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.

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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?

Answer:

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 $a_{23}, a_{31}, a_{12}$

Answer:

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,
a_ij is the representation of elements lying in the ith row and jth column.
For a₂₃:
On comparing a_ij 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) $a_{ij} = |- 2i + 3j|$

Answer:

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$
$$ \Rightarrow $ 1 $ \leq $ i $ \leq $ 2 [$\because$ m = 2]$
And,
$\\1 $ \leq $ j $ \leq $ n\\ \\$ \Rightarrow $ 1 $ \leq $ j $ \leq $ 2 [$\because$ n = 2]$
Put i = 1 and j = 1.
$\\ \mathrm{a}_{11}=\frac{(1-2(1))^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{(1-2)^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{(-1)^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{1}{2} \\ \text { Put } \mathrm{i}=1 \text { and } \mathrm{j}=2 \\ \mathrm{a}_{12}=\frac{(1-2(2))^{2}}{2} \\ \Rightarrow \mathrm{a}_{12}=\frac{(1-4)^{2}}{2}$
$\\ \Rightarrow a_{12}=\frac{(-3)^{2}}{2} \\ \Rightarrow a_{12}=\frac{9}{2} \\ \text { Put } i=2 \text { and } j=1 \\ a_{21}=\frac{(2-2(1))^{2}}{2} \\ \Rightarrow a_{21}=\frac{(2-2)^{2}}{2} \\ \Rightarrow a_{21}=\frac{0}{2} \\ \Rightarrow a_{21}=0 \\ \text { Put } i=2 \text { and } j=2 \\ a_{22}=\frac{(2-2(2))^{2}}{2}$
$\\ \Rightarrow \mathrm{a}_{22}=\frac{(2-4)^{2}}{2} \\ \Rightarrow \mathrm{a}_{22}=\frac{(-2)^{2}}{2} \\ \Rightarrow \mathrm{a}_{22}=\frac{4}{2} \\ \Rightarrow \mathrm{a}_{22}=2$
Let the matrix formed be named A.
$\begin{aligned} &A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]\\ &\text { By substituting the values of } a_{11}, a_{12}, a_{21} \text { and } a_{22}, \end{aligned}$ the matrix formed is
$A=\left[\begin{array}{ll} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{array}\right]$
(ii). We need to construct a matrix, $a_{2 \times 2}$ , where
$a_{ij} = $ \vert $ -2i + 3j$ \vert $$
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_{11} = $ \vert $ -2(1) + 3(1)$ \vert $ \\$ \Rightarrow $ a_{11} = $ \vert $ -2 + 3$ \vert $ \\$ \Rightarrow $ a_{11} = $ \vert $ 1$ \vert $ \\$ \Rightarrow $ a_{11} = 1$
Put i = 1 and j = 2.
$\\a_{12} = $ \vert $ -2(1) + 3(2)$ \vert $ \\$ \Rightarrow $ a_{12} = $ \vert $ -2 + 6$ \vert $ \\$ \Rightarrow $ a_{12} = $ \vert $ 4$ \vert $ \\$ \Rightarrow $ a_{12} = 4$
Put i = 2 and j = 1.
$\\a_{21} = $ \vert $ -2(2) + 3(1)$ \vert $ \\$ \Rightarrow $ a_{21} = $ \vert $ -4 + 3$ \vert $ \\$ \Rightarrow $ a_{21} = $ \vert $ -1$ \vert $ \\$ \Rightarrow $ a_{21} = 1$
Put i = 2 and j = 2. .
$\\a_{22} = $ \vert $ -2(2) + 3(2)$ \vert $ \\$ \Rightarrow $ a_{22} = $ \vert $ -4 + 6$ \vert $ \\$ \Rightarrow $ a_{22} = $ \vert $ 2$ \vert $ \\$ \Rightarrow $ a_{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 $a_{ij} = e^{ix}\sin jx$

Answer:

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:
$a_{ij} = e\textsuperscript{i.x} sin jx$
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.
$\\a_{11} = e\textsuperscript{(1)x} sin (1)x \\$ \Rightarrow $ a_{11} = e\textsuperscript{x} sin x$
Put i = 1 and j = 2.

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

Put i = 2 and j = 1.

