NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

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 that 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 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 the matrix,

If the given matrix has $m n$ elements, then the dimension of the order can be given by $m * n$, where $m$ and $n$ are natural numbers.

So, if a matrix has 28 elements, which is $m n=28$, then the following possible orders can be found:

$\because m n=28$

Take m and n to be any number, so that, when they are multiplied, we get 28.

So, let $\mathrm{m}=1$ and $\mathrm{n}=28$.

Then, $m \times n=1 \times 28(=28)$

$\Rightarrow 1 \times 28$ is a possible order of the matrix with 28 elements

Take $\mathrm{m}=2$ and $\mathrm{n}=14$.

Then, $m \times n=2 \times 14(=28)$

$\Rightarrow 2 \times 14$ is a possible order of the matrix with 28 elements.

Take $\mathrm{m}=4$ and $\mathrm{n}=7$.

Then, $m \times n=4 \times 7(=28)$

$\Rightarrow 4 \times 7$ is a possible order of the matrix with 28 elements.

Take $\mathrm{m}=7$ and $\mathrm{n}=4$.

Then, $m \times n=7 \times 4(=28)$

$\Rightarrow 7 \times 4$ is a possible order of the matrix with 28 elements.

Take $\mathrm{m}=14$ and $\mathrm{n}=2$.

Then, $m \times n=14 \times 2(=28)$

$\Rightarrow 14 \times 2$ is a possible order of the matrix having 28 elements.

Take $\mathrm{m}=28$ and $\mathrm{n}=1$.

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 that 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 a similar way as given above:

Here, $m n=13$.

Take $m$ and $n$ to be any numbers so that when multiplied, we get 13.

Take $\mathrm{m}=1$ and $\mathrm{n}=13$.

Then, $m \times n=1 \times 13(=13)$

$\Rightarrow 1 \times 13$ is a possible order of the matrix with 13 elements.

Take $\mathrm{m}=13$ and $\mathrm{n}=1$.

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:

i) We have, $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a} & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right]$ the order of matrix A is $3 \times 3$

ii) We have, $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a} & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right]$ the number of elements are $3 \times 3=9$

iii) We have, $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{a} & 1 & x \\ 2 & \sqrt{3} & x^2-y \\ 0 & 5 & \frac{-2}{5}\end{array}\right]$

Since, $\mathrm{a}_{\mathrm{ij}}$ is the element lying in the $\mathrm{i}^{\text {th }}$ row an $\mathrm{j}^{\text {th }}$ column We have $a_{23}=x^2-y, a_{31}=0, a_{12}=1$.

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:

i) Let $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]_{2 \times 2}$

Given that $\mathrm{a}_{\mathrm{ij}}=\frac{(\mathrm{i}-2 \mathrm{j})^2}{2}$

$\begin{aligned} & \mathrm{a}_{11}=\frac{(1-2 \times 1)^2}{2}=\frac{1}{2} \\ & \mathrm{a}_{12}=\frac{(1-2 \times 2)^2}{2}=\frac{9}{2} \\ & \mathrm{a}_{21}=\frac{(2-2 \times 1)^2}{2}=0 \\ & \mathrm{a}_{22}=\frac{(2-2 \times 2)^2}{2}=2\end{aligned}$

Hence, the matrix $\mathrm{A}=\left[\begin{array}{cc}\frac{1}{2} & \frac{9}{2} \\ 0 & 2\end{array}\right]$

ii) Let $A=\left[\begin{array}{ll}a_{13} & a_{12} \\ a_{21} & a_{22}\end{array}\right]_{2 \times 2}$

Given that ${ }^` \mathrm{a}_{\mathrm{ij}}=|-2 \mathrm{i}+3 \mathrm{j}|$

$\begin{aligned} & a_{11}=|-2 \times 1+3 \times 1|=1 \\ & a_{12}=|-2 \times 1+3 \times 2|=4 \\ & a_{21}=|-2 \times 2+3 \times 1|=-1 \\ & a_{22}=|-2 \times 2+3 \times 2|=2\end{aligned}$

Hence, the matrix $A=\left[\begin{array}{cc}1 & 4 \\ -1 & 2\end{array}\right]$

Question 4

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

Answer:

