \times-1)+(0 \times 0) \\\Rightarrow(1,0) \cdot(-1,0)=-1+0 \\\Rightarrow(1,0) \cdot(-1,0)=-1 \\\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
So,
$B^{2}=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]$
Now, compute $A^{2}-B^{2}.$
$\\ A^{2}-B^{2}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] \\\Rightarrow A^{2}-B^{2}=\left[\begin{array}{cc}1-(-1) & 1-0 \\ 1-0 & 2-(-1)\end{array}\right] \\\Rightarrow A^{2}-B^{2}=\left[\begin{array}{cc}1+1 & 1 \\ 1 & 2+1\end{array}\right] \\\Rightarrow A^{2}-B^{2}=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]$
Evidently,
$(A+B)(A-B)=\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right]_{\text {and }} A^{2}-B^{2}=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right] are not equal.$
Thus, we get, $(A+B)(A-B) \neq A^{2}-B^{2}.$

Question:10

Find the value of x if
$\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=0$

The given matrix equation is,
$\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=0$
We need to determine the value of x.
Let us compute L.H.S: $\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]$
$\begin{array}{l} \text { Let, } A=\left[\begin{array}{lll} 1 & \text { X } & 1 \end{array}\right] \\ \text { B }=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right] \text { and } \\ C=\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right] \end{array}$
Multiplication of any two matrices is only possible when the number of columns in A is equal to the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
First, let us compute
$\text { A. } B=\left[\begin{array}{lll} 1 & \text { x } & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]=D(\text { say })$
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then sum them up.
$\\(1, x, 1).(1, 2, 15) = (1 \times 1) + (x \times 2) + (1 \times 15) \\ \Rightarrow (1, x, 1).(1, 2, 15) = 1 + 2x + 15 \\ \Rightarrow (1, x, 1).(1, 2, 15) = 2x + 16$
$\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]=[(2 x+16)$
Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then sum them up.
$\\(1, x, 1).(3, 5, 3) = (1 \times 3) + (x \times 5) + (1 \times 3) \\ \Rightarrow (1, x, 1).(3, 5, 3) = 3 + 5x + 3 \\ \Rightarrow (1, x, 1).(3, 5, 3) = 5x + 6$
$\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]=[(2 x+16) \quad(5 x+6) \quad]$
Multiply 1st row of matrix A by matching members of 3rd column of matrix B, then sum them up.
$\\(1, x, 1).(2, 1, 2) = (1 \times 2) + (x \times 1) + (1 \times 2) \\ \Rightarrow (1, x, 1).(2, 1, 2) = 2 + x + 2 \\ \Rightarrow (1, x, 1).(2, 1, 2) = x + 4$
$\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\end{array}\right]=[(2 x+16) \quad(5 x+6) \quad(x+4)]$
So,
$D=[(2 x+16) \quad(5 x+6) \quad(x+4)]$
Now compute
$\text { D. } C=[(2 x+16) \quad(5 x+6) \quad(x+4)]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]$
Multiply 1st row of matrix D by matching members of 1st column of matrix C, then sum them up.
$\\(2x + 16, 5x + 6, x + 4).(1, 2, x) = ((2x + 16) \times 1) + ((5x + 6) \times 2) + ((x + 4) \times x) \\ \Rightarrow (2x + 16, 5x + 6, x + 4).(1, 2, x) = (2x + 16) + (10x + 12) + (x\textsuperscript{2} + 4x) \\ \Rightarrow (2x + 16, 5x + 6, x + 4).(1, 2, x) = x\textsuperscript{2} + 2x + 10x + 4x + 16 + 12 \\ \Rightarrow (2x + 16, 5x + 6, x + 4).(1, 2, x) = x\textsuperscript{2} + 16x + 28$
\begin{aligned} &[(2 x+16) \quad(5 x+6) \quad(x+4)]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=\left[x^{2}+16 x+28\right]\\ &\text { So, we get, }\\ &\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=\left[x^{2}+16 x+28\right] \end{aligned}
Now, put L.H.S = R.H.S
$[x\textsuperscript{2} + 16x + 28] = [0]$
This means,
$\\x\textsuperscript{2} + 16x + 28 = 0 \\ \Rightarrow x\textsuperscript{2} + 14x + 2x + 28 = 0 \\ \Rightarrow x(x + 14) + 2(x + 14) = 0 \\ \Rightarrow (x + 2)(x + 14) = 0 \\ \Rightarrow (x + 2) = 0 or (x + 14) = 0 \\ \Rightarrow x = -2 or x = -14$
Thus, $x = -2, -14.$

Question:11

Show that $\begin{bmatrix} 5 &3 \\-1 &-2 \end{bmatrix}$ satisfies the equation $A^2 - 3A - 7I = 0$ and hence find $A^{-1}$.

We have the given matrix A, such that
$\begin{bmatrix} 5 &3 \\-1 &-2 \end{bmatrix}$
(i). We need to show that the matrix A satisfies the equation $A\textsuperscript{2} -3A - 7I = 0.$
(ii). Also, we need to find $A\textsuperscript{-1}.$
(i). Take L.H.S: $A\textsuperscript{2} - 3A - 7I$
First, compute $A\textsuperscript{2}.$
$A\textsuperscript{2} = A.A$
$A^{2}=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]$
By convention, if we have to multiple matrix A and B then the number of columns in matrix A should be equal to the number of rows in matrix B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
Multiply 1st row of matrix A by matching members of 1st column of matrix A, then sum them up.
$\\(5, 3).(5, -1) = (5 \times 5) + (3 \times -1) \\ \Rightarrow (5, 3).(5, -1) = 25 + (-3) \\ \Rightarrow (5, 3).(5, -1) = 25 - 3 \\ \Rightarrow (5, 3).(5, -1) = 22$
$\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix A, then sum them up.
$\\(5, 3).(3, -2) = (5 \times 3) + (3 \times -2) \\ \Rightarrow (5, 3).(3, -2) = 15 + (-6) \\ \Rightarrow (5, 3).(3, -2) = 15 - 6 \\ \Rightarrow (5, 3).(3, -2) = 9$
$\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & 9\\ & \end{bmatrix}$
Multiply 2nd row of matrix A by matching members of 1st column of matrix A, then sum them up.
$\\ (-1, -2).(5, -1) = (-1 \times 5) + (-2 \times -1) \\ \Rightarrow (-1, -2).(5, -1) = -5 + 2 \\ \Rightarrow (-1, -2).(5, -1) = -3$
$\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & 9\\ -3& \end{bmatrix}$
Multiply 2nd row of matrix A by matching members of 2nd column of matrix A, then sum them up.
$\\ (-1, -2).(3, -2) = (-1 \times 3) + (-2 \times -2) \\ \Rightarrow (-1, -2).(3, -2) = -3 + 4 \\ \Rightarrow (-1, -2).(3, -2) = 1$
$\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] = \begin{bmatrix} 22 & 9\\ -3& 1\end{bmatrix}$
$A^2 = \begin{bmatrix} 22 & 9\\ -3& 1\end{bmatrix}$
Substitute values of $A\textsuperscript{2}$ and A in $A\textsuperscript{2} - 3A - 7I.$
$A^{2}-3 A-7 I=\left[\begin{array}{ll} 22 & 9 \\ -3 & 1 \end{array}\right]-3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-7I$
Also, since matrix A is of the order 2 × 2, then I will be the identity matrix of order 2 × 2 such that,
$\begin{array}{l} \Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-3\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ \Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-\left[\begin{array}{cc} 3 \times 5 & 3 \times 3 \\ 3 \times-1 & 3 \times-2 \end{array}\right]-\left[\begin{array}{cc} 7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1 \end{array}\right] \\\\ \Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22 & 9 \\ -3 & 1 \end{array}\right]-\left[\begin{array}{cc} 15 & 9 \\ -3 & -6 \end{array}\right]-\left[\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right] \\ \\\Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22-15-7 & 9-9-0 \\ -3-(-3)-0 & 1-(-6)-7 \end{array}\right] \\ \\\Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 22-22 & 0 \\ -3+3 & 1+6-7 \end{array}\right] \\ \end{array}$
$\Rightarrow A^{2}-3 A-7 I=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]$
Hence proved,
L.H.S = R.H.S
Thus, we have shown that matrix A satisfy $A\textsuperscript{2} - 3A - 7I = 0.$
(ii). Now, let us find $A\textsuperscript{-1}.$
We know that, inverse of matrix A is $A\textsuperscript{-1}.$ is true only when
$A \times A\textsuperscript{-1} = A\textsuperscript{-1} \times A = I$
Where, I = Identity matrix
We get,
$A\textsuperscript{2} - 3A - 7I = 0$
Multiply $A\textsuperscript{-1}.$ on both sides, we get
$\\A\textsuperscript{-1}(A\textsuperscript{2} - 3A - 7I) = A\textsuperscript{-1} \times 0 \\ \Rightarrow A\textsuperscript{-1}.A\textsuperscript{2} - A\textsuperscript{-1}.3A - A\textsuperscript{-1}.7I = 0 \\ \Rightarrow A\textsuperscript{-1}.A.A - 3A\textsuperscript{-1}.A - 7A\textsuperscript{-1}.I = 0 \\ \Rightarrow (A\textsuperscript{-1}A)A - 3(A\textsuperscript{-1}A) - 7(A\textsuperscript{-1}I) = 0 \\ \text{And as } A\textsuperscript{-1}A = I \: \: and\: \: A\textsuperscript{-1}I = A\textsuperscript{-1} \\ \Rightarrow IA - 3I - 7A\textsuperscript{-1} = 0 \\ \text{Since, IA = A} \\ \Rightarrow A - 3I - 7A\textsuperscript{-1} = 0 \\ \Rightarrow 7A\textsuperscript{-1} = A - 3I$
\begin{aligned} &\Rightarrow \mathrm{A}^{-1}=\frac{1}{7}(\mathrm{~A}-3 \mathrm{I})\\ &\Rightarrow A^{-1}=\frac{1}{7}\left(\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-3\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\right)_{[\because} A=\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right] \text { and } I=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow A^{-1}=\frac{1}{7}\left(\left[\begin{array}{cc} 5 & 3 \\ -1 & -2 \end{array}\right]-\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]\right)\\ &\Rightarrow A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 5-3 & 3-0 \\ -1-0 & -2-3 \end{array}\right]\\ &\Rightarrow A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]\\ ,&A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & 3 \\ -1 & -5 \end{array}\right]\\ \end{aligned}

Question:12

Find the matrix A satisfying the matrix equation:

$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$

The given matrix equation is,
$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
We need to find matrix A.
Let matrix A be of order 2 × 2, and can be represented as
$A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$
Then, we have
\begin{aligned} &\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\text { Take L.H.S: }\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \mathrm{A}\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]\\ &\text { So, first let us calculate }\\ &\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right]=\mathrm{X} . \mathrm{Y}(\text { say }) \end{aligned}
If A and B are two given matrices and we have to multiply them, then the number of columns in matrix A should be equal to the number of rows in matrix B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
Multiply 1st row of matrix X by matching members of 1st column of matrix Y, then finally end by summing them up.
$\\(2, 1).(a, c) = (2 \times a) + (1 \times c) \\ \Rightarrow (2, 1).(a, c) = 2a + c$
$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right]=\left[\begin{array}{l} 2 \mathrm{a}+\mathrm{c} \end{array}\right]$
Multiply 1st row of matrix X by matching members of 2nd column of matrix Y, then finally end by summing them up.
$\\(2, 1).(b, d) = (2 \times b) + (1 \times d) \\ \Rightarrow (2, 1).(b, d) = 2b + d$
$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} 2 a+c & 2 b+d \end{array}\right]$
Multiply 2nd row of matrix X by matching members of 1st column of matrix Y, then finally end by summing them up.
$\\(3, 2).(a, c) = (3 \times a) + (2 \times c) \\ \Rightarrow (3, 2).(a, c) = 3a + 2c$
$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & \end{array}\right]$
Multiply 2nd row of matrix X by matching members of 2nd column of matrix Y, then finally end by summing them up.
$\\(3, 2).(b, d) = (3 \times b) + (2 \times d) \\ \Rightarrow (3, 2).(b, d) = 3b + 2d$
$\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]$
Let X.Y = Z
Now, we need to find $\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix}$
$Z.Q=\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix}$
Where, let $Q=\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix}$
Multiply 1st row of matrix Z by matching members of 1st column of matrix Q, then finally end by summing them up.
$\\(2a + c, 2b + d).(-3, 5) = ((2a + c) \times -3) + ((2b + d) \times 5) \\ \Rightarrow (2a + c, 2b + d).(-3, 5) = -6a - 3c + 10b + 5d \\ \Rightarrow (2a + c, 2b + d).(-3, 5) = -6a + 10b - 3c + 5d$
$\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d & \\ & \end{bmatrix}$
Multiply 1st row of matrix Z by matching members of 2nd column of matrix Q, then finally end by summing them up.
$\\(2a + c, 2b + d).(2, -3) = ((2a + c) \times 2) + ((2b + d) \times -3) \\ \Rightarrow (2a + c, 2b + d).(2, -3) = 4a + 2c - 6b - 3d \\ \Rightarrow (2a + c, 2b + d).(2, -3) = 4a - 6b + 2c - 3d$
$\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d &4a-6b+2c-3d \\ & \end{bmatrix}$
Multiply 2nd row of matrix Z by matching members of 1st column of matrix Q, then finally end by summing them up.
$\\(3a + 2c, 3b + 2d).(-3, 5) = ((3a + 2c) \times -3) + ((3b + 2d) \times 5) \\ \Rightarrow (3a + 2c, 3b + 2d).(-3, 5) = -9a - 6c + 15b + 10d \\ \Rightarrow (3a + 2c, 3b + 2d).(-3, 5) = -9a + 15b - 6c + 10d$
$\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d &4a-6b+2c-3d \\ -9a + 15b - 6c + 10d& \end{bmatrix}$
Multiply 2nd row of matrix Z by matching members of 2nd column of matrix Q, then finally end by summing them up.
$\\(3a + 2c, 3b + 2d).(2, -3) = ((3a + 2c) \times 2) + ((3b + 2d) \times -3) \\ \Rightarrow (3a + 2c, 3b + 2d).(2, -3) = 6a + 4c - 9b - 6d \\ \Rightarrow (3a + 2c, 3b + 2d).(2, -3) = 6a - 9b + 4c - 6d$
$\left[\begin{array}{cc} 2 a+c & 2 b+d \\ 3 a+2 c & 3b+2d \end{array}\right]\begin{bmatrix} -3 &2 \\5 &-3 \end{bmatrix} = \begin{bmatrix} -6a+10b-3c+5d &4a-6b+2c-3d \\ -9a + 15b - 6c + 10d& 6a - 9b + 4c - 6d\end{bmatrix}$
So, we have
${\left[\begin{array}{lll} 2 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{cc} -3 & 2 \\ 5 & -3 \end{array}\right]=\left[\begin{array}{ll} -6 a+10 b-3 c+5 d & 4 a-6 b+2 c-3 d \\ -9 a+15 b-6 c+10 d & 6 a-9 b+4 c-6 d \end{array}\right]} \\$
Now, for L . H . S=R . H . S
${\left[\begin{array}{ll} -6 a+10 b-3 c+5 d & 4 a-6 b+2 c-3 d \\ -9 a+15 b-6 c+10 d & 6 a-9 b+4 c-6 d \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]}$
If the matrices have the same order then we can write them as,
$\\-6a + 10b - 3c + 5d = 1 \ldots (i) \\4a - 6b + 2c - 3d = 0 \ldots (ii) \\-9a + 15b - 6c + 10d = 0 \ldots (iii) \\6a - 9b + 4c - 6d = 1 \ldots (iv)$
We have to find four variables: a, b, c, d and four equations
So, on adding equations (i) and (iv), we get
$\\(-6a + 10b - 3c + 5d) + (6a - 9b + 4c - 6d) = 1 + 1 \\ \Rightarrow -6a + 6a + 10b - 9b - 3c + 4c + 5d - 6d = 2 \\ \Rightarrow 0 + b + c - d = 2 \\ \Rightarrow d = b + c - 2 \ldots (v)$
Now, adding equations (ii) and (iii), we get
$\\(4a - 6b + 2c - 3d) + (-9a + 15b - 6c + 10d) = 0 + 0 \\ \Rightarrow 4a - 9a - 6b + 15b + 2c - 6c - 3d + 10d = 0 \\ \Rightarrow -5a + 9b - 4c + 7d = 0 \ldots (vi)$
By adding equations (iv) and (vi), we get
$\\(6a - 9b + 4c - 6d) + (-5a + 9b - 4c + 7d) = 1 + 0 \\ \Rightarrow 6a - 5a - 9b + 9b + 4c - 4c - 6d + 7d = 1 \\ \Rightarrow a + 0 + 0 + d = 1 \\ \Rightarrow d = 1 - a \ldots (vii)$
Substituting the value of d from equation (vii) in (v), we get
$\\(1 - a) = b + c - 2 \\ \Rightarrow b + c - 2 - 1 = -a \\ \Rightarrow b + c - 3 = -a \\ \Rightarrow a = 3 - b - c \ldots (viii)$
Now, by substituting values of a and d from equations (vii) and (viii) in equation (iii), we get
$\\-9(3 - b - c) + 15b - 6c + 10(1 - a) = 0 \\ \Rightarrow -9(3 - b - c) + 15b - 6c + 10(1 - (3 - b - c)) = 0 [\because a = 3 - b - c] \\ \Rightarrow -27 + 9b + 9c + 15b - 6c + 10(1 - 3 + b + c) = 0 \\ \Rightarrow -27 + 9b + 9c + 15b - 6c + 10(-2 + b + c) = 0 \\ \Rightarrow -27 + 9b + 9c + 15b - 6c - 20 + 10b + 10c = 0 \\ \Rightarrow 9b + 15b + 10b + 9c - 6c + 10c - 27 - 20 = 0 \\ \Rightarrow 34b + 13c - 47 = 0 \\ \Rightarrow 34b + 13c = 47 \ldots (ix)$
Also, substituting values of a and d from equations (vii) and (viii) in equation (ii), we get
$\\4(3 - b - c) - 6b + 2c - 3(1 - a) = 0 \\ \Rightarrow 12 - 4b - 4c - 6b + 2c - 3(1 - (3 - b - c)) = 0 \\ \Rightarrow 12 - 4b - 4c - 6b + 2c - 3(1 - 3 + b + c) = 0 \\ \Rightarrow 12 - 4b - 4c - 6b + 2c - 3(-2 + b + c) = 0 \\ \Rightarrow 12 - 4b - 4c - 6b + 2c + 6 - 3b - 3c = 0 \\ \Rightarrow -4b - 6b - 3b - 4c + 2c - 3c + 12 + 6 = 0 \\ \Rightarrow -13b - 5c + 18 = 0 \\ \Rightarrow 13b + 5c = 18 \ldots (x)$
On multiplication of equation (ix) by 5 and equation (x) by 13, we get
$\\(ix) \Rightarrow 5(34b + 13c) = 5 \times 47 \\ \Rightarrow 170b + 65c = 235 \ldots (xi) \\(x) \Rightarrow 13(13b + 5c) = 13 \times 18 \\ \Rightarrow 169b + 65c = 234 \ldots (xii)$
By subtracting equations (xi) and (xii), we get
$\\(170b + 65c) - (169b + 65c) = 235 - 234 \\ \Rightarrow 170b - 169b + 65c - 65c = 1 \\ \Rightarrow b = 1$
By substituting b = 1 in equation (x), we get
$\\13(1) + 5c = 18 \\ \Rightarrow 13 + 5c = 18 \\ \Rightarrow 5c = 18 - 13 \\ \Rightarrow 5c = 5$
$\\ \Rightarrow c = 1$
By substituting b = 1 and c = 1 in equation (viii), we get
$\\a = 3 - b - c \\ \Rightarrow a = 3 - 1 - 1 \\ \Rightarrow a = 3 - 2 \\ \Rightarrow a = 1$
By substituting a = 1 in equation (vii), we get
$\\d = 1 - a \\ \Rightarrow d = 1 - 1 \\ \Rightarrow d = 0$
Thus, the matrix A is
$A= \begin{bmatrix} 1 & 1\\1 & 0 \end{bmatrix}$

Question:13

Find A, if $\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right] \mathrm{A}=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]$.