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

Put i = 2 and j = 2.
$\\a_{22} = e\textsuperscript{(2)x} sin (2)x \\$ \Rightarrow $ a_{22} = e\textsuperscript{2x} sin 2x$
For i = 3 and j = 1.
$\\a_{31} = e\textsuperscript{(3)x} sin (1)x \\$ \Rightarrow $ a_{31} = e\textsuperscript{3x} sin x$
For i = 3 and j = 2.
$\\a_{32} = e\textsuperscript{(3)x} sin (2)x \\$ \Rightarrow $ a_{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
$\begin{array}{l} A=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right] \\ B=\left[\begin{array}{cc} 2 a+2 & b^{2}+2 \\ 8 & b^{2}-5 b \end{array}\right] \end{array}$

Answer:

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,
$\\a_{11} = b_{11} \\a_{12} = b_{12} \\a_{21} = b_{21} \\a_{22} = b_{22}$
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

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

Answer:

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

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

Answer:

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:
$\mathrm{x}\left[\begin{array}{cc} 2 \mathrm{x} & 2 \\ 3 & \mathrm{x} \end{array}\right]+2\left[\begin{array}{cc} 8 & 5 \mathrm{x} \\ 4 & 4 \mathrm{x} \end{array}\right]=2\left[\begin{array}{cc} \left(\mathrm{x}^{2}+8\right) & 24 \\ (10) & 6 \mathrm{x} \end{array}\right]$

Answer:

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]$$
We need to find the value of x by comparing the elements in the two matrices.
If,
$\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]=\left[\begin{array}{ll} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array}\right]$
Then,
$\\a_{11} = b_{11} \\a_{12} = b_{12} \\a_{21} = b_{21} \\a_{22} = b_{22}$
So,
$\\2x\textsuperscript{2} + 16 = 2(x\textsuperscript{2} + 8) $ \ldots $ (i) \\2x + 10x = 48 $ \ldots $ (ii) \\3x + 8 = 20 $ \ldots $ (iii) \\x\textsuperscript{2} + 8x = 12x $ \ldots $ (iv)$
We have got equations (i), (ii), (iii) and (iv) to solve for x.
So, take equation (i).
$\\2x\textsuperscript{2} + 16 = 2x\textsuperscript{2} + 16$
We cannot find the value of x from this equation as they are similar.
Now, take equation (ii).
2x + 10x = 48
$$ \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 $(A + B) (A - B) \neq A^2 - B^2.$

Answer:

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

Answer:

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 $A^{-1}$ .

Answer:

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:

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

Answer:

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

Answer:

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 $(BA)^{2} \neq B^{2} A^{2}$

Answer:

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
$A=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{ll} 4 & 1 \\ 2 & 3 \\ 1 & 2 \end{array}\right]$

Answer:

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.

Answer:

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 $(AB)' = B'A'$ ?

Answer:

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

Answer:

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
$2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]$

Answer:

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.

Answer:

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.

Answer:

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) $A(B + C) = AB + AC$

Answer:

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

Answer:

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

Answer:

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)

Answer:

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,
AD = A(B + C)
$\Rightarrow \mathrm{AD}=\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right]$
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.

Answer:

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’)

Answer:

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’.

Answer:

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 A^T 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.

Answer:

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.

Answer:

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 $(A + B)^2 = A^2 + 2AB + B^2$ .

Answer:

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

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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

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

Answer:

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

Answer:

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 $(A + B)^2 = A^2 + B^2.$

Answer:

$\begin{aligned} &\text { As, LHS }=(A+B)^{2}=\left(\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\right)^{2}\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]^{2}=\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 1-\mathrm{x} \\ 1+\mathrm{x} & 0 \end{array}\right]\\ &\text { By the rule of matrix multiplication we can write LHS as - }\\ &\text { LHS }=\left[\begin{array}{cc} 0+(1-x)(1+x) & 0 \\ 0 & (1+x)(1-x) \end{array}\right]\\ &\Rightarrow \mathrm{LHS}=\left[\begin{array}{cc} 1-\mathrm{x}^{2} & 0 \\ 0 & 1-\mathrm{x}^{2} \end{array}\right]\\ &\text { Given } x^{2}=-1\\ &\therefore \mathrm{LHS}=\left[\begin{array}{cc} 1-(-1) & 0 \\ 0 & 1-(-1) \end{array}\right]=\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\\ \end{aligned}$
$\begin{aligned} &\mathrm{RHS}=\mathrm{A}^{2}+\mathrm{B}^{2}=\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]^{2}+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]^{2}\\ &\Rightarrow \mathrm{RHS}=\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -\mathrm{x} \\ \mathrm{x} & 0 \end{array}\right]+\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \end{aligned}$
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 $A=\left[\begin{array}{ccc} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{array}\right]$

Answer:

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.