$\operatorname{Let} A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right]_{3 \times 2}$ Given that $\mathrm{a}_{\mathrm{ij}}=\mathrm{e}^{\mathrm{i} . \mathrm{x}} \sin \mathrm{j} \mathrm{x}$

$\begin{aligned} & a_{11}=e^x \sin x \\ & a_{12}=e^x \sin 2 x \\ & a_{21}=e^{2 x} \sin x \\ & a_{22}=e^{2 x} \sin 2 x \\ & a_{31}=e^{3 x} \sin x \\ & a_{32}=e^{3 x} \sin 2 x\end{aligned}$

Hence, the matrix $\mathrm{A}=\left[\begin{array}{cc}\mathrm{e}^x \sin x & \mathrm{e}^x \sin 2 x \\ \mathrm{e}^{2 x} \sin x & \mathrm{e}^{2 x} \sin 2 x \\ \mathrm{e}^{3 x} s \sin x & \mathrm{e}^{3 x} \sin 2 x\end{array}\right]$

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:

Given that $\mathrm{A}=\mathrm{B}$

$\Rightarrow\left[\begin{array}{cc}\mathrm{a}+4 & 3 \mathrm{~b} \\ 8 & -6\end{array}\right]=\left[\begin{array}{cc}2 \mathrm{a}+2 & \mathrm{~b}^2+2 \\ 8 & \mathrm{~b}^2-5 \mathrm{~b}\end{array}\right]$

Equating the corresponding elements, we get

$\begin{aligned} & a+4=2 a+2 \\ & 3 b=b^2+2 \\ & b^2-5 b=-6 \\ & \Rightarrow 2 a-a=2 \\ & b^2-3 b+2=0 \\ & b^2-5 b+6=0 \\ & \therefore a=2 \\ & \therefore b^2-3 b+2=0\end{aligned}$

$\begin{aligned} & \Rightarrow b^2-2 b-b+2=0 \\ & \Rightarrow b(b-2)-1(b-2)=0 \\ & \Rightarrow(b-1)(b-2)=0 \\ & \therefore b=1,2 \\ & \therefore b^2-5 b+6=0 \\ & b^2-3 b-2 b+6=0 \\ & \Rightarrow b(b-3)-2(b-3)=0 \\ & \Rightarrow(b-2)(b-3)=0 \\ & \Rightarrow b=2,3\end{aligned}$

But here 2 is common.

Hence, the value of $a=2$ and $b=2$.

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:

We have, $\mathrm{A}=\left[\begin{array}{cc}\sqrt{3} & 1 \\ 2 & 3\end{array}\right]_{2 \times 2}$, and $\mathrm{B}=\left[\begin{array}{lll}x & y & z \\ a & \mathrm{~b} & 6\end{array}\right]_{2 \times 3}$

Here, A and B are of different orders.

Two matrices A and B are confirmable for addition only if the order of both matrices A and B is the same.

Hence, the sum of matrices A and B is not possible.

Question 7

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

Answer:

Given that $\mathrm{X}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]$ and $\mathrm{Y}=\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]$

i) $\begin{aligned} & \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]+\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right] \\ & =\left[\begin{array}{ccc}3+2 & 1+1 & -1-1 \\ 5+7 & -2+2 & -3+4\end{array}\right] \\ & =\left[\begin{array}{ccc}5 & 2 & -2 \\ 12 & 0 & 1\end{array}\right]\end{aligned}$

ii) $\begin{aligned} & 2 X-3 Y=2\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]-3\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right] \\ & =\left[\begin{array}{ccc}2 \times 3 & 2 \times 1 & -2 \times 1 \\ 2 \times 5 & -2 \times 2 & -2 \times 3\end{array}\right]-\left[\begin{array}{ccc}3 \times 2 & 1 \times 3 & -1 \times 3 \\ 3 \times 7 & 3 \times 2 & 3 \times 4\end{array}\right] \\ & =\left[\begin{array}{ccc}6 & 2 & -2 \\ 10 & -4 & -6\end{array}\right]-\left[\begin{array}{ccc}6 & 3 & -3 \\ 21 & 6 & 12\end{array}\right] \\ & =\left[\begin{array}{ccc}6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & -1 & 1 \\ -11 & -10 & -18\end{array}\right]\end{aligned}$

iii) $\begin{aligned} & \mathrm{X}+\mathrm{Y}+\mathrm{Z}=0 \\ & \Rightarrow\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]+\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]+\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{d} & \mathrm{e} & \mathrm{f}\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\end{aligned}$