We have the matrix,
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right] \mathrm{A}=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]$
We need to find the matrix A.
Let us check what the order of the given matrices is.
We know that order of a matrix is the number of rows and columns in a matrix.
If a given matrix has M rows and N columns, the order of matrix is M × N.
Order of $\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]= X(say)$
Number of rows = 3
$\Rightarrow M = 3$
Number of column = 1
$\Rightarrow N = 1$
Then, order of matrix X $=M \times N$
$\Rightarrow$ Order of matrix X $= 3 \times 1$
Order of $\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]=\mathrm{Y}(\text { say })$
Number of rows = 3
$\Rightarrow$ M = 3
Number of columns = 3
$\Rightarrow$ N = 3
Then, order of matrix Y $= M \times N$
$\Rightarrow$ Order of matrix Y $= 3 \times 3$
We must note that, when a matrix of order $1 \times 3$ is multiplied to the matrix X, only then matrix Y is produced.
Let matrix A be of order $1 \times 3$, and can be represented as
$A=\left[\begin{array}{lll}a & b & c\end{array}\right] Then, we have \left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right] Take \left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}a & b & c\end{array}\right]$
In order to carry out the multiplication of two matrices, A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
So, we get,
$\text { X. } A=\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]$
Multiply 1st row of matrix X by matching member of 1st column of matrix A, then finally end by summing it up.
(4)(a) = 4a
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & \\ & \end{bmatrix}$
Multiply 1st row of matrix X by matching member of 2nd column of matrix A, then finally end by summing it up.
(4)(b) = 4b
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b \\ & \end{bmatrix}$
Multiply 1st row of matrix X by matching member of 3rd column of matrix A, then finally end by summing it up.
(4)(c) = 4c
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ & \end{bmatrix}$
Multiply 2nd row of matrix X by matching member of 1st column of matrix A, then finally end by summing it up.
(1)(a) = a
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& \end{bmatrix}$
Multiply 2nd row of matrix X by matching member of 2nd column of matrix A, then finally end by summing it up.
(1)(b) = b
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b\end{bmatrix}$
Multiply 2nd row of matrix X by matching member of 3rd column of matrix A, then finally end by summing it up.
(1)(c) = c
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\end{bmatrix}$
Multiply 3rd row of matrix X by matching member of 1st column of matrix A, then finally end by summing it up.
(3)(a) = 3a
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\\ 3a & & \end{bmatrix}$
Multiply 3rd row of matrix X by matching member of 2nd column of matrix A, then finally end by summing it up.
(3)(b) = 3b
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\\ 3a & 3b & \end{bmatrix}$
Multiply 3rd row of matrix X by matching member of 3rd column of matrix A, then finally end by summing it up.
(3)(c) = 3c
$\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\\ 3a & 3b &3c \end{bmatrix}$
Now, L.H.S = R.H.S
$\begin{array}{l} \Rightarrow\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & \mathrm{b} & \mathrm{c} \end{array}\right]=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 4 \mathrm{a} & 4 \mathrm{~b} & 4 \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \mathrm{c} \\ 3 \mathrm{a} & 3 \mathrm{~b} & 3 \mathrm{c} \end{array}\right]=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right] \end{array}$
Since, the matrices have the same order, we can say,
$\\4a = -4 \ldots (i) \\4b = 8 \ldots (ii) \\4c = 4 \ldots (iii) \\a = -1 \ldots (iv) \\b = 2 \ldots (v) \\c = 1 \ldots (vi) \\3a = -3 \ldots (vii) \\3b = 6 \ldots (viii) \\3c = 3 \ldots (ix)$
From equation (i), we can determine the value of a,
4a = -4
$\Rightarrow a = -1$
From equation (ii), we can determine the value of b,
4b = 8
$\Rightarrow b = 2$
From equation (iii), we can determine the value of c,
4c = 4
$\Rightarrow c = 1$
And it will satisfy other equations (iv), (v), (vi), (vii), (viii) and (ix) too.
Thus, the matrix A is
$A= \begin{bmatrix} -1 &2 &1 \end{bmatrix}$

Question:14

If \begin{aligned} &A=\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ccc} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right] \end{aligned} then verify $(BA)^{2} \neq B^{2} A^{2}$

We have the following matrices,
\begin{aligned} \\ &A=\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ccc} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right] \end{aligned}
We need to verify $(BA)\textsuperscript{2} \neq B\textsuperscript{2}A\textsuperscript{2}.$
Take L.H.S: $(BA)\textsuperscript{2}$
First, compute BA.
$\text { B. } A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]$
We understand what a order of matrix is,
If a matrix has M rows and N columns, the order of matrix is M × N.
Order of matrix B:
Number of rows = 2
⇒ M = 2
Number of columns = 3
⇒ N = 3
Then, order of matrix = M × N
⇒ Order of matrix B = 2 × 3
Order of matrix A:
Number of rows = 3
⇒ M = 3
Number of columns = 2
⇒ N = 2
Then, order of matrix = M × N
⇒ Order of matrix A = 3 × 2
If we have two given matrices A and B which need to be multiplied, then the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
So, A and B can be multiplied.
$\text { B. } A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]$

Multiply 1st row of matrix B by matching member of 1st column of matrix A, then finally end by summing them up.
$\\(2, 1, 2)(3, 1, 2) = (2 \times 3) + (1 \times 1) + (2 \times 2) \\ \Rightarrow (2, 1, 2)(3, 1, 2) = 6 + 1 + 4 \\ \Rightarrow (2, 1, 2)(3, 1, 2) = 11$
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 & \\ & \end{bmatrix}$
Multiply 1st row of matrix B by matching member of 2nd column of matrix A, then finally end by summing them up.
$\\(2, 1, 2)(-4, 1, 0) = (2 \times -4) + (1 \times 1) + (2 \times 0) \\ \Rightarrow (2, 1, 2)(-4, 1, 0) = -8 + 1 + 0 \\ \Rightarrow (2, 1, 2)(-4, 1, 0) = -7$
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 &-7 \\ & \end{bmatrix}$
Multiply 2nd row of matrix B by matching member of 1st column of matrix A, then finally end by summing them up.
$\\(1, 2, 4)(3, 1, 2) = (1 \times 3) + (2 \times 1) + (4 \times 2) \\ \Rightarrow (1, 2, 4)(3, 1, 2) = 3 + 2 + 8 \\ \Rightarrow (1, 2, 4)(3, 1, 2) = 13$
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 &-7 \\ 13& \end{bmatrix}$
Multiply 2nd row of matrix B by matching member of 2nd column of matrix A, then finally end by summing them up.
$\\ (1, 2, 4)(-4, 1, 0) = (1 \times -4) + (2 \times 1) + (4 \times 0) \\ \Rightarrow (1, 2, 4)(-4, 1, 0) = -4 + 2 + 0 \\ \Rightarrow (1, 2, 4)(-4, 1, 0) = -2$
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 11 &-7 \\ 13& -2\end{bmatrix}$
So,
$(BA)\textsuperscript{2} = (BA).(BA)$
\begin{aligned} &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{ll} 11 & -7 \\ 13 & -2 \end{array}\right]\left[\begin{array}{ll} 11 & -7 \\ 13 & -2 \end{array}\right]\\ &\text { Similarly, }\\ &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{ll} (11 \times 11+(-7) \times 13) & (11 \times-7+(-7) \times(-2)) \\ (13 \times 11+(-2) \times 13) & (13 \times-7+(-2) \times(-2)) \end{array}\right]\\ &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{cc} 121-91 & -77+14 \\ 143-26 & -91+4 \end{array}\right]\\ &\Rightarrow(\mathrm{BA})^{2}=\left[\begin{array}{cc} 30 & -63 \\ 117 & -87 \end{array}\right] \end{aligned}
Take R.H.S: $B\textsuperscript{2}A\textsuperscript{2}$
Let us first compute $B\textsuperscript{2}.$
$B\textsuperscript{2} = B.B$
$\Rightarrow \mathrm{B}^{2}=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$
For multiplication of two matrices, say A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
Note that in matrix B, number of columns is not equal to the number of rows.
Which means, we can’t find $B\textsuperscript{2}.$
$\Rightarrow L.H.S \neq R.H.S$
Thus, we have verified that, $(BA)\textsuperscript{2} \neq B\textsuperscript{2}A\textsuperscript{2}.$

Question:15

If possible, find BA and AB, where
$A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]$

We are given matrices A and B, such that
$A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]$
We are required to find BA and AB, if possible.
To carry out the multiplication of matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
Let us check for BA.
$\mathrm{BA}=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$
If a matrix has M rows and N columns, the order of matrix is M × N.
Order of B:
Number of rows = 3
⇒ M = 3
Number of columns = 2
⇒ N = 2
Then, order of matrix B = M × N
⇒ Order of matrix B = 3 × 2
Order of A:
Number of rows = 2
⇒ M = 2
Number of columns =3
⇒ N = 3
Then, order of matrix A = M × N
⇒ Order of matrix A = 2 × 3
Here,
Number of columns in matrix B = Number of rows in matrix A = 2
So, BA is possible.
Let us check for AB.
$\mathrm{AB}=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]$
Here,
Number of columns in matrix A = Number of rows in matrix B = 3
So, AB is also possible.
Let us find out BA.
$\mathrm{BA}=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]$
Multiply 1st row of matrix B by matching members of 1st column of matrix A, then finally end by summing them up.
$\\(4, 1).(2, 1) = (4 \times 2) + (1 \times 1) \\ \Rightarrow (4, 1).(2, 1) = 8 + 1 \\ \Rightarrow (4, 1).(2, 1) = 9$
$\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]= \begin{bmatrix} 9 & \\ & \\ & \end{bmatrix}$
Multiply 1st row of matrix B by matching members of 2nd column of matrix A, then finally end by summing them up
$\\(4, 1).(1, 2) = (4 \times 1) + (1 \times 2) \\ \Rightarrow (4, 1).(1, 2) = 4 + 2 \\ \Rightarrow (4, 1).(1, 2) = 6$
$\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]= \begin{bmatrix} 9 & 6 \\ & \\ & \end{bmatrix}$
Similarly, let us calculate in the matrix itself.
$\begin{array}{l} {\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]} \\ \Rightarrow\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 9 & 6 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{lll} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{array}\right] \end{array}$
Now, let us find out AB.
$A B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally end by summing it up.
$\\(2, 1, 2).(4, 2, 1) = (2 \times 4) + (1 \times 2) + (2 \times 1) \\ \Rightarrow (2, 1, 2).(4, 2, 1) = 8 + 2 + 2 \\ \Rightarrow (2, 1, 2).(4, 2, 1) = 12$
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]= \begin{bmatrix} 12 & \\ & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then finally end by summing it up.
$\\(2, 1, 2).(1, 3, 2) = (2 \times 1) + (1 \times 3) + (2 \times 2) \\ \Rightarrow (2, 1, 2).(1, 3, 2) = 2 + 3 + 4 \\ \Rightarrow (2, 1, 2).(1, 3, 2) = 9$
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]= \begin{bmatrix} 12 & 9\\ & \\ & \end{bmatrix}$
Similarly, let us calculate in the matrix itself.
$\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]=\begin{bmatrix} 12 &9 \\ (1 \times 4)+(2 \times 2)+(4 \times 1) & (1 \times 1)+(2 \times 3)+(4 \times 2) \end{bmatrix}$
\begin{aligned} &\Rightarrow\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 12 & 9 \\ 4+4+4 & 1+6+8 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right]\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 12 & 9 \\ 12 & 15 \end{array}\right]\\ &A B=\left[\begin{array}{cc} 12 & 9 \\ 12 & 15 \end{array}\right]_{\text {and }} B A=\left[\begin{array}{ccc} 9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10 \end{array}\right] \end{aligned}

Question:16

Show by an example that for A ≠ O, B ≠ O, AB = O.

We know,
To multiply the given matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
We are given that,
A ≠ 0 and B ≠ 0
We need to show that, AB = 0.
For multiplication of A and B,
Number of columns of matrix A = Number of rows of matrix B = 2 (let)
Matrices A and B are square matrices of order 2 × 2.
For AB to become 0, one of the column of matrix A and other row of matrix B must be 0.
For example,
\begin{aligned} &A=\left[\begin{array}{ll} 0 & 1 \\ 0 & 4 \end{array}\right]\\ &B=\left[\begin{array}{cc} 3 & -1 \\ 0 & 0 \end{array}\right]\\ &\text { Check: Multiply AB. }\\ &A B=\left[\begin{array}{ll} 0 & 1 \\ 0 & 4 \end{array}\right]\left[\begin{array}{cc} 3 & -1 \\ 0 & 0 \end{array}\right] \end{aligned}
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally end by summing them up.
$\\(0, 1).(3, 0) = (0 \times 3) + (1 \times 0) \\ \Rightarrow (0, 1).(3, 0) = 0 + 0 = 0$
$\left[\begin{array}{ll}0 & 1 \\ 0 & 4\end{array}\right]\left[\begin{array}{cc}3 & -1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & \end{array}\right]\\\\ Similarly, let us do it for the rest of the elements.\\\\ \left[\begin{array}{cc}0 & 1 \\ 0 & 4\end{array}\right]\left[\begin{array}{cc}3 & -1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{cc}0 & (0 \times-1)+(1 \times 0) \\ (0 \times 3)+(4 \times 0) & (0 \times-1)+(4 \times 0)\end{array}\right]\\\\ \Rightarrow\left[\begin{array}{ll}0 & 1 \\ 0 & 4\end{array}\right]\left[\begin{array}{cc}3 & -1 \\ 0 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\\\\ Hence proved.$

Question:17

Given $A=\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]$ . Is $(AB)' = B'A'$?

We have two given matrices A and B,
$A=\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]$
We need to verify whether $(AB)' = B'A'$
Let us see what a transpose is.
In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as $A^T$.
Take L.H.S = $(AB)'$
So, let us compute AB.
$A B=\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally end by summing it up.
$\\(2, 4, 0)(1, 2, 1) = (2 \times 1) + (4 \times 2) + (0 \times 1) \\ \Rightarrow (2, 4, 0)(1, 2, 1) = 2 + 8 + 0 \\ \Rightarrow (2, 4, 0)(1, 2, 1) = 10$
$\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then finally end by summing them up.
$\\(2, 4, 0)(4, 8, 3) = (2 \times 4) + (4 \times 8) + (0 \times 3) \\ \Rightarrow (2, 4, 0)(4, 8, 3) = 8 + 32 + 0 \\ \Rightarrow (2, 4, 0)(4, 8, 3) = 40$
$\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\ & \end{bmatrix}$
Similarly, let us fill for the rest of the elements.
$\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\(3\times1)+(9\times2)+(6\times1) & (3\times4)+(9\times8)+(6\times3) \end{bmatrix}$
$\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\3+18+6 & 12+72+18 \end{bmatrix}$
$\left[\begin{array}{lll} 2 & 4 & 0 \\ 3 & 9 & 6 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{array}\right]= \begin{bmatrix} 10 &40 \\27 & 102 \end{bmatrix}$
So, $AB= \begin{bmatrix} 10 &40 \\27 & 102 \end{bmatrix}$
Now, for transpose of AB, rows will become columns.
$(AB)'= \begin{bmatrix} 10 &27 \\40 & 102 \end{bmatrix}$
Now, take R.H.S = B’A’
If $B = \begin{bmatrix} 1 &4 \\2 &8 \\1 &3 \end{bmatrix}$
Then, if (1, 4) are the elements of 1st row, it will become elements of 1st column, and so on.
$B' = \begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix}$
Also,
$A = \begin{bmatrix} 2 &4 &0 \\3 &9 &6 \end{bmatrix}$
Then, if (2, 4, 0) are the elements of 1st row, it will become elements of 1st column, and so on.
$A'= \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix}$
Now, multiply B’A’.
$B' A'= \begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix}$$\begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix}$
Multiply 1st row of matrix B’ by matching members of 1st column of matrix A’, then finally end by summing them up.
$\\(1, 2, 1)(2, 4, 0) = (1 \times 2) + (2 \times 4) + (1 \times 0) \\ \Rightarrow (1, 2, 1)(2, 4, 0) = 2 + 8 + 0 \\ \Rightarrow (1, 2, 1)(2, 4, 0) = 10$.

$\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 & & \\ & & \\ & & \end{bmatrix}$
Multiply 1st row of matrix B’ by matching members of 2nd column of matrix A’, then finally end by summing it up.
$\\(1, 2, 1)(3, 9, 6) = (1 \times 3) + (2 \times 9) + (1 \times 6) \\ \Rightarrow (1, 2, 1)(3, 9, 6) = 3 + 18 + 6 \\ \Rightarrow (1, 2, 1)(3, 9, 6) = 27$
$\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 & \\ & & \\ & & \end{bmatrix}$
Filling up the rest of the elements in the similar manner
$\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 \\ 4*2+8*4+3*0 &4*3+8*9+ 3*6 \end{bmatrix}$
$\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 \\8+32+0 &12+72+ 18 \end{bmatrix}$
$\begin{bmatrix} 1 &2 &1 \\4 &8 &3 \end{bmatrix} \begin{bmatrix} 2 &3 \\4 &9 \\0 &6 \end{bmatrix} = \begin{bmatrix} 10 &27 \\40 &102\end{bmatrix}$
⇒ L.H.S = R.H.S
Therefore, $(AB)' = B'A'$

Question:18

Solve for x and y:
$\mathrm{x}\left[\begin{array}{l} 2 \\ 1 \end{array}\right]+\mathrm{y}\left[\begin{array}{l} 3 \\ 5 \end{array}\right]+\left[\begin{array}{c} -8 \\ -11 \end{array}\right]=0$

We are given with the following matrix equation,
$x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{l}-8 \\ -11\end{array}\right]=0$
We need to find x and y
$\\x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{l}-8 \\ -11\end{array}\right]=0 \\\Rightarrow\left[\begin{array}{l}2 \mathrm{x} \\ \mathrm{x}\end{array}\right]+\left[\begin{array}{l}3 \mathrm{y} \\ 5 \mathrm{y}\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=0$
These matrices can be added easily as they are of same order.
$\Rightarrow\left[\begin{array}{l}2 x+3 y-8 \\ x+5 y-11\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right]$
If two matrices are equal, then their corresponding elements of the same matrices are also equal.
This implies,
$\\2x + 3y - 8 = 0 \ldots (i) \\x + 5y - 11 = 0 \ldots (ii)$
We have two variables, x and y; and two equations. It can be solved.
By rearranging equation (i), we get
$2x + 3y = 8 \ldots (iii)$
By rearranging equation (ii), then multiplying it by 2 on both sides, we get
$\\x + 5y = 11 \\2(x + 5y) = 2 \times 11 \\ \Rightarrow 2x + 10y = 22 \ldots (iv)$
By subtracting equation (iii) from (iv), we get
$\\(2x + 10y) - (2x + 3y) = 22 - 8 \\ \Rightarrow 2x + 10y - 2x - 3y = 14 \\ \Rightarrow 2x - 2x + 10y - 3y = 14 \\ \Rightarrow 7y = 14$
$\\ \Rightarrow y = 2$
By substituting y = 2 in equation (iii), we get
$\\2x + 3(2) = 8 \\ \Rightarrow 2x + 6 = 8 \\ \Rightarrow 2x = 8 - 6 \\ \Rightarrow 2x = 2$
$\Rightarrow x = 1$
Thus, x = 1 and y = 2

Question:19

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y
$2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]$

We have the given matrix equations,
$2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]$
By subtracting equation (i) from (ii), we get
$\begin{array}{l} (3 X+2 Y)-(2 X+3 Y)=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]-\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \\ \Rightarrow 3 X+2 Y-2 X-3 Y=\left[\begin{array}{cc} -2-2 & 2-3 \\ 1-4 & -5-0 \end{array}\right] \\ \Rightarrow 3 X-2 X+2 Y-3 Y=\left[\begin{array}{cc} -4 & -1 \\ -3 & -5 \end{array}\right] \\ \Rightarrow X-Y=\left[\begin{array}{ll} -4 & -1 \\ -3 & -5 \end{array}\right] \end{array}$
By adding equations (i) and (ii), we get
\begin{aligned} &(3 X+2 Y)+(2 X+3 Y)=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]+\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]\\ &\Rightarrow 3 \mathrm{X}+2 \mathrm{Y}+2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{cc} -2+2 & 2+3 \\ 1+4 & -5+0 \end{array}\right]\\ &\Rightarrow 3 \mathrm{X}+2 \mathrm{X}+2 \mathrm{Y}+3 \mathrm{Y}=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow 5 X+5 Y=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow 5(X+Y)=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow X+Y=\frac{1}{5}\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{ll} \frac{1}{5} \times 0 & \frac{1}{5} \times 5 \\ \frac{1}{5} \times 5 & \frac{1}{5} \times-5 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right] \end{aligned}
By adding equations (iii) and (iv), we get
$\\ (\mathrm{X}-\mathrm{Y})+(\mathrm{X}+\mathrm{Y})=\left[\begin{array}{ll} -4 & -1 \\ -3 & -5 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right] \\ \Rightarrow \mathrm{X}-\mathrm{Y}+\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll} -4+0 & -1+1 \\ -3+1 & -5-1 \end{array}\right] \\ \Rightarrow \mathrm{X}+\mathrm{X}-\mathrm{Y}+\mathrm{Y}=\left[\begin{array}{ll} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow 2 \mathrm{X}=\left[\begin{array}{lc} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\frac{1}{2}\left[\begin{array}{lc} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\left[\begin{array}{ll} \frac{1}{2} \times-4 & \frac{1}{2} \times 0 \\ \frac{1}{2} \times-2 & \frac{1}{2} \times-6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]$
Substituting the matrix A in equation (iv), we get
$\begin{array}{l} {\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]+\mathrm{Y}=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right]} \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right]-\left[\begin{array}{cc} -2 & 0 \\ -1 & -3 \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 0-(-2) & 1-0 \\ 1-(-1) & -1-(-3) \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 2 & 1 \\ 1+1 & -1+3 \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 2 & 1 \\ 2 & 2 \end{array}\right] \\ \mathrm{X}=\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]_{\text {and }} \mathrm{Y}=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right] \end{array}$

Question:20

If $A = [3\: \: 5], B = [7\: \: 3]$, then find a non-zero matrix C such that AC = BC.