Answer:

Question:37.1

Find inverse, by elementary row operations (if possible), of the following matrices.
$\begin{bmatrix} 1 &3 \\-5 & 7 \end{bmatrix}$
Answer:

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.
$\begin{bmatrix} 1 &-3 \\-2 & 6 \end{bmatrix}$

Answer:

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.

Answer:

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.

Answer:

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 $A^3$ .

Answer:

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

Answer:

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

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

Answer:

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 $A^2 + 2A + 7I$

Answer:

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 $\alpha$

Answer:

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.

Answer:

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

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

Answer:

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 $(I + A)^3 = 7A + I$ .

Answer:

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.

Answer:

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 $(AB)^n = A^n B^n$ .

Answer:

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 $A'= A^{-1}$

Answer:

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
$\begin{bmatrix} 2 &-1 & 3\\-5 &3 &1\\-3 &2 &3 \end{bmatrix}$

Answer:

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
$\begin{bmatrix} 2 &3 &-3 \\-1 &-2 &2 \\1 &1 &-1 \end{bmatrix}$

Answer:

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
$\begin{bmatrix} 2 &0 &-1 \\5 &1 &0 \\0 &1 &3 \end{bmatrix}$

Answer:

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.

Answer:

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

Answer:

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

Answer:

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

Answer:

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

If $A=\frac{1}{\pi}\left[\begin{array}{cc} \sin ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & \cot ^{-1}(\pi \mathrm{x}) \end{array}\right], \mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc} -\cos ^{-1}(\mathrm{x} \pi) & \tan ^{-1} \frac{\mathrm{x}}{\pi} \\ \sin ^{-1} \frac{\mathrm{x}}{\pi} & -\tan ^{-1}(\pi \mathrm{x}) \end{array}\right]$ then A - B is equal to
A. I
B. O
C. 2I
D. $\frac{1}{2}I$

Answer:

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

Answer:

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. $\begin{bmatrix} 1&0 \\0 &1 \end{bmatrix}$
Answer:

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

Question:59

If matrix $A=[a_{ij}]_{2\times 2}$ , where aij = 1 if i ≠ j
aij = 0 if i = j, then $A^2$ is equal to
A. I
B. A
C. 0
D. None of these

Answer:

We are given that,
$a_{11} = 0 , a_{12} = 1 , a_{21} = 1 $ and a_{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

Answer:

$\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]\\ &\text { Then, }\\ &A^{\prime}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]^{T}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]=A \end{aligned}$
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

Answer:

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

Answer:

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

Answer:

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

If A is a square matrix such that $A^2 = I$ , then $(A - I)^3 + (A + I)^3 - 7A$ is equal to
A. A
B. I - A
C. I + A
D. 3A

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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.

Answer:

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.

Answer:

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 ________.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

$A\textsuperscript{3}$ 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 _________.

Answer:

$A\textsuperscript{2}$ 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)]’ = ________.

Answer:

(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)

Answer:

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 _________.

Answer:

(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 _______.

Answer:

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 ______.

Answer:

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 $A^{-1}$ ______.

Answer:

$A^{-1}$ 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.

Answer:

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

Matrices of any order can be added.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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’.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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
Addition of matrices
Multiplication of a matrix by a scalar
Properties of matrix addition
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

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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

Must Read NCERT Solution subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

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.

3. Can I download the solutions for this chapter?

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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I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

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.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

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Read Complete Answer

Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

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.

I hope this was helpful!

Good Luck

Read Complete Answer

Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

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.

Read Complete Answer

Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

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.

Read Complete Answer

Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

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.

Read Complete Answer

Muskan Dhalwal 17 August,2022

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

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What can you learn in NCERT Exemplar Class 12 Maths Solutions Chapter 3?

NCERT Exemplar Class 12 Maths Solutions

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NCERT Exemplar Class 12 Solutions

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