Where $Z=\left[\begin{array}{lll}a & b & c \\ d & e & f\end{array}\right]$

$\begin{aligned} & \Rightarrow\left[\begin{array}{ccc}3+2+\mathrm{a} & 1+1+\mathrm{b} & -1-1+\mathrm{c} \\ 5+7+\mathrm{d} & -2+2+\mathrm{e} & -3+4+\mathrm{f}\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc}5+\mathrm{a} & 2+\mathrm{b} & -2+\mathrm{c} \\ 12+\mathrm{d} & \mathrm{e} & 1+\mathrm{f}\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]\end{aligned}$

Equating the corresponding elements, we get

$\begin{aligned} & 5+\mathrm{a}=0 \\ & \Rightarrow \mathrm{a}=-5,2+\mathrm{b}=0 \\ & \Rightarrow \mathrm{~b}=-2-2+\mathrm{c}=0 \\ & \Rightarrow \mathrm{c}=2 \\ & 12+\mathrm{d}=0 \\ & \Rightarrow \mathrm{~d}=-12, \mathrm{e}=0,1+\mathrm{f}=0 \\ & \Rightarrow \mathrm{f}=-1\end{aligned}$

Hence, the matrix $Z=\left[\begin{array}{ccc}-5 & -2 & 2 \\ -12 & 0 & -1\end{array}\right]$

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:

The given equation can be written as

$\begin{aligned} & {\left[\begin{array}{cc}2 x^2 & 2 x \\ 3 x & x^2\end{array}\right]+\left[\begin{array}{cc}16 & 10 x \\ 8 & 8 x\end{array}\right]=\left[\begin{array}{cc}\left(2 x^2+16\right) & 18 \\ 20 & 12 x\end{array}\right]} \\ & \Rightarrow\left[\begin{array}{cc}2 x^2+16 & 12 x \\ 3 x+8 & x^2+8 x\end{array}\right]=\left[\begin{array}{cc}2 x^2+16 & 48 \\ 20 & 12 x\end{array}\right]\end{aligned}$

Equating the corresponding elements we get

$\begin{aligned} & 12 \mathrm{x}=48 \\ & 3 \mathrm{x}+8=20 \\ & \mathrm{x}^2+8 \mathrm{x}=12 \mathrm{x} \\ & \therefore \mathrm{x}=\frac{48}{12}=4 \\ & 3 \mathrm{x}=20-8=12 \\ & \Rightarrow \mathrm{x}^2=12 \mathrm{x}-8 \mathrm{x}=4 \mathrm{x} \\ & \Rightarrow \mathrm{x}^2-4 \mathrm{x}=0 \\ & \mathrm{x}=0, \mathrm{x}=4 \\ & \therefore \mathrm{x}=4\end{aligned}$

Hence, the non-zero 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:

Given that $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$

$\begin{aligned} & \mathrm{A}+\mathrm{B}=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]+\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ll}0+0 & 1-1 \\ 1+1 & 1+0\end{array}\right] \\ & \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right] \\ & \mathrm{A}-\mathrm{B}=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & \Rightarrow \mathrm{A}-\mathrm{B}=\left[\begin{array}{ll}0-0 & 1+1 \\ 1-1 & 1-0\end{array}\right] \\ & \Rightarrow \mathrm{A}-\mathrm{B}=\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right]\end{aligned}$

$\begin{aligned} & \therefore(A+B) \cdot(A-B)=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right],\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{ll}0+0 & 0+0 \\ 0+0 & 4+1\end{array}\right] \\ & =\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right]\end{aligned}$

Now, R.H.S. $=\mathrm{A}^2-\mathrm{B}^2$

$\begin{aligned} & =\mathrm{A} \cdot \mathrm{A}-\mathrm{B} \cdot \mathrm{B} \\ & =\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & =\left[\begin{array}{ll}0+1 & 0+1 \\ 0+1 & 1+1\end{array}\right]-\left[\begin{array}{cc}0-1 & 0+0 \\ 0+0 & -1+0\end{array}\right]\end{aligned}$

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

Hence, $\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right] \neq\left[\begin{array}{cc}2 & 10 \\ 1 & 3\end{array}\right]$