We have the given matrices A and B, such that
$A = [3\: \: 5], B = [7\: \: 3]$
We need to find matrix C, such that AC = BC.
Let C be a non-zero matrix of order 2 × 1, such that
$C=\begin{bmatrix} X\\Y \end{bmatrix}$
But order of C can be 2 × 1, 2 × 2, 2 × 3, 2 × 4, …
[ if we have to multiply two given matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
∴, number of columns in matrix A = number of rows in matrix C = 2]
Take AC.
$AC=\begin{bmatrix} 3 &5 \end{bmatrix}\begin{bmatrix} X\\Y \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 1st column of matrix C, then finally sum them up.
\begin{aligned} &(3,5)(x, y)=(3 \times x)+(5 \times y)\\ &\Rightarrow(3,5)(x, y)=3 x+5 y\\ &\left[\begin{array}{ll} 3 & 5 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \end{array}\right]=[3 \mathrm{x}+5 \mathrm{y}]\\ &\Rightarrow A C=[3 x+5 y]\\ &\text { Now, take BC. }\\ &\mathrm{BC}=\left[\begin{array}{ll} 7 & 3 \end{array}\right]\left[\begin{array}{l} \mathrm{X} \\ \mathrm{y} \end{array}\right] \end{aligned}
Multiply 1st row of matrix B by matching members of 1st column of matrix C, then finally sum them up,
$\begin{array}{l} (7,3)(x, y)=(7 \times x)+(3 \times y) \\ \Rightarrow(7,3)(x, y)=7 x+3 y \\ {[7 \quad 3]\left[\begin{array}{l} x \\ y \end{array}\right]=[7 x+3 y]} \\ \Rightarrow B C=[7 x+3 y] \end{array}$

And,
AC = BC
$\\ \Rightarrow [3x + 5y] = [7x + 3y] \\ \Rightarrow 3x + 5y = 7x + 3y \\ \Rightarrow 7x - 3x = 5y - 3y \\ \Rightarrow 4x = 2y \\ \Rightarrow y = 2x$
Then, we have,
$C=\left[\begin{array}{l} x \\ 2 x \end{array}\right]$
since, $\mathrm{C} is of orders, 2 \times 1,2 \times 2,2 \times 3, \ldots$
$C=\left[\begin{array}{c}x \\ 2 x\end{array}\right]=\left[\begin{array}{cc}x & x \\ 2 x & 2 x\end{array}\right]=\left[\begin{array}{ccc}x & x & x \\ 2 x & 2 x & 2 x\end{array}\right]=\cdots$
In general,
$C=\left[\begin{array}{l} \mathrm{k} \\ 2 \mathrm{k} \end{array}\right]=\left[\begin{array}{ll} \mathrm{k} & \mathrm{k} \\ 2 \mathrm{k} & 2 \mathrm{k} \end{array}\right]=\left[\begin{array}{ccc} \mathrm{k} & \mathrm{k} & \mathrm{k} \\ 2 \mathrm{k} & 2 \mathrm{k} & 2 \mathrm{k} \end{array}\right]=\cdots$
Where, k can be any real number.

Question:21

Given an example of matrices A, B and C such that AB = AC, where A is non-zero matrix, but B ≠ C.

We need to form matrices A, B and C such that AB = AC, where A is a non-zero matrix, but B ≠ C.
Take,
\begin{aligned} &A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]\\ &C=\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]\\ &\text { First, compute AB. }\\ &A B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right] \end{aligned}
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.
$\\(1, 0)(1, 2) = (1 \times 1) + (0 \times 2) \\ \Rightarrow (1, 0)(1, 2) = 1 + 0 \\ \Rightarrow (1, 0)(1, 2) = 1$
$\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 1 & \\ & \end{bmatrix}$
Similarly, let us do the same for other elements.
$\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & (1 \times 3)+(0 \times 0) \\ (0 \times 1)+(0 \times 2) & (0 \times 3)+(0 \times 0) \end{array}\right]$
$AB=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 1 &3 \\0 & 0\end{bmatrix}$
Now, let us compute AC.
$AC=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix C, then finally sum them up.
$\\(1, 0)(1, 2) = (1 \times 1) + (0 \times 2) \\ \Rightarrow (1, 0)(1, 2) = 1 + 0 \\ \Rightarrow (1, 0)(1, 2) = 1$
$\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]=\begin{bmatrix} 1 & \\ & \end{bmatrix}$
Similarly, let us do the same for other elements.
$\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 2 \end{array}\right]=\left[\begin{array}{cc} 1 & (1 \times 3)+(0 \times 2) \\ (0 \times 1)+(0 \times 2) & (0 \times 3)+(0 \times 2) \end{array}\right]$
$AC=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]=\begin{bmatrix} 1 &3 \\0 & 0\end{bmatrix}$
Clearly, AB = AC. but B ≠ C.
Hence, we have found an example which fulfills the required criteria.

Question:22

If $A=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]_{\text {and }} C=\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]$, verify:
(i) $(AB) C = A (BC)$
(ii) $A(B + C) = AB + AC$

We have the given matrices A, B and C, such that
$A=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]_{\text {and }} C=\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]$
To multiply two given matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
(i). We need to verify: (AB)C = A(BC)
Take L.H.S = (AB)C
First, compute AB.
$AB=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.
$\\(1, 2)(2, 3) = (1 \times 2) + (2 \times 3) \\ \Rightarrow (1, 2)(2, 3) = 2 + 6 \\ \Rightarrow (1, 2)(2, 3) = 8$
$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\begin{bmatrix} 8 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then finally sum them up.
$\\(1, 2)(3, -4) = (1 \times 3) + (2 \times -4) \\ \Rightarrow (1, 2)(3, -4) = 3 - 8 \\ \Rightarrow (1, 2)(3, -4) = -5$
$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\begin{bmatrix} 8 &-5 \\ & \end{bmatrix}$
Similarly, let us repeat for the rest of the elements.
\begin{aligned} &\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ (-2 \times 2)+(1 \times 3) & (-2 \times 3)+(1 \times-4) \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ -4+3 & -6-4 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\\ &\text { Let } D=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\\ &\text { Now, compute for DC. }[\because(A B) C=D C]\\ &\mathrm{DC}=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] \end{aligned}
Multiply 1st row of matrix D by matching members of 1st column of matrix C, then finally sum them up.
$\\(8, -5)(1, -1) = (8 \times 1) + (-5 \times -1) \\ \Rightarrow (8, -5)(1, -1) = 8 + 5 \\ \Rightarrow (8, -5)(1, -1) = 13$
$\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] = \begin{bmatrix} 13 & \\ & \end{bmatrix}$
Multiply 1st row of matrix D by matching members of 2nd column of matrix C, then finally sum them up.
$\\(8, -5)(0, 0) = (8 \times 0) + (-5 \times 0) \\ \Rightarrow (8, -5)(0, 0) = 0 + 0 \\ \Rightarrow (8, -5)(0, 0) = 0$
$\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] = \begin{bmatrix} 13 & 0\\ & \end{bmatrix}$
Similarly, let us repeat for the rest of the elements.
$\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ (-1 \times 1)+(-10 \times-1) & (-1 \times 0)+(-10 \times 0)\end{array}\right]\\ \\\Rightarrow\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ -1+10 & 0\end{array}\right]\\ \\\Rightarrow\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right] \\So, (\mathrm{AB}) \mathrm{C}=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right] \\Take R.H.S: \mathrm{A}(\mathrm{BC}) \\First, compute BC. \\\mathrm{BC}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
Multiply 1st row of matrix B by matching members of 1st column of matrix C, then finally sum them up.
$\\(2, 3)(1, -1) = (2 \times 1) + (3 \times -1) \\ \Rightarrow (2, 3)(1, -1) = 2 - 3 \\ \Rightarrow (2, 3)(1, -1) = -1$
$\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]= \begin{bmatrix} -1 & \\ & \end{bmatrix}$
Multiply 1st row of matrix B by matching members of 2nd column of matrix C, then finally sum them up.
$\\(2, 3)(0, 0) = (2 \times 0) + (3 \times 0) \\ \Rightarrow (2, 3)(0, 0) = 0 + 0 \\ \Rightarrow (2, 3)(0, 0) = 0$
$\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]= \begin{bmatrix} -1 & 0\\ & \end{bmatrix}$
Similarly, let us repeat for the rest of the elements.
\begin{aligned} &\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ (3 \times 1)+(-4 \times-1) & (3 \times 0)+(-4 \times 0) \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 3+4 & 0 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]\\ &\text { Let } E=\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] .\\ &\text { Now, compute for AE. }\\ &A E=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] \end{aligned}
Multiply 1st row of matrix A by matching members of 1st column of matrix E, then finally sum them up.
$\\(1, 2)(-1, 7) = (1 \times -1) + (2 \times 7) \\ \Rightarrow (1, 2)(-1, 7) = -1 + 14 \\ \Rightarrow (1, 2)(-1, 7) = 13$
$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] =\begin{bmatrix} 13 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix E, then finally sum them up.
$\\(1, 2)(0, 0) = (1 \times 0) + (2 \times 0) \\ \Rightarrow (1, 2)(0, 0) = 0 + 0 \\ \Rightarrow (1, 2)(0, 0) = 0$
$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] =\begin{bmatrix} 13 &0 \\ & \end{bmatrix}$
Similarly, let us fill for other elements.
${\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ (-2 \times-1)+(1 \times 7) & (-2 \times 0)+(1 \times 0) \end{array}\right]} \\ \Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ 2+7 & 0 \end{array}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ 9 & 0 \end{array}\right] \\ \text { So, } \\ A(B C)=\left[\begin{array}{cc} 13 & 0 \\ 9 & 0 \end{array}\right] \\ \text { Thus, }(A B) C=A(B C)$
We need to verify: A(B + C) = AB + AC
ii)Take L.H.S: A(B + C)
Now, by Adding B + C, we get,
$\\ \mathrm{B}+\mathrm{C}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right] \\\Rightarrow B+C=\left[\begin{array}{cc}2+1 & 3+0 \\ 3-1 & -4+0\end{array}\right] \\\Rightarrow \mathrm{B}+\mathrm{C}=\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] \\Let B+C=F, such that F=\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] \\Now, by multiplying A and F, we get,\\ A F=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix F, then finally sum yhem up.
$\\(1, 2)(3, 2) = (1 \times 3) + (2 \times 2) \\ \Rightarrow (1, 2)(3, 2) = 3 + 4 \\ \Rightarrow (1, 2)(3, 2) = 7$
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] = \begin{bmatrix} 7 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix F, then finally sum them up.
$\\(1, 2)(3, -4) = (1 \times 3) + (2 \times -4) \\ \Rightarrow (1, 2)(3, -4) = 3 - 8 \\ \Rightarrow (1, 2)(3, -4) = -5$
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] = \begin{bmatrix} 7 &-5 \\ & \end{bmatrix}$
Similarly, let us fill for other elements.
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ (-2 \times 3)+(1 \times 2) & (-2 \times 3)+(1 \times-4)\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -6+2 & -6-4\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right] \\So, A(B+C)=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right] \\Now, take R.H.S: \mathrm{AB}+\mathrm{AC} \\Compute AB. \\A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.
$\\(1, 2)(2, 3) = (1 \times 2) + (2 \times 3) \\ \Rightarrow (1, 2)(2, 3) = 2 + 6 \\ \Rightarrow (1, 2)(2, 3) = 8$
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] = \begin{bmatrix} 8 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then finally sum them up.
$\\(1, 2)(3, -4) = (1 \times 3) + (2 \times -4) \\ \Rightarrow (1, 2)(3, -4) = 3 - 8 \\ \Rightarrow (1, 2)(3, -4) = -5$
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] = \begin{bmatrix} 8 &-5 \\ & \end{bmatrix}$
Similarly, let us fill for other elements.
$\\\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ (-2 \times 2)+(1 \times 3) & (-2 \times 3)+(1 \times-4)\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -4+3 & -6-4\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right] \\So, A B=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right] \\Now, compute AC. \\A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix C, then finally sum them up..
$\\(1, 2)(1, -1) = (1 \times 1) + (2 \times -1) \\ \Rightarrow (1, 2)(1, -1) = 1 - 2 \\ \Rightarrow (1, 2)(1, -1) = -1$
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right] =\begin{bmatrix} -1 & \\ & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix C, then finally sum them up.
$\\(1, 2)(0, 0) = (1 \times 0) + (2 \times 0) \\ \Rightarrow (1, 2)(0, 0) = 0 + 0 \\ \Rightarrow (1, 2)(0, 0) = 0$
$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right] =\begin{bmatrix} -1 &0 \\ & \end{bmatrix}$
Similarly, let us fill for the other elements.
$\\\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ (-2 \times 1)+(1 \times-1) & (-2 \times 0)+(1 \times 0)\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ -2-1 & 0\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right] \\So, \\A C=\left[\begin{array}{ll}-1 & 0 \\ -3 & 0\end{array}\right] \\Now, by Adding A B+A C. \\A B+A C=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]$
If two matrices have the same order, they can be added or subtracted.
\begin{aligned} &\Rightarrow A B+A C=\left[\begin{array}{cc} 8-1 & -5+0 \\ -1-3 & -10+0 \end{array}\right]\\ &\Rightarrow \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{cc} 7 & -5 \\ -4 & -10 \end{array}\right]\\ &\text { Hence proved, L.H.S }=\mathrm{R.H.S.}\\ &\text { Thus, } A(B+C)=A B+A C . \end{aligned}

Question:23

$\text{If } P=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right], Q=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]$ prove that $P Q=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right]=Q P$.

We have the following given matrices P and Q, such that
$P=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right], Q=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]$
We have to prove that:
$\quad\\ P Q=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right]=Q P$
Proof: First, we shall compute PQ.
$\mathrm{PQ}=\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]$

For carrying out the multiplication of two matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.
Order of P = 3 × 3
And order of Q = 3 × 3
Number of columns of matrix P = Number of rows of matrix Q = 3
So, P and Q can be multiplied.
So, multiply 1st row of matrix P by matching members of 1st column of matrix Q, then finally sum them up.
(x, 0, 0)(a, 0, 0) = (x × a) + (0 × 0) + (0 × 0)
⇒ (x, 0, 0)(a, 0, 0) = xa
$\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right] = \begin{bmatrix} xa & & \\ & & \\ & & \end{bmatrix}$
Multiply 1st row of matrix P by matching members of 2nd column of matrix Q, then finally sum them up
$\\(x, 0, 0)(a, 0, 0) = (x \times a) + (0 \times 0) + (0 \times 0) \\ \Rightarrow (x, 0, 0)(a, 0, 0) = xa$
$\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right] = \begin{bmatrix} xa & 0& \\ & & \\ & & \end{bmatrix}$
Similarly, let us fill for other elements.
$\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]=\left[\begin{array}{ccc} \text { Xa } & 0 & (\mathrm{x} \times 0)+(0 \times 0)+(0 \times \mathrm{c}) \\ (0 \times \mathrm{a})+(\mathrm{y} \times 0)+(0 \times 0) & (0 \times 0)+(\mathrm{y} \times \mathrm{b})+(0 \times 0) & (0 \times 0)+(\mathrm{y} \times 0)+(0 \times \mathrm{c}) \\ (0 \times \mathrm{a})+(0 \times 0)+(\mathrm{z} \times 0) & (0 \times 0)+(0 \times \mathrm{b})+(\mathrm{z} \times 0) & (0 \times 0)+(0 \times 0)+(\mathrm{z} \times \mathrm{c}) \end{array}\right]$
$\Rightarrow\left[\begin{array}{ccc}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]\left[\begin{array}{ccc}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{xa} & 0 & 0+0+0 \\ 0+0+0 & 0+\mathrm{yb}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+\mathrm{z} \mathrm{c}\end{array}\right] \Rightarrow\left[\begin{array}{lll}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \mathrm{C}\end{array}\right] \\So,\\ \mathrm{PQ}=\left[\begin{array}{ccc}\mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \mathrm{C}\end{array}\right] \ldots$
Now, we shall compute QP.
$\mathrm{QP}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\left[\begin{array}{lll}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]\\ Multiply 1^{\text {st }} row of matrix \mathrm{Q} by matching members of 1^{\text {st }} column of matrix \mathrm{P}, then finally sum them up.\\ (\mathrm{a}, 0,0)(\mathrm{x}, 0,0)=(\mathrm{a} \times \mathrm{x})+(0 \times 0)+(0 \times 0) \\\Rightarrow(a, 0,0)(x, 0,0)=x a+0+0 \\\Rightarrow(a, 0,0)(x, 0,0)=x a \left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\left[\begin{array}{lll}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]=[\mathrm{xa}$
Similarly, let us fill the other elements.
\begin{aligned} &\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\\ &=\left[\begin{array}{ccc} \text { xa } & (a \times 0)+(0 \times y)+(0 \times 0) & (a \times 0)+(0 \times 0)+(0 \times z) \\ (0 \times x)+(b \times 0)+(0 \times 0) & (0 \times 0)+(b \times y)+(0 \times 0) & (0 \times 0)+(b \times 0)+(0 \times z) \\ (0 \times x)+(0 \times 0)+(c \times 0) & (0 \times 0)+(0 \times y)+(c \times 0) & (0 \times 0)+(0 \times 0)+(c \times z) \end{array}\right]\\ \end{aligned}
\begin{aligned} &\Rightarrow\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]=\left[\begin{array}{ccc} \mathrm{xa} & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+\mathrm{yb}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+\mathrm{z} \mathrm{c} \end{array}\right]\\ &\Rightarrow\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]=\left[\begin{array}{ccc} \mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \mathrm{c} \end{array}\right] \end{aligned}
\begin{aligned} &\text { So, }\\ &\begin{aligned} \mathrm{QP}=&\left[\begin{array}{ccc} \mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right] \\ \mathrm{Thus}, &PQ=\left[\begin{array}{ccc} \mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{zc} \end{array}\right]=\mathrm{QP} \end{aligned} \end{aligned}

