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NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

Edited By Ravindra Pindel | Updated on Sep 15, 2022 - 2:01 p.m. 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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More About NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

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

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

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