Hence, $(A+B) \cdot(A-B) \neq A^2-B^2$

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:

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

$\begin{aligned} & \Rightarrow\left[\begin{array}{lll}1+2 x+15 & 3+5 x+3 & 2+x+2\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=0 \\ & \Rightarrow\left[\begin{array}{lll}2 x+16 & 5 x+6 & x+4\end{array}\right]\left[\begin{array}{l}1 \\ 2 \\ x\end{array}\right]=0 \\ & \Rightarrow\left[\begin{array}{ll}2 \mathrm{x}+16+10 \mathrm{x}+12+\mathrm{x}^2+4 \mathrm{x}\end{array}\right]=0 \\ & \Rightarrow \mathrm{x}^2+16 \mathrm{x}+28=0 \\ & \Rightarrow \mathrm{x}^2+14 \mathrm{x}+2 \mathrm{x}+28=0 \\ & \Rightarrow \mathrm{x}(\mathrm{x}+14)+2(\mathrm{x}+14)=0 \\ & \Rightarrow(\mathrm{x}+2)(\mathrm{x}+14)=0 \\ & \mathrm{x}+2=0 \quad \text { or } \quad \mathrm{x}+14=0 \\ & \therefore \mathrm{x}=-2 \quad \text { or } \quad \mathrm{x}=-14\end{aligned}$

Hence, the values of $x$ are -2 and -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:

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

$\begin{aligned} & \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A} \\ & =\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}25-3 & 15-6 \\ -5+2 & -3+4\end{array}\right] \\ & =\left[\begin{array}{cc}22 & 9 \\ -3 & 1\end{array}\right] \\ & \mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}=\mathrm{O}\end{aligned}$

L.H.S. $\left[\begin{array}{cc}2 & 9 \\ -3 & 1\end{array}\right]-3\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]-7\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$

$\begin{aligned} & \Rightarrow\left[\begin{array}{cc}22 & 9 \\ -3 & 1\end{array}\right]-\left[\begin{array}{cc}15 & 9 \\ -3 & -6\end{array}\right]-\left[\begin{array}{cc}7 & 0 \\ 0 & 7\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}22-15-7 & 9-9-0 \\ -3+3-0 & 1+6-7\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \text { R.H.S. }\end{aligned}$

We are given $\mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}=0$

$\begin{aligned} & \Rightarrow \mathrm{A}^{-1}\left[\mathrm{~A}^2-3 \mathrm{~A}-7 \mathrm{I}\right]=\mathrm{A}^{-1} \mathrm{O} \ldots .\left[\text { Pre-multiplying both sides by } \mathrm{A}^{-1}\right] \\ & \Rightarrow \mathrm{A}^{-1} \mathrm{~A} \cdot \mathrm{~A}-3 \mathrm{~A}^{-1} \cdot \mathrm{~A}-7 \mathrm{~A}^{-1} \mathrm{I}=\mathrm{O} \ldots . .\left[\mathrm{A}^{-1} \mathrm{O}=\mathrm{O}\right] \\ & \Rightarrow \mathrm{I} \cdot \mathrm{A}-3 \mathrm{I}-7 \mathrm{~A}-1 \mathrm{I}=\mathrm{O} \\ & \Rightarrow \mathrm{A}-3 \mathrm{I}-7 \mathrm{~A}^{-1}=\mathrm{O} \\ & \Rightarrow-7 \mathrm{~A}^{-1}=3 \mathrm{I}-\mathrm{A} \\ & \Rightarrow \mathrm{A}^{-1}=\frac{1}{-7}[3 \mathrm{I}-\mathrm{A}]\end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{A}^{-1}=\frac{1}{-7}\left[3\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)-\left(\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right)\right] \\ & =\frac{1}{-7}\left[3\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)-\left(\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right)\right] \\ & =1(-7)\left[\begin{array}{cc}3-5 & 0-3 \\ 0+1 & 3+2\end{array}\right] \\ & =\frac{1}{-7}\left[\begin{array}{cc}-2 & -3 \\ 1 & 5\end{array}\right]\end{aligned}$

Hence, $\mathrm{A}^{-1}=-\frac{1}{7}\left[\begin{array}{cc}-2 & -3 \\ 1 & 5\end{array}\right]$

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:

We have $\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] \mathrm{A}\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ or $\mathrm{PAQ}=\mathrm{I}$,