Question:24

If $\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\mathrm{A}$ , find A

We are given the following matrix equation,
$\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\mathrm{A}$
We need to determine the value of A.
Take L.H.S: $\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]$
\begin{aligned} &\begin{array}{l} \text { Let us solve } \\ \left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]=\mathrm{XY}(\text { say }) \\ , \text { where } \end{array}\\ &X=\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\\ &Y=\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\\ &\text { Then, }\\ &X Y=\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] \end{aligned}
Order of X = 1 × 3
Order of Y = 3 × 3
Then, the order of matrix Z(say) = 1 × 3 [Let Z = XY]
Multiply 1st row of matrix X by matching members of 1st column of matrix Y, then finally sum them up..
$\\(2, 1, 3)(-1, -1, 0) = (2 \times -1) + (1 \times -1) + (3 \times 0) \\ \Rightarrow (2, 1, 3)(-1, -1, 0) = -2 - 1 + 0 \\ \Rightarrow (2, 1, 3)(-1, -1, 0) = -3$
$\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] = \begin{bmatrix} -3 & & \end{bmatrix}$
Multiply 1st row of matrix X by matching members of 2nd column of matrix Y, then finally sum them up.
$\\(2, 1, 3)(0, 1, 1) = (2 \times 0) + (1 \times 1) + (3 \times 1) \\ \Rightarrow (2, 1, 3)(0, 1, 1) = 0 + 1 + 3 \\ \Rightarrow (2, 1, 3)(0, 1, 1) = 4$
$\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] = \begin{bmatrix} -3 & 4& \end{bmatrix}$
Multiply 1st row of matrix X by matching members of 3rd column of matric Y, then finally sum them up.
$\\(2, 1, 3)(-1, 0, 1) = (2 \times -1) + (1 \times 0) + (3 \times 1) \\ \Rightarrow (2, 1, 3)(-1, 0, 1) = -2 + 0 + 3 \\ \Rightarrow (2, 1, 3)(-1, 0, 1) = 1$
$\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] = \begin{bmatrix} -3 & 4&1 \end{bmatrix}$
So,we have,
$\begin{array}{l} \mathrm{Z}=\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right] \\ \text { Now, multiplying } \mathrm{Z} \text { by }\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\mathrm{Q}(\text { say }) \\ \mathrm{ZQ}=\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \end{array}$
Order of Z = 1 × 3
Order of Q = 3 × 1
Then, order of the resulting matrix = 1 × 1
Multiply 1st row of matrix Z by matching members of 1st column of matrix Q, then finally sum them up.
$\\(-3, 4, 1)(1, 0, -1) = (-3 \times 1) + (4 \times 0) + (1 \times -1) \\ \Rightarrow (-3, 4, 1)(1, 0, -1) = -3 + 0 - 1 \\ \Rightarrow (-3, 4, 1)(1, 0, -1) = -4$
$\\ \left[\begin{array}{lll}-3 & 4 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=[-4] \\Now, since \left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=\mathrm{A} \\Thus, A=[-4]$

Question:25

If $\mathrm{A}=\left[\begin{array}{ll} 2 & 1 \end{array}\right], \mathrm{B}=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]_{\text {and }} \mathrm{C}=\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]$ verify that A(B+C)=(AB+AC)

We are given the following matrices A, B and C, such that
$\mathrm{A}=\left[\begin{array}{ll} 2 & 1 \end{array}\right], \mathrm{B}=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]_{\text {and }} \mathrm{C}=\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]$
We need to verify that, A(B + C) = AB + AC.
Take L.H.S: A(B + C)
By Solving (B + C).
\begin{aligned} &B+C=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]+\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]\\ &\text { since, the above matrices have the same order, they can be added. }\\ &\Rightarrow B+C=\left[\begin{array}{lll} 5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2 \end{array}\right]\\ &\Rightarrow B+C=\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] \end{aligned}
Now, multiply A by (B + C).
Let (B + C) = D.
We get,
$\Rightarrow \mathrm{AD}=\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right]$
Order of A = 1 × 2
Order of D = 2 × 3
Then, order of the matrix is = 1 × 3
Multiply 1st row of matrix A by matching members of 1st column of matrix D, then finally sum them up.
$\\(2, 1)(4, 9) = (2 \times 4) + (1 \times 9) \\ \Rightarrow (2, 1)(4, 9) = 8 + 9 \\ \Rightarrow (2, 1)(4, 9) = 17$
$\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] = \begin{bmatrix} 17 & \end{bmatrix}$
Multiply 1st row of matrix A by matching members of 2nd column of matrix D, then finally sum them up.
$\\(2, 1)(5, 7) = (2 \times 5) + (1 \times 7) \\ \Rightarrow (2, 1)(5, 7) = 10 + 7 \\ \Rightarrow (2, 1)(5, 7) = 17$
$\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] = \begin{bmatrix} 17 & 17\end{bmatrix}$
Multiply 1st row of matrix A by matching members of 3rd column of matrix D, then finally sum them up.
$\\ (2,1)(5,8)=(2 \times 5)+(1 \times 8) \\\Rightarrow(2,1)(5,8)=10+8 \\\Rightarrow(2,1)(5,8)=18 \\\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}4 & 5 & 5 \\ 9 & 7 & 8\end{array}\right]=\left[\begin{array}{lll}17 & 17 & 18\end{array}\right]$
So,
$A(B+C)=\left[\begin{array}{lll}17 & 17 & 18\end{array}\right]$
Now, take R.H.S: $A B+A C$
Let us compute A B.
$A B=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$
Order of A = 1 × 2
Order of B = 2 × 3
Then, order of AB = 1 × 3
Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.
\begin{aligned} &(2,1)(5,8)=(2 \times 5)+(1 \times 8)\\ &\Rightarrow(2,1)(5,8)=10+8\\ &\Rightarrow(2,1)(5,8)=18\\ &\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]=[18\\ \end{aligned}
Similarly, repeat steps to fill for the rest of the elements.
$\begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} 5 &3 &4 \\8 &7 & 6 \end{bmatrix} = \begin{bmatrix} 18 &2*3+1*7 &2*4+1*6 \end{bmatrix}$
$\begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} 5 &3 &4 \\8 &7 & 6 \end{bmatrix} = \begin{bmatrix} 18 &13 &14 \end{bmatrix}$
Now, let us compute AC.
$A C=\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]$
Order of AC = 1 × 3
Multiply 1st row of matrix A by matching members of 1st column of matrix C, then finally sum them up.
$\\$$\\(2, 1)(-1, 1) = (2 \times -1) + (1 \times 1) \\ \Rightarrow (2, 1)(-1, 1) = -2 + 1 \\ \Rightarrow (2, 1)(-1, 1) = -1$
$\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right] = \begin{bmatrix} -1 & & \end{bmatrix}$
Similarly, repeat steps to fill for other elements.
$\begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} -1 &2 &1 \\1 &0 & 2 \end{bmatrix} = \begin{bmatrix} -1 &2*2+1*0 &2*1+1*2 \end{bmatrix}$
$\begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} -1 &2 &1 \\1 &0 & 2 \end{bmatrix} = \begin{bmatrix} -1 &4 &4 \end{bmatrix}$
\begin{aligned} &\text { Now, } \mathrm{Add} , \mathrm{AB}+\mathrm{AC} .\\ &\begin{array}{lll} \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll} 18 & 13 & 14 \end{array}\right]+\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right] \end{array}\\ &\Rightarrow \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll} 18-1 & 13+4 & 14+4 \end{array}\right]\\ &\Rightarrow \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll} 17 & 17 & 18 \end{array}\right]\\ &\text { Thus, }\\ &\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC} . \end{aligned}

Question:26

If $A = \begin{bmatrix} 1 &0 &-1 \\2 & 1 & 3\\0 &1 &1 \end{bmatrix}$, then verify that $A^2 + A = A(A + I)$, where I is 3 × 3 unit matrix.

We are given the following matrix A, such that
$A = \begin{bmatrix} 1 &0 &-1 \\2 & 1 & 3\\0 &1 &1 \end{bmatrix}$.
We need to verify $A^2 + A = A(A + I)$
Take L.H.S: $A\textsuperscript{2} + A.$
Solve for $A\textsuperscript{2}.$
$A\textsuperscript{2} = A.A$
$\Rightarrow A^{2}=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]$
Multiply 1st row of matrix A by matching members of 1st column of matrix A, then finally sum them up.
$\\(1, 0, -1)(1, 2, 0) = (1 \times 1) + (0 \times 2) + (-1 \times 0) \\ \Rightarrow (1, 0, -1)(1, 2, 0) = 1 + 0 + 0 \\ \Rightarrow (1, 0, -1)(1, 2, 0) = 1$
$\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]= \begin{bmatrix} 1 & & \\ & & \\ & & \end{bmatrix}$
Similarly, repeat steps to find other elements.
$\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right] =\left[\begin{array}{ccc} 1 & (1 \times 0)+(0 \times 1)+(-1 \times 1) & (1 \times-1)+(0 \times 3)+(-1 \times 1) \\ (2 \times 1)+(1 \times 2)+(3 \times 0) & (2 \times 0)+(1 \times 1)+(3 \times 1) & (2 x-1)+(1 \times 3)+(3 \times 1) \\ (0 \times 1)+(1 \times 2)+(1 \times 0) & (0 \times 0)+(1 \times 1)+(1 \times 1) & (0 \times-1)+(1 \times 3)+(1 \times 1) \end{array}\right]$
\begin{aligned} &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{array}\right]\\ &\text { Now, add } A^{2} \text { and } A \text { , }\\ &A^{2}+A=\left[\begin{array}{ccc} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{array}\right]+\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\\ \end{aligned}
\begin{aligned} &\Rightarrow A^{2}+A=\left[\begin{array}{ccc} 1+1 & -1+0 & -2-1 \\ 4+2 & 4+1 & 4+3 \\ 2+0 & 2+1 & 4+1 \end{array}\right]\\ &\Rightarrow A^{2}+A=\left[\begin{array}{ccc} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{array}\right] \end{aligned}
Take R.H.S: A(A + I)
First, let us solve for (A + I).
\begin{aligned} &A+1=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]+\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\Rightarrow A+1=\left[\begin{array}{ccc} 1+1 & 0+0 & -1+0 \\ 2+0 & 1+1 & 3+0 \\ 0+0 & 1+0 & 1+1 \end{array}\right]\\ &\Rightarrow A+I=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{array}\right]\\ &\text { Multiply }(\mathrm{A}+1) \text { from } \mathrm{A} \text { . }\\ &A(A+I)=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{array}\right]\\ &\Rightarrow A(A+1) \end{aligned}
$\begin{array}{l} =\left[\begin{array}{ccc} (1 \times 2)+(0 \times 2)+(-1 \times 0) & (1 \times 0)+(0 \times 2)+(-1 \times 1) & (1 \times-1)+(0 \times 3)+(-1 \times 2) \\ (2 \times 2)+(1 \times 2)+(3 \times 0) & (2 \times 0)+(1 \times 2)+(3 \times 1) & (2 \times-1)+(1 \times 3)+(3 \times 2) \\ (0 \times 2)+(1 \times 2)+(1 \times 0) & (0 \times 0)+(1 \times 2)+(1 \times 1) & (0 \times-1)+(1 \times 3)+(1 \times 2) \end{array}\right] \\ \Rightarrow A(A+1)=\left[\begin{array}{ccc} 2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2 \end{array}\right] \\ \Rightarrow A(A+1)=\left[\begin{array}{ccc} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{array}\right] \end{array}$
Since, L.H.S = R.H.S.
Hence proved, $A^2 + A = A(A + I)$

Question:27

If $A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]$, then verify that:
(i) (A’)’ = A
(ii) (AB)’ = B’A’
(iii) (kA)’ = (kA’)

We are given with the following matrices A and B, such that
$A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]_{\text {and }} B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]$
(i). We need to verify that, (A’)’ = A.
Take L.H.S: (A’)’
In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, thatis it switches the row and column indices of the matrix by producing another matrix denoted as $A\textsuperscript{T}$ or A’.
So, in transpose of a matrix,
The rows of the matrix become the columns of the matrix.
So, If $A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]$
(0, -1, 2) and (4, 3, -4) are 1st and 2nd rows respectively, will become 1st and 2nd columns respectively.
Then $A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]$

Also, if $A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]$
Similarly, (0, 4), (-1, 3) and (2, -4) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
Then $\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$
Note, that
$\left(A^{\prime}\right)^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]=A$
Thus, verified that $\left(A^{\prime}\right)^{\prime}=A$
(ii). We need to verify that, (AB)’ = B’A’.
Take L.H.S: (AB)’
Compute AB.
$AB=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]$
Order of A = 2 × 3
Order of B = 3 × 2
Then, order of AB = 2 × 2
Multiplying 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.
(0, -1, 2)(4, 1, 2) = (0 × 4) + (-1 × 1) + (2 × 2)
⇒ (0, -1, 2)(4, 1, 2) = 0 - 1 + 4
⇒ (0, -1, 2)(4, 1, 2) = 3
$\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right] = \begin{bmatrix} 3 & \\ & \end{bmatrix}$
Similarly, repeat the steps to fill for the other elements.
$\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \left[\begin{array}{ll} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right] =\left[\begin{array}{cc} 3 & (0 \times 0)+(-1 \times 3)+(2 \times 6) \\ (4 \times 4)+(3 \times 1)+(-4 \times 2) & (4 \times 0)+(3 \times 3)+(-4 \times 6) \end{array}\right]$
$\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]=\left[\begin{array}{cc} 3 & 0-3+12 \\ 16+3-8 & 0+9-24 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 3 \\ 2 & 6 \end{array}\right]=\left[\begin{array}{cc} 3 & 9 \\ 11 & -15 \end{array}\right] \\ \Rightarrow \mathrm{AB}=\left[\begin{array}{cc} 3 & 9 \\ 11 & -15 \end{array}\right] \end{array}$
Transpose of AB is (AB)’.
(3, 9) and (11, -15) are 1st and 2nd rows respectively, will become 1st and 2nd columns respectively.
$(\mathrm{AB})^{\prime}=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right]$
Take R.H.S:$\mathrm{B}^{\prime} \mathrm{A}^{\prime}$
$\mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$
(4, 0), (1, 3) and (2, 6) are 1st, 2nd and 3rd rows of matrix B respectively, will become 1st, 2nd and 3rd columns respectively.
$\Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right]$
Also, if $A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$
(0, -1, 2) and (4, 3, -4) are 1st and 2nd rows respectively, will become 1st and 2nd columns respectively.
$\Rightarrow A^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]$
By multiplying$B^{\prime} by A^{\prime}$we get,
$\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]$
Order of B’ = 2 × 3
Order of A’ = 3 × 2
Then, order of B’A’ = 2 × 2
Multiply 1st row of matrix B’ by matching members of 1st column of matrix A’, then finally sum them up.
(4, 1, 2)(0, -1, 2) = (4 × 0) + (1 × -1) + (2 × 2)
⇒ (4, 1, 2)(0, -1, 2) = 0 - 1 + 4
⇒ (4, 1, 2)(0, -1, 2) = 3
$\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right] = \begin{bmatrix} 3 & \\ & \end{bmatrix}$
Similarly, repeat the same steps to fill the rest of the elements.
$\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right] = \left[\begin{array}{cc} 3 & (4 \times 4)+(1 \times 3)+(2 \times-4) \\ (0 \times 0)+(3 \times-1)+(6 \times 2) & (0 \times 4)+(3 \times 3)+(6 \times-4) \end{array}\right]$
$\begin{array}{l} \Rightarrow\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 & 16+3-8 \\ 0-3+12 & 0+9-24 \end{array}\right] \\ \Rightarrow\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \\ \Rightarrow \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \\ \text { since, } \mathrm{L.H.S}=\mathrm{R.H.S} \\ \text { Thus, }(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime} \end{array}$
(iii). We need to verify that, (kA)’ = kA’.
Take L.H.S: (kA)’
We know that,
$A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$
By Multiplying k on both sides, we get, (k is a scalar quantity)
$\begin{array}{l} k A=k \times\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \\ \Rightarrow k A=\left[\begin{array}{ccc} k \times 0 & k \times-1 & k \times 2 \\ k \times 4 & k \times 3 & k \times-4 \end{array}\right] \\ \Rightarrow k A=\left[\begin{array}{ccc} 0 & -k & 2 k \\ 4 k & 3 k & -4 k \end{array}\right] \end{array}$
Now, to find transpose of kA,
(0, -k, 2k) and (4k, 3k, -4k) are 1st and 2nd rows of matrix kA respectively, will become 1st and 2nd columns respectively.
\begin{aligned} &\Rightarrow(\mathrm{kA})^{\prime}=\left[\begin{array}{cc} 0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -4 \mathrm{k} \end{array}\right]\\ &\text { Take R.H.S: kA }\\ &A=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right] \end{aligned}
Then, for transpose of A,
(0, -1, 2) and (4, 3, -4) are 1st and 2nd rows of matrix A respectively, will become 1st and 2nd columns respectively.
$A'=\begin{bmatrix} 0 &4 \\-1 &3 \\2 &-4 \end{bmatrix}$
By Multiplying k on both sides, we get,
$\\ \mathrm{kA}^{\prime}=\mathrm{k}\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right] \\\Rightarrow \mathrm{kA}^{\prime}=\left[\begin{array}{cc}\mathrm{k} \times 0 & \mathrm{k} \times 4 \\ \mathrm{k} \times-1 & \mathrm{k} \times 3 \\ \mathrm{k} \times 2 & \mathrm{k} \times-4\end{array}\right] \\\Rightarrow \mathrm{kA}^{\prime}=\left[\begin{array}{cc}0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -4 \mathrm{k}\end{array}\right]$
As, L.H.S = R.H.S.
Hence proved, $(kA)' = kA'$.

Question:28

If $A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right], B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]$ then verify that:

(i) (2A + B)’ = 2A’ + B’
(ii) (A - B)’ = A’ - B’.