Where $\mathrm{P}=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]$ and $\mathrm{Q}=\left[\begin{array}{cc}-3 & 2 \\ 5 & -3\end{array}\right]$

$\begin{aligned} & \therefore \mathrm{P}^{-1} \mathrm{PAQ}=\mathrm{P}^{-1} \mathrm{I} \\ & \Rightarrow \mathrm{IQA}=\mathrm{P}^{-1} \\ & \Rightarrow \mathrm{AQ}=\mathrm{P}^{-1} \\ & \Rightarrow \mathrm{AQQ}^{-1}=\mathrm{P}^{-1} \mathrm{Q}^{-1} \\ & \Rightarrow \mathrm{AI}=\mathrm{P}^{-1} \mathrm{Q}^{-1} \\ & \Rightarrow \mathrm{~A}=\mathrm{P}^{-1} \mathrm{Q}^{-1}\end{aligned}$

Now adj. $\mathrm{P}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$ and $|\mathrm{P}|=1$

$\therefore \mathrm{P}^{-1}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$

$\begin{aligned} & \therefore \mathrm{Q}^{-1}=\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right] \\ & \Rightarrow \mathrm{A}=\mathrm{P}^{-1} \mathrm{Q}^{-1} \\ & =\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}6-5 & 4-3 \\ -9+10 & -6+6\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]\end{aligned}$

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

Let $\mathrm{A}=\left[\begin{array}{lll}x & y & z\end{array}\right]$

$\begin{aligned} & \therefore\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}x & y & z\end{array}\right]=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc}4 x & 4 y & 4 z \\ x & y & z \\ 3 x & 3 y & 3 z\end{array}\right]\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]\end{aligned}$

Comparing elements of both sides

$\begin{aligned} & 4 x=-4 \\ & \Rightarrow x=-1 \\ & 4 y=8 \\ & y=2\end{aligned}$

And 4z = 4

⇒ z = 1

$\therefore \mathrm{A}=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]$

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:

Here, $B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}$ and $A=\left[\begin{array}{cc}3 & -4 \\ 1 & 1 \\ 2 & 0\end{array}\right]_{3 \times 2}$

$\begin{aligned} & \therefore \mathrm{BA}=\left[\begin{array}{ll}6+1+4 & -8+1+0 \\ 3+2+8 & -4+2+0\end{array}\right]_{2 \times 2} \\ & \Rightarrow \mathrm{BA}=\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right]\end{aligned}$

L.H.S: $(\mathrm{BA})^2=(\mathrm{BA}) \cdot(\mathrm{BA})$

$\begin{aligned} & =\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right]\left[\begin{array}{ll}11 & -7 \\ 13 & -2\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}121-91 & -77+14 \\ 143-26 & -91+4\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}30 & -63 \\ 117 & -87\end{array}\right]\end{aligned}$

$\begin{aligned} & \text { R.H.S: B }{ }^2=\mathrm{B} \cdot \mathrm{B} \\ & =\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3} \cdot\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}\end{aligned}$

Here, the number of columns of the first

i.e., 3 is not equal to the number of rows of the second matrix i.e., 2.

So, $\mathrm{B}^2$ is not possible.

Similarly, $\mathrm{A}^2$ is also not possible.

Hence, $(\mathrm{BA})^2 \cdot \mathrm{~B}^2 \mathrm{~A}^2$

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:

$\begin{aligned} & \mathrm{BA}=\left[\begin{array}{ll}4 & 1 \\ 2 & 3 \\ 1 & 2\end{array}\right]_{3 \times 2}\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3} \\ & \mathrm{BA}=\left[\begin{array}{lll}8+1 & 4+2 & 8+4 \\ 4+3 & 2+6 & 4+12 \\ 2+2 & 1+4 & 2+8\end{array}\right]_{3 \times 3} \\ & =\left[\begin{array}{lll}9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10\end{array}\right]_{3 \times 3}\end{aligned}$

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

$\begin{aligned} & =\left[\begin{array}{cc}8+2+2 & 2+3+4 \\ 4+4+4 & 1+6+8\end{array}\right]_{2 \times 2} \\ & =\left[\begin{array}{cc}12 & 9 \\ 12 & 15\end{array}\right]_{2 \times 2}\end{aligned}$

Hence, $\mathrm{BA}=\left[\begin{array}{lll}9 & 6 & 12 \\ 7 & 8 & 16 \\ 4 & 5 & 10\end{array}\right]$ and $\mathrm{AB}=\left[\begin{array}{cc}12 & 9 \\ 12 & 15\end{array}\right]$.

Question 16

An example is shown for A ≠ O, B ≠ O, AB = O.

Answer:

Let $A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

$\begin{aligned} & \mathrm{AB}=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] \\ & \Rightarrow \mathrm{AB}=\left[\begin{array}{cc}1-1 & 1-1 \\ -1+1 & -1+1\end{array}\right] \\ & =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=\mathrm{O}\end{aligned}$

Hence, $A=\left[\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

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:

$\begin{aligned} & \text { Here, } \mathrm{A}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right] \\ & \mathrm{AB}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}2+8+0 & 8+32+0 \\ 3+18+6 & 12+72+18\end{array}\right] \\ & =\left[\begin{array}{cc}10 & 40 \\ 27 & 102\end{array}\right]\end{aligned}$

$\begin{aligned} & \text { L.H.S. }(A B)^{\prime}=\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right] \\ & \text { Now } B=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]\end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{B}^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right] \\ & \mathrm{A}=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right] \\ & \Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc}2 & 3 \\ 4 & 90 \\ 0 & 6\end{array}\right]\end{aligned}$

R.H.S. $\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 4 & 9 \\ 0 & 6\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}2+8+0 & 3+18+6 \\ 8+32+0 & 1272+18\end{array}\right] \\ & =\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right] \\ & =\text { L.H.S. }\end{aligned}$

Hence, L.H.S. = R.H.S.

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:

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

L.H.S. $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[\begin{array}{c}-8 \\ -11\end{array}\right]=\mathrm{O}$

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

Comparing the corresponding elements of both sides, we get,

$\begin{aligned} & 2 x+3 y-8=0 \\ & \Rightarrow 2 x+3 y=8 \\ & x+5 y-11=0 \\ & \Rightarrow x+5 y=11\end{aligned}$

Multiplying equation (1) by 1 and equation (2) by 2, and then subtracting, we get,

$\begin{aligned} 2 x+3 y & =8 \\ 2 x+10 y & =22 \\ (-)-(-) & -(-) \\ -7 y & =-14\end{aligned}$

$\therefore \mathrm{y}=2$

$\begin{aligned} & x+5 \times 2=11 \\ & \Rightarrow x+10=11 \\ & x=11-10=1\end{aligned}$

Hence, the values of x and y are 1 and 2, respectively.

Question 19

If X and Y are 2 × 2 matrices, 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, $\mathrm{A}=\left[\begin{array}{ll}3 & 5\end{array}\right]_{1 \times 2}$ and $\mathrm{B}=\left[\begin{array}{ll}7 & 3\end{array}\right]_{1 \times 2}$

For $\mathrm{AC}=\mathrm{BC}$

We have order of $\mathrm{C}=2 \times \mathrm{n}$

For $\mathrm{n}=1$

Let $\mathrm{C}=\left[\begin{array}{l}x \\ y\end{array}\right]$

$\therefore \mathrm{AC}=\left[\begin{array}{ll}3 & 5\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[(3 x+5 y]$

And $\mathrm{BC}=\left[\begin{array}{ll}7 & 3\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=[3 \mathrm{x}+5 \mathrm{y}]$

For $\mathrm{AC}=\mathrm{BC}$,

$\begin{aligned} & {[3 \mathrm{x}+5 \mathrm{y}]=[7 \mathrm{x}+3 \mathrm{y}]} \\ & \Rightarrow 3 \mathrm{x}+5 \mathrm{y}=7 \mathrm{x}+3 \mathrm{y} \\ & \Rightarrow 4 \mathrm{x}=2 \mathrm{y} \\ & \Rightarrow \mathrm{x}=\frac{1}{2} y \\ & \Rightarrow \mathrm{y}=2 \mathrm{x} \\ & \therefore \mathrm{C}=\left[\begin{array}{c}x \\ 2 x\end{array}\right]\end{aligned}$

We see that on taking C of order $2 \times 1,2 \times 2,2 \times 3, \ldots$, we get

$\mathrm{C}=\left[\begin{array}{c}x \\ 2 x\end{array}\right],\left[\begin{array}{cc}x & x \\ 2 x & 2 x\end{array}\right],\left[\begin{array}{ccc}x & x & x \\ 2 x & 2 x & 2 x\end{array}\right] \ldots$

In general,

$\mathrm{C}=\left[\begin{array}{c}\mathrm{k} \\ 2 \mathrm{k}\end{array}\right],\left[\begin{array}{cc}\mathrm{k} & \mathrm{k} \\ 2 \mathrm{k} & 2 \mathrm{k}\end{array}\right]$ etc $\ldots$

Where k is any real number.