We are given the following matrices A and B, such that
$A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right], B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]$
In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as AT or A’.
So, in transpose of a matrix,
The rows of the matrix become the columns of the matrix. .
(i). We need to verify that, (2A + B)’ = 2A’ + B’.
Take L.H.S: (2A + B)’
By substituting the matrices A and B, in (2A + B)’, we get,
\begin{aligned} &(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(2\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right]+\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{lll} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 1 \\ 2 \times 5 & 2 \times 6 \end{array}\right]+\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{cc} 2 & 4 \\ 8 & 2 \\ 10 & 12 \end{array}\right]+\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{cc} 2+1 & 4+2 \\ 8+6 & 2+4 \\ 10+7 & 12+3 \end{array}\right]\right)^{\prime}\\ &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left(\left[\begin{array}{cc} 3 & 6 \\ 14 & 6 \\ 17 & 15 \end{array}\right]\right)^{\prime} \end{aligned}
For transpose of (2A + B),
(3, 6), (14, 6) and (17, 15) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
\begin{aligned} &\Rightarrow(2 \mathrm{~A}+\mathrm{B})^{\prime}= \begin{bmatrix} 3 &14 &17 \\6 &6 &15 \end{bmatrix} \end{aligned}
Take R.H.S: 2A’ + B’
If $A=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right]$
(1, 2), (4, 1) and (5, 6) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
$\Rightarrow A^{\prime}=\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]$
Multiply both sides by 2 we get,
$2 \mathrm{~A}^{\prime}=2\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 1 & 6 \end{array}\right]$
$\\\Rightarrow 2 A^{\prime}=\left[\begin{array}{lll}2 \times 1 & 2 \times 4 & 2 \times 5 \\ 2 \times 2 & 2 \times 1 & 2 \times 6\end{array}\right] \\\Rightarrow 2 \mathrm{~A}^{\prime}=\left[\begin{array}{lll}2 & 8 & 10 \\ 4 & 2 & 12\end{array}\right]$
Also,
If
$B=\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]$
(1, 2), (6, 4) and (7, 3) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
\begin{aligned} &\Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right]\\ &\text { Now, add } 2 \mathrm{~A}^{\prime} \text { and } \mathrm{B}^{\prime}\\ &2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}=\left[\begin{array}{lll} 2 & 8 & 10 \\ 4 & 2 & 12 \end{array}\right]+\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right]\\ &\Rightarrow 2 A^{\prime}+B^{\prime}=\left[\begin{array}{lll} 2+1 & 8+6 & 10+7 \\ 4+2 & 2+4 & 12+3 \end{array}\right]\\ &\Rightarrow 2 A^{\prime}+B^{\prime}=\left[\begin{array}{ccc} 3 & 14 & 17 \\ 6 & 6 & 15 \end{array}\right] \end{aligned}
Since, L.H.S = R.H.S
Thus, (2A + B)’ = 2A’ + B’.
(ii). We need to verify that, (A - B)’ = A’ - B’.
Take L.H.S: (A - B)’
By substituting the matrices A and B in (A - B)’, we get,
$\begin{array}{l} (A-B)^{\prime}=\left(\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \\ 5 & 6 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 6 & 4 \\ 7 & 3 \end{array}\right]\right)^{\prime} \\ \Rightarrow(A-B)^{\prime}=\left(\left[\begin{array}{ll} 1-1 & 2-2 \\ 4-6 & 1-4 \\ 5-7 & 6-3 \end{array}\right]\right)^{\prime} \\ \Rightarrow(A-B)^{\prime}=\left(\left[\begin{array}{cc} 0 & 0 \\ -2 & -3 \\ -2 & 3 \end{array}\right]\right)^{\prime} \end{array}$
To find transpose of (A - B),
(0, 0), (-2, -3) and (-2, 3) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
$\Rightarrow(A-B)^{\prime}=\left[\begin{array}{ccc} 0 & -2 & -2 \\ 0 & -3 & 3 \end{array}\right]$
Take R.H.S: $\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$
$A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$
$(1,2),(4,1) and (5,6) are 1^{\text {st }}, 2^{\text {nd }} and 3^{\text {rd }}$ rows respectively, will become $1^{\text {st }}, 2^{\text {nd }} and 3^{\text {rd }}$ columns respectively.
$\Rightarrow A^{\prime}=\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]$
Also,
$B=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$
(1, 2), (6, 4) and (7, 3) are 1st, 2nd and 3rd rows respectively, will become 1st, 2nd and 3rd columns respectively.
$\Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right]$
When Subtracting $\mathrm{B}^{\prime} \text{ from } \mathrm{A}^{\prime}$, we get,
$A^{\prime}-B^{\prime}=\left[\begin{array}{lll} 1 & 4 & 5 \\ 2 & 1 & 6 \end{array}\right]-\left[\begin{array}{lll} 1 & 6 & 7 \\ 2 & 4 & 3 \end{array}\right]$
$\begin{array}{l} \Rightarrow A^{\prime}-B^{\prime}=\left[\begin{array}{rrr} 1-1 & 4-6 & 5-7 \\ 2-2 & 1-4 & 6-3 \end{array}\right] \\ \Rightarrow A^{\prime}-B^{\prime}=\left[\begin{array}{rrr} 0 & -2 & -2 \\ 0 & -3 & 3 \end{array}\right] \\ \text { As, L.H.S = R.H.S } \\ \text { Hence proved, }(A-B)^{\prime}=A^{\prime}-B \end{array}$

Question:29

Show that A’A and AA’ are both symmetric matrices for any matrix A.

We know that,
In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose. Formally, because equal matrices have equal dimensions, only square matrices can be symmetric.
And we know that, transpose of AB is given by
(AB)’ = B’A’
Using this result, and by taking transpose of A’A we have,
Transpose of A’A = (A’A)T = (A’A)’
Using, transpose of A’A = (A’A)’
⇒ (A’A)’ = A’(A’)’
And also,
(A’)’ = A
So,
(A’A)’ = A’A
Since, (A’A)’ = A’A
This means, A’A is symmetric matrix for any matrix A.
Now, take transpose of AA’.
Transpose of AA’ = (AA’)’
⇒ (AA’)’ = (A’)’A’ [ (AB)’ = B’A’]
⇒ (AA’)’ = AA’ [(A’)’ = A]
Since, (AA’)’ = AA’
This means, AA’ is symmetric matrix for any matrix A.
Thus, A’A and AA’ are symmetric matrix for any matrix A.

Question:30

Let A and B be square matrices of the order 3 × 3. Is $(AB)^2 = A^2B^2$? Give reasons.

We have been given that,
A and B are square matrices of the order $3 \times 3.$
We need to check whether $(AB)\textsuperscript{2} = A\textsuperscript{2}B\textsuperscript{2}$ is true or not.
Take $(AB)\textsuperscript{2}$.
$(AB)\textsuperscript{2} = (AB)(AB)$
[$\because$ A and B are of order ($3 \times 3$) each, A and B can be multiplied; A and B be any matrices of order ($3 \times 3$)]
$\Rightarrow (AB)\textsuperscript{2} = ABAB$
[$\because$ (AB)(AB) = ABAB]
$\Rightarrow (AB)\textsuperscript{2} = AABB$ [ if BA = AB]
$\Rightarrow (AB)\textsuperscript{2} = A\textsuperscript{2}B\textsuperscript{2}$
$(AB)\textsuperscript{2} = A\textsuperscript{2}B\textsuperscript{2}$ is possible if BA = AB.

Question:31

Show that if A and B are square matrices such that AB = BA, then $(A + B)^2 = A^2 + 2AB + B^2$.

According to matrix multiplication we can say that:
$(A + B)\textsuperscript{2} = (A+B)(A+B) = A\textsuperscript{2} + AB + BA + B\textsuperscript{2}$
We know that matrix multiplication is not commutative but it is given that: AB = BA
$\\ \therefore (A + B)\textsuperscript{2} = A\textsuperscript{2} + AB + AB + B\textsuperscript{2} \\ \Rightarrow (A + B)\textsuperscript{2} = A\textsuperscript{2} + 2AB + B\textsuperscript{2} \ldots$ is proved

Question:32.1

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.
Show that:
A + (B + C) = (A + B) + C

Given,
$A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$
\begin{aligned} &\text { LHS }=A+(B+C)=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left(\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\right)\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left(\left[\begin{array}{ll} 4+2 & 0+0 \\ 1+1 & 5-2 \end{array}\right]\right) \end{aligned}
\begin{aligned} &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left[\begin{array}{ll} 6 & 0 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]\\ &R H S=(A+B)+C=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]+\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\right)+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left(\left[\begin{array}{cc} 1+4 & 2+0 \\ -1+1 & 3+5 \end{array}\right]\right)+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll} 5 & 2 \\ 0 & 8 \end{array}\right]+\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]\\ &\text { Clearly LHS }=R H S=\left[\begin{array}{ll} 7 & 2 \\ 1 & 6 \end{array}\right]\\ &\text { Hence, we have }\\ &A+(B+C)=(A+B)+C \text { ...proved } \end{aligned}

Question:32.2

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.
Show that:
A(BC) = (AB)C

We have to prove that: A(BC) = (AB)C
$\begin{array}{l} \text { LHS = } \mathrm{A}(\mathrm{BC})=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left(\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\right) \\ \Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left(\begin{array}{cc} 4 \times 2+0 \times 1 & 4 \times 0+0 \times(-2) \\ 1 \times 2+5 \times 1 & 1 \times 0+5 \times(-2)]) \end{array}\right. \\ \Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right] \\ LHS= {\left[\begin{array}{cc} 22 & -20 \\ 13 & -30 \end{array}\right]} \end{array}$
\begin{aligned} &\mathrm{RHS}=(\mathrm{AB}) \mathrm{C}=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]\right)\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\text { By matrix multiplication as done for LHS }\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc} 6 & 10 \\ -1 & 15 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll} 22 & -20 \\ 13 & -30 \end{array}\right]\\ &\text { Evidently, LHS = RHS }=\left[\begin{array}{rr} 22 & -20 \\ 13 & -30 \end{array}\right]\\ &\therefore \mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C} \ldots \text { .proved } \end{aligned}

Question:32.3

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.
Show that:(a + b)B = aB + bB

To prove: (a + b)B = aB + bB
Given, a = 4 and b = -2
$\\LHS =(4+(-2)) B=2\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right] \\\mathrm{RHS}=\mathrm{aB}+\mathrm{bB}=4\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]-2\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \\\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc}16 & 0 \\ 4 & 20\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]$
It is clear that, $\mathrm{LHS}=\mathrm{RHS}=\left[\begin{array}{cc}8 & 0 \\ 2 & 10\end{array}\right]$
Hence, we have,
(a + b)B = aB + bB …proved

Question:32.4

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.

Show that:
a(C - A) = aC -aA

We have to prove: a(C - A) = aC -aA
As,
$\\LHS =a(C-A)=4\left(\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\right) \\\Rightarrow \mathrm{LHS}=4\left(\left[\begin{array}{cc}2-1 & 0-2 \\ 1-(-1) & -2-3\end{array}\right]\right)=4\left[\begin{array}{cc}1 & -2 \\ 2 & -5\end{array}\right]=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right] \\\mathrm{RHS}=\mathrm{aC}-\mathrm{aA}={ }^{4}\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]-4\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right] \\\Rightarrow a C-a A=\left[\begin{array}{cc}8 & 0 \\ 4 & -8\end{array}\right]-\left[\begin{array}{cc}4 & 8 \\ -4 & 12\end{array}\right]=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$
Clearly $LHS =\mathrm{RHS}=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$
Hence, we have
$a(C-A)=a C-a A \ldots proved$

Question:32.5

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.
Show that: $(AT)^{}T = A$

To prove: $(AT)^{}T = A$
In transpose of a matrix, the rows of the matrix become the columns.
$\\\text { LHS }=\left(A^{T}\right)^{T}\\=\left(\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^{\mathrm{T}}\right)^{\mathrm{T}}=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]=\mathrm{A}=\mathrm{RHS}$
Hence, proved.

Question:32.6

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.
Show that:
$(bA)^T = bA^T$

a) To prove: $(bA)^T = bA^T$
As, LHS = $(bA)^T = (-2A)^T=(-2\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right])^T$
$(bA)^T = (-2A)^T=\left[\begin{array}{cc} -2 & -4 \\ 2 & -6 \end{array}\right]^T=\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]$

\begin{aligned} &\text { Similarly, }\\ &\mathrm{RHS}=-2\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]^{\mathrm{T}}=-2\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{cc} -2 & 2 \\ -4 & -6 \end{array}\right]\\ &\text { Hence proved }L H S=R H S=\left[\begin{array}{cc} -2 &2 \\ -4 & -6 \end{array}\right]\\ &\text { Then, }(b A)^{T}=b A^{\top} \ldots \text { proved } \end{aligned}

Question:32.7

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.
Show that: $(AB)^T = B^T A^T$

To prove: $(AB)^T = B^T A^T$
By multiplying the matrices and taking the transpose, we get,
$\therefore \mathrm{LHS}=\left(\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\right)^{\mathrm{T}} \\\Rightarrow \mathrm{LHS}=\left[\begin{array}{ll}1 \times 4+2 \times 1 & 1 \times 0+2 \times 5 \\ -1 \times 4+3 \times 1 & -1 \times 0+3 \times 5\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}6 & 10 \\ -1 & 15\end{array}\right]^{\mathrm{T}} \\\therefore \mathrm{LHS}=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
As $\mathrm{RHS}=\mathrm{B}^{\top} \mathrm{A}^{\top}$
By taking transpose of matrices and then multiplying, we get,
$\mathrm{RHS}=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]^{\mathrm{T}}\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ll}4 & 1 \\ 0 & 5\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right] \\\Rightarrow \mathrm{RHS}=\left[\begin{array}{ll}4 \times 1+1 \times(2) & 4 \times (-1)+1 \times 3 \\ 0 \times 1+5 \times(2) & 0 \times (-1)+5 \times 3\end{array}\right]=\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
We have, LHS = RHS = $\left[\begin{array}{cc}6 & -1 \\ 10 & 15\end{array}\right]$
Hence $(A B)^{\top}=B^{\top} A^{\top}$... proved

Question:32.8

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.

Show that:
(A - B)C = AC - BC

c) To prove: (A - B)C = AC - BC
$A s, L H S=(A-B) C$
Substituting the values of $\mathrm{A} . \mathrm{B} and \mathrm{C}$ and multiplying according to the rule of matrix multiplication.
$(A-B)=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]-\left[\begin{array}{cc} 4 & 0 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]$
$LHS=(A-B) C=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right]$

$A C=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 4 & -4 \\ 1 & -6 \end{array}\right]$
$B C=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{cc} 8 & 0 \\ 7 & -10 \end{array}\right]$
$RHS=A C-B C=\left[\begin{array}{cc} 4-8 & -4-0 \\ 1-7 & -6+10 \end{array}\right]=\left[\begin{array}{cc} -4 & -4 \\ -6 & 4 \end{array}\right]=LHS$
Hence $(A-B) C=A C-B C$...proved

Question:32.9

Let $A=\left[\begin{array}{cc} 1 & 2 \\ -1 & 3 \end{array}\right], B=\left[\begin{array}{ll} 4 & 0 \\ 1 & 5 \end{array}\right], C=\left[\begin{array}{cc} 2 & 0 \\ 1 & -2 \end{array}\right]$ and a = 4, b = -2.

Show that:
$(A - B)^T = A^T - B^T$

To Prove: $(A - B)^T = A^T - B^T$
$(A-B)=\left[\begin{array}{cc} 1-4 & 2-0 \\ -1-1 & 3-5 \end{array}\right]=\left[\begin{array}{cc} -3 & 2 \\ -2 & -2 \end{array}\right]$
$(A-B)^T=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right]$
$\\A^{T}-B^{T}=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]-\left[\begin{array}{cc} 4 & 1 \\ 0 & 5 \end{array}\right]\\=\left[\begin{array}{cc} -3 & -2 \\ 2 & -2 \end{array}\right]\\=(A-B)^{T}$
Hence $(A-B)^{T}=A^T-B^T$.

Question:33

If $A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right],$ then show that $\quad \mathrm{A}^{2}=\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]$

As $A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right],$
$A^2=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
According to the rule of matrix multiplication:
$\\ \mathrm{A}^{2}=\left[\begin{array}{cc} \cos \theta \times \cos \theta+\sin \theta \times(-\sin \theta) & \cos \theta \times \sin \theta+\sin \theta \times \cos \theta \\ -\cos \theta \times \sin \theta+(-\sin \theta \times \cos \theta) & \cos \theta \times \cos \theta+\sin \theta \times(-\sin \theta) \end{array}\right] \\ \Rightarrow \mathrm{A}^{2}=\left[\begin{array}{cc}\cos ^{2} \theta-\sin ^{2} \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta\end{array}\right]$
We know that:
$\\2 \sin \theta \cos \theta=\sin 2 \theta and \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta \\\therefore A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]_{\ldots}$
Hence.proved.

Question:34

If $A=\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right], B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]$ and $x^2 = -1$, then show that $(A + B)^2 = A^2 + B^2.$

\begin{aligned} &\text { As, LHS }=(A+B)^{2}=\left(\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\right)^{2}\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]^{2}=\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]\\ &\text { By the rule of matrix multiplication we can write LHS as - }\\ &\text { LHS }=\left[\begin{array}{cc} 0+(1-x)(1+x) & 0 \\ 0 & (1+x)(1-x) \end{array}\right]\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1-\mathrm{x}^{2} & 0 \\ 0 & 1-\mathrm{x}^{2} \end{array}\right]\\ &\text { Given } x^{2}=-1\\ &\therefore \mathrm{LHS}=\left[\begin{array}{cc} 1-(-1) & 0 \\ 0 & 1-(-1) \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\\ \end{aligned}
\begin{aligned} &\mathrm{RHS}=\mathrm{A}^{2}+\mathrm{B}^{2}=\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]^{2}+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]^{2}\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{aligned}
By the rule pf matrix multiplication we can write-
$\mathrm{RHS}=\left[\begin{array}{cc}-\mathrm{x}^{2} & 0 \\ 0 & -\mathrm{x}^{2}\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1-\mathrm{x}^{2} & 0 \\ 0 & 1-\mathrm{x}^{2}\end{array}\right]$
Given $x^{2}=-1$
$\therefore \mathrm{RHS}=\left[\begin{array}{cc}1-(-1) & 0 \\ 0 & 1-(-1)\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]$
We have, $\mathrm{RHS}=\mathrm{LHS}=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]$
Hence, $(A+B)^{2}=A^{2}+B^{2}$. -proved

Question:35

Verify that $A^2 = I$ when $A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$

We need to prove that
$\begin{array}{l} A^{2}=I=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \because A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \\ \therefore A^{2}=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right] \end{array}$
According to the rule of matrix multiplication we have-
\begin{aligned} &A^{2}=\left[\begin{array}{ccc} 0 \times 0+1 \times 4+(-1) \times 3 & 0 \times 1+1 \times(-3)+(-1) \times(-3) & 0 \times(-1)+1 \times 4+(-1) \times 4 \\ 4 \times 0+(-3) \times 4+4 \times 3 & 4 \times 1+(-3) \times(-3)+4 \times(-3) & 4 \times(-1)+(-3) \times 4+4 \times 4 \\ 3 \times 0+(-3) \times 4+4 \times 3 & 3 \times 1+(-3) \times(-3)+4 \times(-3) & 3 \times(-1)+(-3) \times 4+4 \times 4 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ccc} 4-3 & -3+3 & 4-4 \\ -12+12 & 4+9-12 & -4-12+16 \\ -12+12 & 3+9-12 & -3+16-12 \end{array}\right]\\ &\therefore \mathrm{A}^{2}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I\\ &\text { Hence Proved } \end{aligned}

Question:36

Prove by Mathematical Induction that $(A')^n = (A^n)'$, where n ∈ N for any square matrix A.

By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.
We are given to prove that $(A')\textsuperscript{n} = (A\textsuperscript{n})'.$
Let P(n) be the statement :$(A')\textsuperscript{n} = (A\textsuperscript{n})'.$
Clearly, $P(1): (A')\textsuperscript{1} = (A\textsuperscript{1})'$
$\\ \Rightarrow P(1) : A' = A' \\ \Rightarrow P(1) is true$
Let P(k) be true.
$\therefore (A')\textsuperscript{k} = (A\textsuperscript{k})' \ldots (1)$
Let’s take P(k+1) now:
$\because (A\textsuperscript{k+1})' = (A\textsuperscript{k}A)'$
We know that according tu the rule of trabnspose of a matrix,
$(AB)\textsuperscript{T} = B\textsuperscript{T}A\textsuperscript{T} \therefore (A\textsuperscript{k}A)' = A'(A\textsuperscript{k})' = A'(A')\textsuperscript{k} = (A')\textsuperscript{k+1}$
Thus,$(A\textsuperscript{k+1})' = (A')\textsuperscript{k+1}$
$\therefore P(k+1) is true.$
Hence proved: $(A')\textsuperscript{n} = (A\textsuperscript{n})'$ is true for all $n \in N.$

Question:37.1

Let $A=\begin{bmatrix} 1 &3 \\-5 & 7 \end{bmatrix}$
To apply elementary row transformations we can say that:
A = IA where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XA
And this X is called inverse of A = A-1
So we get:
\begin{aligned} &\left[\begin{array}{cc} 1 & 3 \\ -5 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { By Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+5 \mathrm{R}_{1}\\ &\Rightarrow\left[\begin{array}{cc} 1 & 3 \\ 0 & 22 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right] \mathrm{A}\\ &\text { By Applying } \mathrm{R}_{2} \rightarrow(1 / 22) \mathrm{R}_{2}\\ &\Rightarrow\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ \frac{5}{22} & \frac{1}{22} \end{array}\right] \mathrm{A}\\ &\text { By Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-3 \mathrm{R}_{2}\\ &\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} \frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{array}\right] \mathrm{A}\\ &\text { As we have an Identity matrix in LHS. }\\ \end{aligned}
\begin{aligned} &\therefore A^{-1}=\left[\begin{array}{ll} \frac{7}{22} & -\frac{3}{22} \\ \frac{5}{22} & \frac{1}{22} \end{array}\right] \end{aligned}

Question:37.2

Find inverse, by elementary row operations (if possible), of the following matrices.
$\begin{bmatrix} 1 &-3 \\-2 & 6 \end{bmatrix}$

Let $B=\begin{bmatrix} 1 &-3 \\-2 & 6 \end{bmatrix}$
To apply elementary row transformations we write:
B = IB where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XB
And this X is called inverse of $B = B^{-1}$
So we get,
$\begin{array}{l} {\left[\begin{array}{cc} 1 & -3 \\ -2 & 6 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \mathrm{B}} \end{array}$
By Applying R2→ R2 + 2R1
$\Rightarrow\left[\begin{array}{cc} 1 & -3 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 2 & 1 \end{array}\right] \mathrm{A}$
We have got all zeroes in one of the row of matrix in LHS.
So by any means we can't make identity matrix in LHS.
∴ inverse of B does not exist.
$B^{-1}$ does not exist.