Question 21

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

Answer:

Let $\mathrm{A}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right] \quad \ldots \ldots .[\because \mathrm{B} \neq \mathrm{C}]$

$\therefore \mathrm{AB}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right] \ldots . .(\mathrm{i})$

And $A C=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right]$

We have $\mathrm{AB}=\mathrm{AC}$ but $\mathrm{B} \neq \mathrm{C}$

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:

i) We have, $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]$, $\mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

$\begin{aligned} & \mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] \\ & =\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\end{aligned}$

And $(\mathrm{AB}) \mathrm{C}=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}8+5 & 0 \\ -1+10 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right] \ldots . .(\mathrm{i})\end{aligned}$

Again, $(\mathrm{BC})=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ll}2-3 & 0 \\ 3+4 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]\end{aligned}$

And $A(B C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}-1+14 & 0 \\ 2+7 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right]\end{aligned}$

From (i) and (ii), we get

$\therefore(\mathrm{AB}) \mathrm{C}=\mathrm{A}(\mathrm{BC})$

ii) We have, $\mathrm{A}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

$\begin{aligned} & \mathrm{B}+\mathrm{C}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]\end{aligned}$

$\Rightarrow \mathrm{A} \cdot(\mathrm{B}+\mathrm{C})=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$

$=\left[\begin{array}{cc}3+4 & 3-8 \\ -6+2 & -6-4\end{array}\right]$

$=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right] \ldots .$. (iii)

$\mathrm{AB}=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right] \\ & =\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\end{aligned}$

And $A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

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

$\therefore \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]$

$=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]$

$\mathrm{A} \cdot(\mathrm{B}+\mathrm{C})=\mathrm{A} \cdot \mathrm{B}+\mathrm{A} \cdot \mathrm{C}$

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:

Given that,

$\begin{aligned} & \mathrm{P}=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right] \text { and } \mathrm{Q}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right] \\ & \mathrm{PQ}=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\end{aligned}$

$\begin{aligned} \mathrm{PQ} & =\left[\begin{array}{ccc}x \mathrm{a}+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y \mathrm{~b}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z \mathrm{c}\end{array}\right] \\ \mathrm{PQ} & =\left[\begin{array}{ccc}x \mathrm{a} & 0 & 0 \\ 0 & y \mathrm{~b} & 0 \\ 0 & 0 & z \mathrm{c}\end{array}\right]\end{aligned}$

Now $\mathrm{QP}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$

$\begin{aligned} & \mathrm{QP}=\left[\begin{array}{ccc}x \mathrm{a}+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+y \mathrm{~b}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+z \mathrm{c}\end{array}\right] \\ & \mathrm{QP}=\left[\begin{array}{ccc}x \mathrm{a} & 0 & 0 \\ 0 & y \mathrm{~b} & 0 \\ 0 & 0 & z \mathrm{c}\end{array}\right]\end{aligned}$

Hence, $\mathrm{PQ}=\mathrm{QP}$.

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

$\therefore\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{c}-1+0+1 \\ -1+0+0 \\ 0+0-1\end{array}\right]_{3 \times 1}=\mathrm{A}$

$\begin{aligned} & \Rightarrow\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]_{1 \times 3}\left[\begin{array}{c}0 \\ -1 \\ -1\end{array}\right]_{3 \times 1}{ }^`=\mathrm{A} \\ & \Rightarrow[0-1-3]=\mathrm{A} \\ & \Rightarrow \mathrm{A}=[-4]\end{aligned}$

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 have $\mathrm{A}=\left[\begin{array}{ll}2 & 1\end{array}\right], \mathrm{B}=\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$