Question:38

If $\left[\begin{array}{cc} x y & 4 \\ z+6 & x+y \end{array}\right]=\left[\begin{array}{cc} 8 & w \\ 0 & 6 \end{array}\right]$ then find values of x, y, z and w.

We are given the following matrices,
$\left[\begin{array}{cc} x y & 4 \\ z+6 & x+y \end{array}\right]=\left[\begin{array}{cc} 8 & w \\ 0 & 6 \end{array}\right]$
Since, both the matrices are equal, so all the elements in them are equal.
$\therefore xy = 8 ; w = 4 ; z + 6 = 0\ and\ x + y = 6$
Hence, we have,
$\\w = 4 \\z = -6 \\\because x + y = 6 \\ \Rightarrow y = 6 - x \\ \therefore x(6-x) = 8 \\ \Rightarrow x\textsuperscript{2} - 6x + 8 = 0 \\ \Rightarrow x\textsuperscript{2} - 4x - 2x + 8 = 0 \\ \Rightarrow x(x - 4) - 2(x - 4) = 0 \\ \Rightarrow (x - 2)(x - 4) = 0 \\ \Rightarrow x = 2 or x = 4$
When x = 2 ; y = 4
And when x = 4 ; y = 2
Thus, we have the values of
x = 2 or 4 ; y = 4 or 2 ; z = -6 and w = 4

Question:39

If $A=\begin{bmatrix} 1 &5 \\7 &12 \end{bmatrix} and B=\begin{bmatrix} 9 &1 \\7 & 8 \end{bmatrix}$ find a matrix C such that 3A + 5B + 2C is a null matrix.

Given that:
3A + 5B + 2C = O = null matrix
We have to determine the value of C,
$\begin{array}{l} \text { As, } 3\left[\begin{array}{lr} 1 & 5 \\ 7 & 12 \end{array}\right]+5\left[\begin{array}{ll} 9 & 1 \\ 7 & 8 \end{array}\right]+2 C=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ll} 3 & 15 \\ 21 & 36 \end{array}\right]+\left[\begin{array}{ll} 45 & 5 \\ 35 & 40 \end{array}\right]+2 C=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ \Rightarrow 2 C+\left[\begin{array}{ll} 48 & 20 \\ 56 & 76 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ \therefore 2 C=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]-\left[\begin{array}{ll} 48 & 20 \\ 56 & 76 \end{array}\right] \\ \Rightarrow 2 C=\left[\begin{array}{ll} -48 & -20 \\ -56 & -76 \end{array}\right] \\ \therefore C=\frac{1}{2}\left[\begin{array}{ll} -48 & -20 \\ -56 & -76 \end{array}\right]=\left[\begin{array}{ll} -24 & -10 \\ -28 & -38 \end{array}\right] \end{array}$

Question:40

If $A=\begin{bmatrix} 3 &-5 \\-4 & 2 \end{bmatrix}$ then find $A^2 - 5A - 14I$. Hence, obtain $A^3$.

Given, $A=\begin{bmatrix} 3 &-5 \\-4 & 2 \end{bmatrix}$
$\therefore A^{2}=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$
According to the rule of matrix multiplication we can write:
$\\ \mathrm{A}^{2}=\left[\begin{array}{cc}3 \times 3+(-5) \times(-4) & 3 \times(-5)+2 \times(-5) \\ -4 \times 3+2 \times(-4) & (-4) \times(-5)+2 \times 2\end{array}\right] \\\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]_{...}(1)$
We have to find: $A^{2}-5 A-14I$
$\begin{array}{l} \therefore A^{2}-5 A-14I=\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-5\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right]-14\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \Rightarrow A^{2}-5 A-14 I=\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-\left[\begin{array}{cc} 15 & -25 \\ -20 & 10 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right] \\ \Rightarrow A^{2}-5 A-14 I=\left[\begin{array}{cc} 29-15-14 & -25+25+0 \\ -20+20+0 & 24-10-14 \end{array}\right] \\ A^{2}-5 A-14 I=\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \end{array}$
We need to find value of $A^3$ using the above equation:
Now we have,
$A\textsuperscript{2} - 5A - 14I = O$
$\Rightarrow A\textsuperscript{2} = 5A + 14I$
By multiplying with A both sides we get,
$\\ \Rightarrow A\textsuperscript{2}.A = 5A.A + 14IA \\\\ \Rightarrow A\textsuperscript{3} = 5A\textsuperscript{2} + 14A$
By Using equation 1 we get:
$\begin{array}{l} \Rightarrow \mathrm{A}^{3}=5\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]+14\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right] \\ \Rightarrow \mathrm{A}^{3}=\left[\begin{array}{cc} 145 & -125 \\ -100 & 120 \end{array}\right]+\left[\begin{array}{cc} 42 & -70 \\ -56 & 28 \end{array}\right] \\ \Rightarrow \mathrm{A}^{3}=\left[\begin{array}{cc} 187 & -195 \\ -156 & 148 \end{array}\right] \end{array}$

Question:41

Find the value of a, b, c and d, if $3\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right]=\left[\begin{array}{cc} \mathrm{a} & 6 \\ -1 & 2 \mathrm{~d} \end{array}\right]+\left[\begin{array}{cc} 4 & \mathrm{a}+\mathrm{b} \\ \mathrm{c}+\mathrm{d} & 3 \end{array}\right]$

We are given the following matrices,
$3\left[\begin{array}{ll} \mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d} \end{array}\right]=\left[\begin{array}{cc} \mathrm{a} & 6 \\ -1 & 2 \mathrm{~d} \end{array}\right]+\left[\begin{array}{cc} 4 & \mathrm{a}+\mathrm{b} \\ \mathrm{c}+\mathrm{d} & 3 \end{array}\right]$
We need to determine the value of a, b, c and d.
$\begin{array}{l} \text { As, } 3\left[\begin{array}{ll} \text { a } & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} a & 6 \\ -1 & 2 d \end{array}\right]+\left[\begin{array}{cc} 4 & a+b \\ c+d & 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ll} 3 a & 3 b \\ 3 c & 3 d \end{array}\right]=\left[\begin{array}{cc} a+4 & 6+a+b \\ -1+c+d & 2 d+3 \end{array}\right] \end{array}$
As both matrices are equal so their corresponding elements must also be equal.
$\\ \therefore 3a = a + 4 \\ \Rightarrow 2a = 4 \\ \Rightarrow a = 2$
Similarly,
$\Rightarrow 2b = 6 + a$
As from above a = 2
$\\3b = 6 + a + b \\ \therefore 2b = 6+2 = 8 \\ \Rightarrow b = 4 \\Also 3d = 2d + 3 \\ \Rightarrow d = 3$
And, we have,
$\\3c = -1 + c + d \\ \Rightarrow 2c = d - 1 \\ \Rightarrow 2c = 3-1 \\ \Rightarrow c = 2/2 = 1 \\Thus, a = 2, b = 4, c = 1 and d = 3.$

Question:42

Find the matrix A such that

$\left[\begin{array}{rr} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right] A=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]$

We are given that,
$\left[\begin{array}{rr} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right] A=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]$
As A is multiplied with a matrix of order 3×2 and gives a resultant matrix of order 3×3
For matrix multiplication to be possible A must have 2 rows and as resultant matrix is of 3rd order A must have 3 columns
∴ A is matrix of order 2×3
Let A = $\left[\begin{array}{lll} \mathrm{a} & \mathrm{b} & c \\ \mathrm{~d} & \mathrm{e} & \mathrm{f} \end{array}\right]$ where a, b, c, d, e and f are unknown variables.
\begin{aligned} &\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & \mathrm{b} & c \\ \mathrm{~d} & \mathrm{e} & \mathrm{f} \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \end{aligned}
∴ According to the rule of matrix multiplication we have-
$\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a & b & c \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]$
By equating the elements of 2 equal matrices, as both the matrices are equal to each other, we get-
a = 1 ; b = -2 and c = -5
also, we have,
$\\2a - d = -1 \Rightarrow d = 2a + 1 = 2 + 1 = 3 \\ \therefore d = 3 \\2b - e = -8 \Rightarrow e = 2b + 8 = -4 + 8 = 4 \\ \therefore e = 4 \\Similarly, f = 2c + 10 = 0$
$\\ \therefore A = \begin{bmatrix} 1 &-2 &-5 \\3 &4 &0 \end{bmatrix}$

Question:43

If $A=\begin{bmatrix} 1 &2 \\4 &1 \end{bmatrix}$ find $A^2 + 2A + 7I$

We are given the following matrix A such that,
$A=\begin{bmatrix} 1 &2 \\4 &1 \end{bmatrix}$
$\begin{array}{l} \because \mathrm{A}^{2}=\mathrm{A} . \mathrm{A} \\ \Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right] \end{array}$
According to the rule of matrix multiplication, we get
\begin{aligned} &A^{2}=\left[\begin{array}{ll} 1 \times 1+2 \times 4 & 1 \times 2+2 \times 1 \\ 4 \times 1+1 \times 4 & 4 \times 2+1 \times 1 \end{array}\right]\\ &\Rightarrow A^{2}=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]\\ &\therefore \mathrm{A}^{2}+2 \mathrm{~A}+71=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]+2\left[\begin{array}{ll} 1 & 2 \\ 4 & 1 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}+2 \mathrm{~A}+7 \mathrm{I}=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]+\left[\begin{array}{ll} 2 & 4 \\ 8 & 2 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]\\ &\Rightarrow A^{2}+2 A+7 I=\left[\begin{array}{ll} 9+2+7 & 4+4+0 \\ 8+8+0 & 9+2+7 \end{array}\right]\\ &\Rightarrow \mathrm{A}^{2}+2 \mathrm{~A}+7I=\left[\begin{array}{cc} 18 & 8 \\ 16 & 18 \end{array}\right] \ldots \mathrm{ans} \end{aligned}

Question:44

If $A= \begin{bmatrix} \cos \alpha &\sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}$ and $A^{-1} = A'$, find value of $\alpha$

Given,$A= \begin{bmatrix} \cos \alpha &\sin \alpha \\-\sin \alpha &\cos \alpha \end{bmatrix}$
We know that in transpose of a matrix, the rows of the matrix become the columns.
\begin{aligned} &\therefore \mathrm{A}^{\prime}=\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\\ &\text { Inverse of a matrix }\\ &A=A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\\ &\text { Clearly }|\mathrm{A}|=\left|\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right|\\ &\therefore|\mathrm{A}|=\cos ^{2} \alpha+\sin ^{2} \alpha=1_{\{\mathrm{using} \text { trigonometric identity }} \end{aligned}
Adj(A) is given by the transpose of the cofactor matrix.
$\\\therefore \operatorname{adj}(\mathrm{A})=\left[\begin{array}{cc}\cos \alpha & -(-\sin \alpha) \\ -\sin \alpha & \cos \alpha\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \\\therefore \mathrm{A}^{-1}=\frac{\operatorname{adj}(\mathrm{A})}{|\mathrm{A}|}=1\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
According to question:
$\\A^{\prime}=A^{-1} \\\therefore\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
since both the matrices are equal irrespective of the value of $\alpha.$
$\therefore \alpha$ can be any real number

Question:45

If the matrix $\begin{bmatrix} 0 &a &3 \\2 & b & -1\\c &1 &0 \end{bmatrix}$ is a skew symmetric matrix, find the values of a, b and c.

A matrix is said to be skew-symmetric if A = -A’
Let, A = $\begin{bmatrix} 0 &a &3 \\2 & b & -1\\c &1 &0 \end{bmatrix}$
As, A is skew symmetric matrix.
∴ A = -A’
$\begin{array}{l} \left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]^{T} \\\\ \left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{array}\right] \\\\ {\left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -2 & -c \\ -a & -b & -1 \\ -3 & 1 & 0 \end{array}\right]} \end{array}$
Equating the respective elements of both matrices, as both the matrices are equal to each other we have,
a = -2 ; c = -3 ; b = -b ⇒ 2b = 0 ⇒ b = 0
Thus, we get,
a = -2 , b = 0 and c = -3

Question:46

If $P(x)= \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$ then show that

$P(x).P(y) = P(x + y) = P(y).P(x)$

We are given that,
$P(x)= \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$
$P(y)= \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix}$
$\begin{array}{l} \therefore P(x) \cdot P(y)=\left[\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]\left[\begin{array}{cc} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \\ \Rightarrow P(x) \cdot P(y)=\left[\begin{array}{cc} \cos x \cos y+\sin x(-\sin y) & \cos x \sin y+\sin x \cos y \\ -\sin x \cos y-\sin y(\cos x) & -\sin x \sin y+\cos x \cos y \end{array}\right] \end{array}$
We know that-
$\\ \\\cos x \cos y + \sin x \sin y = \cos (x - y) \\\cos x \sin y + \sin x \cos y = \sin (x + y) \\and \cos x \cos y - \sin x \sin y = \cos (x + y)$
$\Rightarrow P(x) \cdot P(y)=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]$
By comparing with equation 1 we can say that:
$\\\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]=\mathrm{P}(x+y) \\\therefore P(x) \cdot P(y)=P(x+y)$
Similarly, we can show for $P(y) \cdot P(x)$
$\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos \mathrm{y} & \sin \mathrm{y} \\ -\sin \mathrm{y} & \cos \mathrm{y}\end{array}\right]\left[\begin{array}{cc}\cos \mathrm{x} & \sin \mathrm{x} \\ -\sin \mathrm{x} & \cos \mathrm{x}\end{array}\right]$
By the rule of matrix multiplication, we have -
$\\ P(y) \cdot P(x)=\left[\begin{array}{cc}\cos y \cos x+\sin y(-\sin x) & \cos y \sin x+\sin y \cos x \\ -\sin y \cos x-\sin x \cos y & -\sin y \sin x+\cos x \cos y\end{array}\right] \\\\\Rightarrow P(y) \cdot P(x)=\left[\begin{array}{cc}\cos x \cos y-\sin x \sin y & \sin x \cos y+\cos x \sin y \\ -(\sin x \cos y+\cos x \sin y) & \cos x \cos y-\sin x \sin y\end{array}\right] \\\\\Rightarrow P(y) \cdot P(x)=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]_{-...(3)}\\ \therefore From equation 2 and 3, we have,\\ P(x) \cdot P(y)=P(y) \cdot P(x)=P(x+y)$

Question:47

If A is square matrix such that $A^2 = A$, show that $(I + A)^3 = 7A + I$.

We are given that,
$\\A\textsuperscript{2} = A \\\because (a+b)\textsuperscript{3} = a\textsuperscript{3} + b\textsuperscript{3} + 3a\textsuperscript{2}b + 3ab\textsuperscript{2} \\As, (I + A)\textsuperscript{3} = I\textsuperscript{3} + A\textsuperscript{3} + 3I\textsuperscript{2}A + 3IA\textsuperscript{2} \\\because I is an identity matrix. \\ \therefore I\textsuperscript{3} = I\textsuperscript{2} = I \\ \therefore (I + A)\textsuperscript{3} = I + A\textsuperscript{3} + 3IA + 3IA$
As, I is an identity matrix.
$\\ \therefore IA = AI = A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A\textsuperscript{3} + 6IA \\\because A\textsuperscript{2} = A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A\textsuperscript{2}.A + 6A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A.A + 6A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A\textsuperscript{2} + 6A \\ \Rightarrow (I + A)\textsuperscript{3} = I + A + 6A = I + 7A$
Hence proved,
$(I + A)\textsuperscript{3} = I + 7A$

Question:48

If A, B are square matrices of same order and B is a skew-symmetric matrix, show that A’ BA is skew symmetric.

A matrix is said to be skew-symmetric if A = -A’
Given, B is a skew-symmetric matrix.
$\therefore B = -B'$
Let $C = A'BA \ldots (1)$
We have to prove C is skew-symmetric.
To prove: C = -C’
As $C = A'BA \ldots (1)$
We know that: (AB)’ = B’A’
$\\ \Rightarrow C' = (A'BA)' = A'B'(A')' \\ \Rightarrow C' = A'B'A \{ \because (A')' = A \} \\ \Rightarrow C' = A'(-B)A \\ \Rightarrow C' = -A'BA \ldots (2)$
From equation 1 and 2:
We get,
C’ = -C
Thus, we say that C = A’ BA is a skew-symmetric matrix.

Question:49

If AB = BA for any two square matrices, prove by mathematical induction that $(AB)^n = A^n B^n$.

By principle of mathematical induction we say that if a statement P(n) is true for n = 1 and if we assume P(k) to be true for some random natural number k and usnig it if we prove P(k+1) to be true we can say that P(n) is true for all natural numbers.
We have to prove that $(AB)\textsuperscript{n} = A\textsuperscript{n}B\textsuperscript{n}$
Let P(n) be the statement : $(AB)\textsuperscript{n} = A\textsuperscript{n}B\textsuperscript{n}$
So, $P(1): (AB)\textsuperscript{1} = A\textsuperscript{1}B\textsuperscript{1}$
$\\ \Rightarrow P(1) : AB = AB \\ \Rightarrow P(1) is true$
Let P(k) be true.
$\therefore (AB)\textsuperscript{k} = A\textsuperscript{k}B\textsuperscript{k} \ldots (1)$
Let’s take P(k+1) now:
$\\ \because (AB)\textsuperscript{k+1} = (AB)\textsuperscript{k}(AB) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k}(AB)$
NOTE: As we know that Matrix multiplication is not commutative. So we can’t write directly that
$A\textsuperscript{k}B\textsuperscript{k}(AB) = A\textsuperscript{k+1}B\textsuperscript{k+1}$
But we are given that AB = BA
$\\ \therefore (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k}(AB) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-1}(BAB)$
As, AB = BA
$\\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-1}(ABB) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-1}(AB\textsuperscript{2}) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-2}(BAB\textsuperscript{2}) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-2}(ABB\textsuperscript{2}) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}B\textsuperscript{k-2}(AB\textsuperscript{3})$
We observe that one power of B is decreasing while other is increasing. After certain repetitions decreasing power of B will become I
And at last step:
$\\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}I(AB\textsuperscript{k+1}) \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k}AB\textsuperscript{k+1} \\ \Rightarrow (AB)\textsuperscript{k+1} = A\textsuperscript{k+1}B\textsuperscript{k+1}$
Thus P(k+1) is true when P(k) is true.
$\therefore (AB)\textsuperscript{n} = A\textsuperscript{n} B\textsuperscript{n} \forall n \in N when AB = BA.$

Question:50

Find x, y, z if $A=\begin{bmatrix} 0 &2y &z \\x &y &-z \\x &-y &z \end{bmatrix}$ satisfies $A'= A^{-1}$

We are given the following matrix A such that,
$A=\begin{bmatrix} 0 &2y &z \\x &y &-z \\x &-y &z \end{bmatrix}$
We need to find the values of x, y and z such that $A'= A\textsuperscript{-1}$
If $A' = A\textsuperscript{-1}$
Pre-multiplying A on both sides, we get
$AA' = AA\textsuperscript{-1}$
$\Rightarrow AA'= I$where I is the identity matrix.
\begin{aligned} &\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]^{T}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]\left[\begin{array}{ccc} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\text { By the rule of matrix multiplication we have: }\\ \end{aligned}
$\Rightarrow\left[\begin{array}{ccc} 4 y^{2}+z^{2} & 2 y^{2}-z^{2} & -2 y^{2}+z^{2} \\ 2 y^{2}-z^{2} & x^{2}+y^{2}+z^{2} & x^{2}-y^{2}-z^{2} \\ -2 y^{2}+z^{2} & x^{2}-y^{2}+z^{2} & x^{2}+y^{2}+z^{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
On equating the corresponding elements of matrix as the matrix is equal to each other.
We need basically 3 equations as we have 3 variables to solve for. You can pick any three elements and equate them.
We have the following equations,
$\\4y\textsuperscript{2} + z\textsuperscript{2} = 1 \ldots (1) \\x\textsuperscript{2} + y\textsuperscript{2} + z\textsuperscript{2} = 1 \ldots (2) \\2y\textsuperscript{2} - z\textsuperscript{2} = 0 \ldots (3)$
By Adding equation 2 and 3, we get,
$\\6y\textsuperscript{2} = 1 \\ \Rightarrow y\textsuperscript{2} = 1/6$
$\\ y=\pm \frac{1}{\sqrt{6}}\\ From equation 3, we get, z^{2}=2 y^{2}\\ \Rightarrow z^{2}=2(1 / 6) \\\therefore z^{2}=1 / 3 \\z=\pm \frac{1}{\sqrt{3}} \\From equation 2, we get, \\x^{2}=1-y^{2}-z^2 \\\Rightarrow x^{2}=1-(1 / 6)-(1 / 3) \\\Rightarrow x^{2}=1-1 / 2=1 / 2 \\x=\pm \frac{1}{\sqrt{2}} \\Thus, we get that, \\\mathrm{x}=\pm \frac{1}{\sqrt{2} ;} \mathrm{y}=\pm \frac{1}{\sqrt{6} }\text { and } \mathrm{z}=\pm \frac{1}{\sqrt{3}}$

Question:51.1

If possible, using elementary row transformations, find the inverse of the following matrices
$\begin{bmatrix} 2 &-1 & 3\\-5 &3 &1\\-3 &2 &3 \end{bmatrix}$

Let A = $\begin{bmatrix} 2 &-1 & 3\\-5 &3 &1\\-3 &2 &3 \end{bmatrix}$
To apply elementary row transformations we write:
A = IA where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XA
And this X is called inverse of $A = A^{-1}$
Note: Never apply row and column transformations simultaneously over a matrix.
So we get,
\begin{aligned} &\left[\begin{array}{ccc} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } R_{2} \rightarrow R_{2}+R_{1}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } R_{1} \rightarrow R_{1}+R_{2}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A} \end{aligned}
Applying R2→ R2 - 3R1
\begin{aligned} &\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow(-1) \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \end{aligned}
$\text { Applying } R_{1} \rightarrow R_{1}+R_{2}$
$\left[\begin{array}{ccc} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] A$
$\\\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+10 \mathrm{R}_{3} \text { and } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+17 \mathrm{R}_{3}\\ \Rightarrow\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \text { Applying } \mathrm{R}_{1} \rightarrow(-1) \mathrm{R}_{1} \text { and } \mathrm{R}_{2} \rightarrow(-1) \mathrm{R}_{2}\\$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right] A$
$\text { As we have an Identity Matrix in LHS, }\\ \\\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right]$

Question:51.2

If possible, using elementary row transformations, find the inverse of the following matrices
$\begin{bmatrix} 2 &3 &-3 \\-1 &-2 &2 \\1 &1 &-1 \end{bmatrix}$

Let A = $\begin{bmatrix} 2 &3 &-3 \\-1 &-2 &2 \\1 &1 &-1 \end{bmatrix}$
To apply elementary row transformations we write:
A = IA where I is the identity matrix
We proceed with solving the problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XA
And this X is called inverse of $A = A^{-1}$
Note: Never apply row and column transformations simultaneously over a matrix.
So we get:
$\begin{array}{l} {\left[\begin{array}{ccc} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 2 & 3 & -3 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}}\\ \\ \text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-2 \mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -1 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \end{array}$
As second row of LHS contains all zeros, so we aren’t going to get any matrix in LHS.
∴ Inverse of A does not exist.
Hence, A-1 does not exist.

Question:51.3

If possible, using elementary row transformations, find the inverse of the following matrices
$\begin{bmatrix} 2 &0 &-1 \\5 &1 &0 \\0 &1 &3 \end{bmatrix}$

Let A = $\begin{bmatrix} 2 &0 &-1 \\5 &1 &0 \\0 &1 &3 \end{bmatrix}$
To apply elementary row transformations we write:
A = IA where I is the identity matrix
We proceed with solving our problem in such a way that LHS becomes I and the transformations in I give us a new matrix such that
I = XA
And this X is called inverse of $A = A^{-1}$
Note: Never apply row and column transformations simultaneously over a matrix.
So we get,
\begin{aligned} &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-(5 / 2) \mathrm{R}_{1}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A\\ &\text { Applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -\frac{5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1 \end{array}\right] \mathrm{A}\\ \end{aligned}
\begin{aligned} &\text { Applying } R_{2} \rightarrow R_{2}-5 R_{3}\\ &\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -15 & 6 & -5 \\ \frac{5}{2} & -1 & 1 \end{array}\right]\\ &\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+2 \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 6 & -2 & 2 \\ -15 & 6 & -5 \\ \frac{5}{2} & -1 & 1 \end{array}\right]\\ &\text { Applying } \mathrm{R}_{1} \rightarrow(1 / 2) \mathrm{R}_{1} \text { and } \mathrm{R}_{3} \rightarrow 2 \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ -5 & -2 & 2 \end{array}\right] \mathrm{A}\\ \end{aligned}
\begin{aligned} &\text { As we have Identity matrix in LHS, we get, }\\ &\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ -5 & -2 & 2 \end{array}\right] \end{aligned}

Question:52

Express the matrix $\begin{bmatrix} 2 &3 & 1\\1 &-1 &2 \\4 &1 &2 \end{bmatrix}$ as the sum of a symmetric and a skew symmetric matrix.

If A is any matrix then it can be written as the sum of a symmetric and skew symmetric matrix.
Symmetric matrix is given by 1/2(A + A’)
Skew symmetric is given by 1/2(A - A’)
And A = 1/2(A + A’) + 1/2(A - A’)
Here, A = $\begin{bmatrix} 2 &3 & 1\\1 &-1 &2 \\4 &1 &2 \end{bmatrix}$
Symmetric matrix is given by –
$\Rightarrow \frac{1}{2}\left ( A+A' \right )=\frac{1}{2}\left (\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]+\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]^{\mathrm{T}} \right )$
$\begin{array}{l} \Rightarrow 1 / 2\left(A+A^{\prime}\right)={\frac{1}{2}}\left(\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]+\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right]\right) \\ \end{array}$
$\Rightarrow 1 / 2\left(A+A^{\prime}\right)={\frac{1}{2}}\left(\left[\begin{array}{ccc} 2+2 & 3+1 & 1+4 \\ 1+3 & -1-1 & 2+1 \\ 4+1 & 1+2 & 2+2 \end{array}\right]\right)$
$\Rightarrow 1 / 2\left(A+A^{\prime}\right)=\left[\begin{array}{ccc} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]$
Skew Symmetric matrix is given by –
$\frac{1}{2}\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]-\frac{1}{2}\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]^{\mathrm{T}}$
$\Rightarrow \frac{1}{2}\left ( A-A' \right )=\frac{1}{2}\left (\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]-\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]^{\mathrm{T}} \right )$
$\begin{array}{l} \Rightarrow 1 / 2\left(A-A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right]-\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right]\right) \\ \end{array}$
$\Rightarrow 1 / 2\left(A-A^{\prime}\right)=\frac{1}{2}\left(\left[\begin{array}{ccc} 2-2 & 3-1 & 1-4 \\ 1-3 & -1+1 & 2-1 \\ 4-1 & 1-2 & 2-2 \end{array}\right]\right)$
$\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]$
$\therefore A=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]$

Question:53

The matrix $P=\begin{bmatrix} 0 &0 &4 \\0 &4 &0 \\4 &0 &0 \end{bmatrix}$ is a
A. square matrix
B. diagonal matrix
C. unit matrix
D. none

As P has equal number of rows and columns and thus it matches with the definition of square matrix.
The given matrix does not satisfy the definition of unit and diagonal matrices.
Hence, we can say that,
∴ Option (A) is the only correct answer.

Question:54

Total number of possible matrices of order 3 × 3 with each entry 2 or 0 is
A. 9
B. 27
C. 81
D. 512

D)
As the above matrix has a total 3× 3 = 9 element, then
As each element can take 2 values (0 or 2)
∴ By simple counting principle we can say that total number of possible matrices = total number of ways in which 9 elements can take possible values = $2^9$ = 512
Clearly it matches with option D.
Hence we can say that,
∴ Option (D) is the only correct answer.

Question:55

If $\left[\begin{array}{cc} 2 x+y & 4 x \\ 5 x-7 & 4 x \end{array}\right]=\left[\begin{array}{cc} 7 & 7 y-13 \\ y & x+6 \end{array}\right]$ then the value of x + y is
A. x = 3, y = 1
B. x = 2, y = 3
C. x = 2, y = 4
D. x = 3, y = 3

We are given that,
$\left[\begin{array}{cc} 2 x+y & 4 x \\ 5 x-7 & 4 x \end{array}\right]=\left[\begin{array}{cc} 7 & 7 y-13 \\ y & x+6 \end{array}\right]$
By equating the of two matrices, we get-
$\\4x = x + 6 \\ \Rightarrow 3x = 6 \\ \Rightarrow x = 2$
Also, 2x + y = 7
$\\ \Rightarrow y = 7 - 2x = 7 - 4 = 3 \\ \therefore y = 3$
As only option (B) matches with our answer.
Hence, we can say that,
$\therefore$ Option(B) is the correct answer.

Question:56

We will use Inverse trigonometric function to solve the problem
$cos\textsuperscript{-1} x + sin\textsuperscript{-1} x = \pi /2 \: \: and \: \: cot\textsuperscript{-1} x + tan\textsuperscript{-1} x = \pi /2$
As $A=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & \cot ^{-1}(\pi \mathrm{x}) \end{array}\right], \mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & -\tan ^{-1}(\pi \mathrm{x}) \end{array}\right]$
$\begin{array}{l} \therefore A-B=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(x \pi)+\cos ^{-1}(x \pi) & 0 \\ 0 & \tan ^{-1} \pi x+\cot ^{-1} \pi x \end{array}\right] \\ \Rightarrow A-B=\left[\begin{array}{cc} \frac{\pi}{2} \times \frac{1}{\pi} & 0 \\ 0 & \frac{\pi}{2} \times \frac{1}{\pi} \end{array}\right]=\left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \\ \therefore A-B=\frac{1}{2}\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\frac{1}{2} I \end{array}$
As it matches with option (D)
Hence, we can say that,
∴ option(D) is the only correct answer.

Question:57

If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A - 2B) is
A. m × 3
B. 3 × 3
C. m × n
D. 3 × n

As order of A is 3 × m and order of B is 3 × n
As m = n. So, order of A and B is same = 3 × m
∴ Subtraction can be carried out.
And (5A - 3B) also has same order.
Hence option D is correct

Question:58

If $A= \begin{bmatrix} 0 &1 \\1 &0 \end{bmatrix}$ then $A^2$ is equal to
A. $\begin{bmatrix} 0 &1 \\1 &0 \end{bmatrix}$
B.$\begin{bmatrix} 1&0 \\1 &0 \end{bmatrix}$
C.$\begin{bmatrix} 0&1 \\0 &1 \end{bmatrix}$
D.$\begin{bmatrix} 1&0 \\0 &1 \end{bmatrix}$

\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\\ &\therefore \mathrm{A}^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\\ &\text { By the rule of matrix multiplication, we have, }\\ &\Rightarrow \mathrm{A}^{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \text { which matches with option (D) } \end{aligned}
Hence we can say that,
∴ Option (D) is the correct answer.

Question:59

We are given that,
$a\textsubscript{11} = 0 , a\textsubscript{12} = 1 , a\textsubscript{21} = 1 and a\textsubscript{22} = 0$
$\begin{array}{l} \therefore A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \\ \therefore A^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{array}$
According to the rule of matrix multiplication:
$\begin{array}{l} \therefore A^2=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ \end{array}$ which matches with option (A)
Hence we can say that,
∴ Option (A) is the correct answer.

Question:60

The matrix $\begin{bmatrix} 1 &0 &0 \\0 &2 &0 \\0 &0 &4 \end{bmatrix}$ is a
A. identity matrix
B. symmetric matrix
C. skew symmetric matrix
D. none of these

\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]\\ &\text { Then, }\\ &A^{\prime}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]^{T}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]=A \end{aligned}
As, $A^T = A$
∴ It is symmetric matrix.
Hence we can say that,
∴ Option(B) is the correct answer.

Question:61

The matrix $\begin{bmatrix} 0 &-5 &8 \\5 &0 &12 \\-8 &-12 &0 \end{bmatrix}$ is a
A. diagonal matrix
B. symmetric matrix
C. skew symmetric matrix
D. scalar matrix

Let A = $\begin{bmatrix} 0 &-5 &8 \\5 &0 &12 \\-8 &-12 &0 \end{bmatrix}$
$\mathrm{A}^{\prime}=\left[\begin{array}{ccc} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} 0 & 5 & -8 \\ -5 & 0 & -12 \\ 8 & 12 & 0 \end{array}\right]=-\mathrm{A}$
As $A^T = -A$
∴ It is skew - symmetric matrix.
Hence, we can say that,
∴ Option(C) is the correct answer.

Question:62

If A is matrix of order m × n and B is a matrix such that AB’ and B’A are both defined, then order of matrix B is
A. m × m
B. n × n
C. n × m
D. m × n

As AB’ is defined. So, B’ must have n rows.
∴ B has n columns.
And, B’A is also defined. As, A’ has order n × m
∴ B’A to exist B must have m rows.
∴ m × n is the order of B.
Hence we can say that,
Option (D) is the correct answer.

Question:63

If A and B are matrices of same order, then (AB’ - BA’) is a
A. skew symmetric matrix
B. null matrix
C. symmetric matrix
D. unit matrix

Let C = (AB’ - BA’)
C’ = (AB’ - BA’)’
$\\ \Rightarrow$ C’ = (AB’)’ - (BA’)’
$\\ \Rightarrow$ C’ = (B’)’A’ - (A’)’B’
$\\ \Rightarrow$ C’ = BA’ - AB’
$\\ \Rightarrow$ C’ = -C
∴ C is a skew-symmetric matrix.
Clearly Option (A) matches with our deduction.
Hence we can say that,
∴ Option (A) is the correct.

Question:64

As, $(A - I)^3 + (A + I)^3 - 7A$
Use $a^3 + b^3 = (a + b)(a^2 + ab + b^2)$
Also, $A^2 = I$
$(A - I)^3 + (A + I)^3 - 7A$
$\begin{array}{l} =A^{3}-3 A^{2}+3 A-I^{3}+A^{3}+3 A^{2}+3 A+I^{3}-7 A \\ =2 A^{3}+6 A-7 A \\ =2 A^{2} \cdot A+6 A-7 A \\ =2 I \cdot A+6 A-7 A \\ =2 A+6 A-7 A=8 A-7 A=A \end{array}$
∴ then $(A - I)^3 + (A + I)^3 - 7A= A$
Clearly our answer is similar to option (A)
Hence, we can say that,
∴ option (A) is the correct answer.

Question:65

For any two matrices A and B, we have
A. AB = BA
B. AB ≠ BA
C. AB = O
D. None of the above

For any two matrix:
Not always option A , B and C are true.
Hence we can say that,
∴ Option (D) is the only suitable answer

Question:66

On using elementary column operations C2→ C2 — 2C1 in the following matrix equation
$\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]$ we have:

A.$\left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -2 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]$
B. $\left[\begin{array}{cc}1 & -5 \\ 0 & 4\end{array}\right]=\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\left[\begin{array}{cc}3 & -5 \\ -0 & 5\end{array}\right]$
C.$\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -2 & 4\end{array}\right]$
D. $\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]$

$\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]$
For column transformation, we operate the post matrix.
As,
$\left[\begin{array}{cc} 1 & -3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 1 \\ 2 & 4 \end{array}\right]$
By Applying C2→ C2 — 2C1,
$\left[\begin{array}{cc}1 & -5 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ 2 & 0\end{array}\right]$
Clearly, it matches with option (D).
Hence we can say that,
∴ Option (D) is the correct answer.

Question:67

On using elementary row operation R1→ R1 — 3R2 in the following matrix equation:

$\left[\begin{array}{ll} 4 & 2 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]$

$\begin{array}{l} \text { A. }\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -7 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \\ \\B.{\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} -1 & -3 \\ 1 & 1 \end{array}\right]} \\ \\C.{\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ 1 & -7 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right]} \\\\ D.{\left[\begin{array}{cc} 4 & 2 \\ -5 & -7 \end{array}\right]=\left[\begin{array}{cc} 1 & 2 \\ -3 & -3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right]} \end{array}$

Elementary row transformation is applied on the first matrix of RHS.
$\left[\begin{array}{ll} 4 & 2 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]$
By Applying R1→ R1 — 3R2 we get -
$\left[\begin{array}{ll} 4 & 2 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right]$
$\left[\begin{array}{cc} -5 & -7 \\ 3 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -7 \\ 0 & 3 \end{array}\right]\left[\begin{array}{cc} 2 & 0 \\ 1 & 1 \end{array}\right] \\$
Clearly it matches with option (A)
Hence we can say that,
∴ Option (A) is the correct answer.

Question:68

Fill in the blanks in each of the
______ matrix is both symmetric and skew symmetric matrix.

A Zero matrix
∴ Let A be the symmetric and skew symmetric matrix.
⇒ A’=A (Symmetric)
⇒ A’=-A (Skew-Symmetric)
Considering the above two equations,
⇒ A=-A
⇒ 2A=0
⇒ A=0 (A Zero Matrix)
Hence Zero matrix is both symmetric and skew symmetric matrix.

Question:69

Fill in the blanks in each of the
Sum of two skew symmetric matrices is always _______ matrix.

A skew symmetric matrix
∴ Let A and B are two skew symmetric matrices.
$\\ \Rightarrow A'=-A ..(1) \\ \Rightarrow B'=-B ..(2)$
Now Let A+B=C ..(3)
$\\ \Rightarrow C'=(A+B)'=A'+B' \\ \Rightarrow A'+B'=(-A)+(-B) \\ \Rightarrow (-A)+(-B)=-(A+B)=-C \\ \Rightarrow C'=-C (Skew Symmetric matrix)$

Question:70

Fill in the blanks in each of the
The negative of a matrix is obtained by multiplying it by ________.

The negative of a matrix is obtained by multiplying it by -1.
For example:
\begin{aligned} \text { Let}&A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ \\ \end{aligned}
$\text { So }\left[\begin{array}{ll} -1 & -2 \\ -3 & -4 \end{array}\right]=-1\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ =-A$

Question:71

Fill in the blanks in each of the
The product of any matrix by the scalar _____ is the null matrix.

The null matrix is the one in which all elements are zero.
If we want to make A = $\begin{bmatrix} 1 &2 \\3 &4 \end{bmatrix}$ a null matrix we need to multiply it by 0.
0A = $0\begin{bmatrix} 1 &2 \\3 &4 \end{bmatrix}$
$\begin{bmatrix} 0 &0 \\0 &0 \end{bmatrix}$
Hence, we can say that,
The product of any matrix by the scalar 0 is the null matrix.

Question:72

Fill in the blanks in each of the
A matrix which is not a square matrix is called a _____ matrix.

Rectangular Matrix
As we know in a square matrix is the one in which there are same number of rows and columns.
Eg: A = $\begin{bmatrix} 1 &2 \\3 &4 \end{bmatrix}$
Here there are 2 rows and 2 columns.
The matrix which is not square is called rectangular matrix as it does not have same number of rows and columns.
Eg $\begin{bmatrix} 1 &2 &3 \\4 &5 &6 \end{bmatrix}$
Here number of rows are 2 and columns are 3.

Question:73

Fill in the blanks in each of the
Matrix multiplication is _____ over addition.

Distributive
⇒ Matrix multiplication is distributive over addition.
i.e A(B+C)=AB+AC
and (A+B)C=AC+BC

Question:74

Fill in the blanks in each of the
If A is a symmetric matrix, then $A^3$ is a ______ matrix.

$A\textsuperscript{3}$ is Also a symmetric matrix.
We are given that: A’=A ..(1)
$\\ \Rightarrow (A\textsuperscript{2})'=(AA)'=A'A' \\ \Rightarrow A'A'=(A)(A)=A\textsuperscript{2} \\ \Rightarrow (A\textsuperscript{2})'=A\textsuperscript{2} (symmetric matrix) ..(2) \\ \Rightarrow (A\textsuperscript{3})'=(A(A\textsuperscript{2}))'=(A\textsuperscript{2})'A' \\ \Rightarrow (A\textsuperscript{2})'A'=A\textsuperscript{2}A= A\textsuperscript{3} (Using (1) and (2) ) \\ \Rightarrow (A\textsuperscript{3})'=A\textsuperscript{3} (symmetric matrix)$

Question:75

Fill in the blanks in each of the
If A is a skew symmetric matrix, then $A^2$ is a _________.

$A\textsuperscript{2}$ is a symmetric matrix.
We are given that: A'=-A
$\\ \Rightarrow (A\textsuperscript{2})'=(AA)'=A'A' \\ \Rightarrow A'A'=(-A)(-A)=A\textsuperscript{2} \\ \Rightarrow (A\textsuperscript{2})'=A\textsuperscript{2} (symmetric matrix)$

Question:76

Fill in the blanks in each of the
If A and B are square matrices of the same order, then
(i) (AB)’ = ________.
(ii) (kA)’ = ________. (k is any scalar)
(iii) [k (A - B)]’ = ________.

(i) (AB)’ = ________.
(AB)’ = B’A’
Let A be matrix of order m× n and B be of n× p.
A’ is of order n× m and B’ is of order p× n.
Hence, we get, B’ A’ is of order p× m.
So, AB is of order m× p.
And (AB)’ is of order p× m.
We can see (AB)’ and B’ A’ are of same order p× m.
Hence proved, (AB)’ = B’ A’
(ii) (kA)’ = ________. (k is any scalar)
If a scalar “k” is multiplied to any matrix the new matrix becomes
K times of the old matrix.
$\begin{array}{l} \text { Eg: } A=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \\ 2 A=2\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 6 & 8 \end{array}\right] \\ (2 A)=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right] \\ A^{\prime}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right] \end{array}$
Now 2A’ = $2\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]$
Hence (2A)’ =2A’
Hence (kA)’ = k(A)’
(iii) [k (A - B)]’ = ________.
\begin{aligned} &A=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]\\ &A'=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]\\ &2A'=\left[\begin{array}{ll} 10 & 14 \\ 14 & 12 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]\\ &B^{\prime}=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\\ &2 B^{\prime}={2}\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\\ &=\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]\\ &A-B=\left[\begin{array}{ll} 4 & 5 \\ 4 & 2 \end{array}\right] \end{aligned}
$\begin{array}{l} \text { Now Let } k=2 \\ 2(A-B)=2\left[\begin{array}{ll} 4 & 5 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 8 & 10 \\ 8 & 4 \end{array}\right] \\ {[2(A-B)]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 4 \end{array}\right]} \\ 2 A^{\prime}-2 B'=\left[\begin{array}{ll} 10 & 14 \\ 14 & 12 \end{array}\right]-\left[\begin{array}{ll} 2 & 6 \\ 4 & 8 \end{array}\right]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 4 \end{array}\right] \\ A^{\prime}-B'=\left[\begin{array}{ll} 5 & 7 \\ 7 & 6 \end{array}\right]-\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{ll} 4 & 4 \\ 5 & 3 \end{array}\right] \\ 2\left(A^{\prime}-B^{\prime}\right)=2\left[\begin{array}{ll} 4 & 4 \\ 5 & 3 \end{array}\right]=\left[\begin{array}{cc} 8 & 8 \\ 10 & 6 \end{array}\right] \\ \text { Hence we can see }[k(A-B)]^{\prime}=k(A)^{\prime}-k(B)^{\prime}=k\left(A^{\prime}-B^{\prime}\right) \end{array}$

Question:77

Fill in the blanks in each of the
If A is skew symmetric, then kA is a ______. (k is any scalar)

A skew symmetric matrix.
We are given that, A’=-A
⇒ (kA)’=k(A)’=k(-A)
⇒ (kA)’=-(kA)

Question:78

Fill in the blanks in each of the
If A and B are symmetric matrices, then
(i) AB - BA is a _________.
(ii) BA - 2AB is a _________.

(i) AB - BA is a Skew Symmetric matrix
We are given that A’=A and B’=B
⇒ (AB-BA)’=(AB)’-(BA)’
⇒ (AB)’-(BA)’=B’A’-A’B’
⇒ B’A’-A’B’=BA-AB=-(AB-BA)
⇒ (AB-BA)’=-(AB-BA) (skew symmetric matrix)
\begin{aligned} &\text { For example, Let }\\ &A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{AB}=\left[\begin{array}{ll} 7 & 5 \\ 7 & 8 \end{array}\right] \text { and } \mathrm{BA}=\left[\begin{array}{ll} 7 & 7 \\ 5 & 8 \end{array}\right]\\ &\Rightarrow A B-B A=\left[\begin{array}{cc} 0 & -2 \\ 2 & 0 \end{array}\right]\\ &\Rightarrow(A B-B A)^{\prime}=\left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right]\\ &\Rightarrow=(A B-B A)=\left[\begin{array}{cc} 0 & 2 \\ -2 & 0 \end{array}\right] \end{aligned}
(ii) BA - 2AB is a Neither Symmetric nor Skew Symmetric matrix
Given A’=A and B’=B
⇒ (BA-2AB)’=(BA)’-(2AB)’
⇒ (BA)’-(2AB)’=A’B’-2B’A’
⇒ A’B’-2B’A’=AB-2BA=-(2BA-AB)
⇒ (BA-2AB)’=-(2BA-AB)
\begin{aligned} &\text { For example Let }\\ &A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right]\\ &B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{AB}=\left[\begin{array}{ll} 7 & 5 \\ 7 & 8 \end{array}\right] \text { and } B A=\left[\begin{array}{ll} 7 & 7 \\ 5 & 8 \end{array}\right]\\ &\Rightarrow B A-2 A B=\left[\begin{array}{cc} 7 & -3 \\ -9 & 8 \end{array}\right] \end{aligned}

Question:79

Fill in the blanks in each of the
If A is symmetric matrix, then B’AB is _______.

B’AB is a symmetric matrix.
Solution:
Given A is symmetric matrix
⇒ A’=A ..(1)
Now in B’AB,
Let AB=C ..(2)
⇒ B’AB=B’C
Now Using Property (AB)’=B’A’
⇒ (B’C)’=C’(B’)’ (As (B’)’=B)
⇒ C’(B’)’=C’B
⇒ C’B=(AB)’B (Using Property (AB)’=B’A’)
⇒ (AB)’B=B’A’B (Using (1))
⇒ B’A’B= B’AB
⇒ Hence (B’AB)’= B’AB

Question:80

Fill in the blanks in each of the
If A and B are symmetric matrices of same order, then AB is symmetric if and only if ______.

Given A and B are symmetric matrices,
⇒ A’=A ..(1)
⇒ B’=B ..(2)
Let AB is a Symmetric matrix:-
⇒ (AB)’=AB
Using Property (AB)’=B’A’
⇒ B’A’=AB
⇒ Now using (1) and (2)
⇒ BA=AB
Hence A and B matrix commute.

Question:81

Fill in the blanks in each of the
In applying one or more now operations while finding $A^{-1}$ by elementary row operations, we obtain all zeros in one or more, then $A^{-1}$ ______.

$A^{-1}$ Does not exist,
$A^{-1}=\frac{1}{|A|} a d j(A)$
And |A|=0 if there are one or more rows or columns with all zero elements.

Question:82

Which of the following statements are True or False
A matrix denotes a number.

False
A matrix is an ordered rectangular array of numbers of functions.
Only a matrix of order (1×1) denotes a number.
For example, $[8]_{1\times 1}=8$

Question:83

Which of the following statements are True or False

False
Matrices having same order can be added.
For example
$\begin{array}{l} \text { Let } A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right] \\ B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \\ \Rightarrow A+B=\left[\begin{array}{ll} 2 & 5 \\ 5 & 3 \end{array}\right] \end{array}$

Question:84

Which of the following statements are True or False
Two matrices are equal if they have same number of rows and same number of columns.

False
Two matrices are equal if they have same number of rows and same number of columns and corresponding elements within each matrix are equal or identical.
For example:
$\Rightarrow A=\left[\begin{array}{ll} 2 & 5 \\ 5 & 3 \end{array}\right], B=\left[\begin{array}{ll} 2 & 5 \\ 5 & 3 \end{array}\right]$
Here both matrices have two rows and two columns.
Also, they both have same elements.

Question:85

Which of the following statements are True or False
Matrices of different order cannot be subtracted.

True
Matrices of only same order can be added or subtracted.
Let A = $\begin{bmatrix} 1 &3 \\3 &2 \end{bmatrix}$
B= $\begin{bmatrix} 1 & 0 \end{bmatrix}$
⇒ A-B= Not possible

Question:86

Which of the following statements are True or False
Matrix addition is associative as well as commutative.

True
1. A+B=B+A (commutative)
2. (A+B)+C= A+(B+C) (associative)

Question:87

Which of the following statements are True or False
Matrix multiplication is commutative.

False
In general matrix multiplication is not commutative
But it’s associative.
⇒ (AB)C=A(BC)

Question:88

Which of the following statements are True or False
A square matrix where every element is unity is called an identity matrix.

False
A square matrix where every element of the leading diagonal is unity and rest elements are zero is called an identity matrix.
i.e $I=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

Question:89

Which of the following statements are True or False
If A and B are two square matrices of the same order, then A + B = B + A.

True
If A and B are two square matrices of the same order, then A + B = B + A ( Property of square matrix)
For example,
$\begin{array}{l} \text {Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right] \\ A+B=\left[\begin{array}{lll} 5 & 7 & 10 \\ 7 & 6 & 10 \\ 10 & 10 & 10 \end{array}\right] \\ \Rightarrow \quad B+A=\left[\begin{array}{lll} 5 & 7 & 10 \\ 7 & 6 & 10 \\ 10 & 10 & 10 \end{array}\right] \\ \end{array}$

Question:90

Which of the following statements are True or False
If A and B are two matrices of the same order, then A - B = B - A.

False
If A and B are two matrices of the same order,
then A - B = -(B - A)
For example,
\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right]\\ &A-B=\left[\begin{array}{lll} -3 & -3 & -4 \\ -3 & -4 & -2 \\ -4 & -2 & -8 \end{array}\right]\\ &B-A=\left[\begin{array}{lll} 3 & 3 & 4 \\ 3 & 4 & 2 \\ 4 & 2 & 8 \end{array}\right]\\ &\Rightarrow-(B-A)=\left[\begin{array}{ccc} -3 & -3 & -4 \\ -3 & -4 & -2 \\ -4 & -2 & -8 \end{array}\right] \end{aligned}

Question:91

Which of the following statements are True or False
If matrix AB = O, then A = O or B = O or both A and B are null matrices.

False
Its not necessary that for multiplication of matrix A and B to be 0 one of them has to be a null matrix.
For example,
$\begin{array}{c} \text {Let } A=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & -1 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right] \\ A \times B=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & -1 \end{array}\right]\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{array}$

Question:92

Which of the following statements are True or False
Transpose of a column matrix is a column matrix.

False
Transpose of a column matrix is a Row matrix and vice-versa.
$\begin{array}{l} \text { Let } A=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \text { (Column Matrix) } \\ \Rightarrow A^{\prime}=\left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right] \text { (Row Matrix) } \end{array}$

Question:93

Which of the following statements are True or False
If A and B are two square matrices of the same order, then AB = BA.

False
Matrix multiplication is not commutative.
For example,
$\begin{array}{l} \text { Let } A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 2 \end{array}\right] \\ B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] \\ \Rightarrow A B=\left[\begin{array}{ll} 7 & 5 \\ 7 & 8 \end{array}\right] \text { and } B A=\left[\begin{array}{ll} 7 & 7 \\ 5 & 8 \end{array}\right] \\ \Rightarrow A B \neq B A \end{array}$

Question:94

Which of the following statements are True or False
If each of the three matrices of the same order are symmetric, then their sum is a symmetric matrix.

True
For example,
$\begin{array}{l} \text { Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right], B=\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right] \end{array}$ and $C = \begin{bmatrix} 3 &9 &1 \\ 9 &2 &8 \\1 &8 &5 \end{bmatrix}$
$\Rightarrow A+B+C=\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 4 \\ 3 & 4 & 1 \end{array}\right]+\left[\begin{array}{lll} 4 & 5 & 7 \\ 5 & 5 & 6 \\ 7 & 6 & 9 \end{array}\right]+\left[\begin{array}{lll} 3 & 9 & 1 \\ 9 & 2 & 8 \\ 1 & 8 & 5 \end{array}\right]$
$A+B+C=\left[\begin{array}{lll} (1+4+3) & (2+5+9) & (3+7+1) \\ (2+5+9) & (1+5+2) & (4+6+8) \\ (3+7+1) & (4+6+8) & (1+9+5) \end{array}\right]$
$A+B+C=\left[\begin{array}{ccc} 8 & 16 & 11 \\ 16 & 8 & 18 \\ 11 & 18 & 15 \end{array}\right]$

Question:95

Which of the following statements are True or False
If A and B are any two matrices of the same order, then (AB)’ = A’B’.

False
If A and B are any two matrices for which AB is defined, then
(AB)’=B’A’.

Question:96

Which of the following statements are True or False
If (AB)’ = B’A’, where A and B are not square matrices, then number of rows in A is equal to number of columns in B and number of columns in A is equal to number of rows in B.

True
If A and B are any two matrices for which AB is defined, and (AB)’=B’A’.
$A is of order m \times n and B is of order p \times q\\ (A B)^{\prime}=B^{\prime} A^{\prime} \\ \left.A_{(m \times n)} B_{(p \times q)}\right. \text { defined if } \Rightarrow n=p$
$A B is of order m \times q$
$(A B)^{\prime} is of order q \times m$
$B' is order of q \times p$
$A^{\prime} is of order n \times m$
$B^{\prime} A^{\prime} is defined if \Rightarrow p=n$
$B'A' is of order q \times m$
Hence given statement is true.

Question:97

Which of the following statements are True or False
If A, B and C are square matrices of same order, then AB = AC always implies that B = C.

False
If AB = AC => B=C
The above condition is only possible if matrix A is invertible
$\\(i.e \vert A \vert \neq 0). \\ \Rightarrow If A is invertible, then \\ \Rightarrow A\textsuperscript{-1}(AB)= A\textsuperscript{-1}(AC) \\ \Rightarrow (A\textsuperscript{-1}A)B = (A\textsuperscript{-1}A)C \\ \Rightarrow IB=IC \\ \Rightarrow B=C$

Question:98

Which of the following statements are True or False
AA’ is always a symmetric matrix for any square matrix A.

True
(AA’)’=(A’)’A’
As we know that (A’)’ = A
(AA’)’=AA’ (Condition of symmetric matrix)

Question:99

Which of the following statements are True or False

If $A=\left[\begin{array}{rrr} 2 & 3 & -1 \\ 1 & 4 & 2 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{array}\right]$ then AB and BA are defined and equal.

False
Here A has an order (2×3) and B has an order (3×2),
Hence AB is defined and will give an output matrix of order (2×2)
And BA is also defined but will give an output matrix of order (3×3).
⇒ AB ≠ BA

Question:100

Which of the following statements are True or False
If A is skew symmetric matrix, then $A^2$ is a symmetric matrix.

True
For skew symmetric matrix A’=-A
$\\ \Rightarrow (A\textsuperscript{2})'=(AA)'=A'A' \\ \Rightarrow A'A'=(-A)(-A)=A\textsuperscript{2} \\ \Rightarrow (A\textsuperscript{2})'=A\textsuperscript{2} (symmetric matrix)$

Question:101

Which of the following statements are True or False
$(AB)^{-1} = A^{-1}.B^{-1}$, where A and B are invertible matrices satisfying cumulative property with respect to multiplication.

True
Given:
$AB=BA$
\begin{aligned} (A B)\left(A^{-1} B^{-1}\right) &=(B A)\left(A^{-1} B^{-1}\right) \\ &=B\left(A A^{-1}\right) B^{-1} \\ &=B \cdot I \cdot B^{-1}=B \cdot B^{-1}=I \\ (A B)^{-1} &=A^{-1} B^{-1} \end{aligned}

Students can make use of NCERT exemplar Class 12 Maths solutions chapter 3 pdf download, to access it offline. We will help the students to understand the matrices and its functions and operations by solving the questions given in NCERT.

NCERT exemplar solutions for Class 12 Maths chapter 3 Matrices Sub-topics covered

## The Sub-Topics That Are Covered Under The Class 12 Maths NCERT Exemplar Solutions Chapter 3 Matrices Are:

• Introduction
• Matrix
• Order of matrix
• Types of matrices
• Operations on matrices
• Multiplication of a matrix by a scalar
• Properties of scalar multiplication of matrix
• Multiplication of matrices
• Properties of multiplication of matrices
• Transpose of matrix
• Properties of the transpose of matrices
• Symmetric and skew-symmetric matrices
• Elementary operation of matrix
• Invertible matrices
• Inverse of matrices by elementary operations
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## What can you learn in NCERT Exemplar Class 12 Maths Solutions Chapter 3?

• In NCERT Exemplar Class 12 Math solutions chapter 3, the students will learn about various types of matrices, and their properties along with how the operations are used on them.
• The transpose of a matrix and its properties are discussed in detail. Along with it the definition, properties, and theorems of symmetric and skew-symmetric matrices.
• In NCERT exemplar Class 12 Maths chapter 3 solutions, students will also learn about the transformations of the matrix and will get a detailed idea about how the matrices are inverted with elementary operations.
• The students will understand the properties of matrices and its types in detail. They will learn about the operations and elementary operation on matrices.
• One can solve higher level questions with much more confidence and ease.
• NCERT Exemplar Class 12 Math chapter 3 solutions are compiled by the experts and highly experienced teachers who have years of experience to back their know-how.
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## NCERT Exemplar Class 12 Maths Solutions

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Applications of Derivatives Chapter 7 Integrals Chapter 8 Applications of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability

## Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 3

• NCERT exemplar solutions for Class 12 Math chapter 3 will help the students to understand the basics. It will help in making the student understand what matrices and its fundamentals are
• The introduction also covers applications of matrices
• NCERT exemplar Class 12 Maths solutions chapter 3 covers the order of a matrix and also how the student can create a matrix and solve it for high order problem solving
• NCERT exemplar Class 12 Maths chapter 3 solutions are prepared in simple language only so that it is easier for the student to grasp the idea assuredly and in a self-explanatory way.

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### NCERT Exemplar Class 12 Solutions

 NCERT Exemplar Class 12 Chemistry Solutions NCERT Exemplar Class 12 Physics Solutions NCERT Exemplar Class 12 Biology Solutions

### Also, check NCERT Solutions for questions given in the book:

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Application of Derivatives Chapter 7 Integrals Chapter 8 Application of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three Dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability

#### Also Check NCERT Books and NCERT Syllabus here

1. Who has prepared the solutions?

The NCERT exemplar solutions for Class 12 Maths chapter 3 are prepared by the team of experts. They refer to various advanced maths books to prepare these precise solutions.

2. Are NCERT exemplar Class 12 Maths solutions chapter 3 reliable?

Yes, NCERT provides precise solutions that are prepared by experts for students to prepare for their boards as well as entrance exams.

Yes, you can easily download NCERT exemplar Class 12 Maths solutions chapter 3 pdf by using the webpage to pdf tools.

4. How many questions are there in this chapter?

: There is one exercise with 146 questions in the Class 12 Maths NCERT exemplar solutions chapter 3.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

Thank you

Hope this information helps you.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9

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2 Jobs Available
##### Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

4 Jobs Available
##### Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
##### Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
##### Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Producer

An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story.

They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

2 Jobs Available
##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion.

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
##### Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available
##### Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.

5 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available
##### Structural Engineer

A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software.

2 Jobs Available
##### Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
##### AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.

4 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available