$\begin{aligned} & \therefore \mathrm{A}(\mathrm{B}+\mathrm{C})=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2\end{array}\right] \\ & =\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}4 & 5 & 5 \\ 9 & 7 & 8\end{array}\right] \\ & =\left[\begin{array}{lll}8+9 & 10+7 & 10+8\end{array}\right] \\ & =\left[\begin{array}{lll}17 & 17 & 18\end{array}\right] \ldots \ldots .(\mathrm{i})\end{aligned}$

Now $A B=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{lll}10+8 & 6+7 & 8+6\end{array}\right] \\ & =\left[\begin{array}{lll}18 & 3 & 14\end{array}\right]\end{aligned}$

And $A C=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 0 & 2\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{lll}-2+1 & 4+0 & 2+2\end{array}\right] \\ & {\left[\begin{array}{lll}-1 & 4 & 4\end{array}\right]} \\ & \therefore \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll}18 & 13 & 14\end{array}\right]+\left[\begin{array}{lll}-1 & 4 & 4\end{array}\right] \\ & =\left[\begin{array}{lll}17 & 17 & 18\end{array}\right] \ldots . .(\mathrm{ii)}\end{aligned}$

From equations (i) and (ii)
$\mathrm{A}(\mathrm{B}+\mathrm{C})=(\mathrm{AB}+\mathrm{AC})$

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

$\begin{aligned} & \therefore \mathrm{A}^2=\mathrm{A} \cdot \mathrm{A} \\ & =\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right] \\ & =\left[\begin{array}{ccc}1+0+0 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]\end{aligned}$

$\therefore \mathrm{A}^2+\mathrm{A}=\left[\begin{array}{ccc}1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -4 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]$

$=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right] \ldots \ldots .(\mathrm{i})$

Now, $A+I=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

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

So, $A(A+I)=\left[\begin{array}{ccc}1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]$

$=\left[\begin{array}{ccc}2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{array}\right]$

$=\left[\begin{array}{ccc}2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5\end{array}\right]$

From (i) and (ii)

We get $\mathrm{A}^2+\mathrm{A}=\mathrm{A}(\mathrm{A}+\mathrm{I})$

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:

i) Given that: $\mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$

$\begin{aligned} & \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]_{2 \times 3}^{\prime} \\ & =\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]_{3 \times 2} \\ & \left(\mathrm{~A}^{\prime}\right)^{\prime}=\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]_{3 \times 2}^{\prime} \\ & =\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]_{2 \times 3} \\ & =\mathrm{A}\end{aligned}$

ii) Given that: $\mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$

L.H.S. $A B=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]_{2 \times 3}\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]_{3 \times 2}$

$\begin{aligned} & =\left[\begin{array}{cc}0-1+4 & 0-3+12 \\ 16+3-8 & 0+9-24\end{array}\right]_{2 \times 2} \\ & =\left[\begin{array}{cc}3 & 9 \\ 11 & -15\end{array}\right]_{2 \times 2}\end{aligned}$

$\begin{aligned} & (\mathrm{AB})^{\prime}=\left[\begin{array}{cc}3 & 11 \\ 9 & -15\end{array}\right]_{2 \times 2} \\ & \text { R.H.S. } \mathrm{B}^{\prime}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]^{\prime}\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{lll}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right] \\ & \mathrm{A}^{\prime}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]^{\prime} \\ & =\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]\end{aligned}$

$\mathrm{B}^{\prime} \mathrm{A}^{\prime}=\left[\begin{array}{lll}4 & 1 & 2 \\ 0 & 3 & 6\end{array}\right]_{2 \times 3}\left[\begin{array}{cc}0 & 4 \\ -1 & 3 \\ 2 & -4\end{array}\right]_{3 \times 2}$

$\begin{aligned} & =\left[\begin{array}{cc}0-1+4 & 16+3-8 \\ 0-3+12 & 0+9-24\end{array}\right]_{2 \times 2} \\ & =\left[\begin{array}{cc}3 & 11 \\ 9 & -15\end{array}\right]_{2 \times 2}\end{aligned}$

L.H.S. = R.H.S.

Hence, $(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}$ is verified.

iii) $\begin{aligned} & \text { Given that: } \mathrm{A}=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right] \\ & \text { L.H.S. kA }=\mathrm{k}\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{ccc}0 & -\mathrm{k} & 2 \mathrm{k} \\ 4 \mathrm{k} & 3 \mathrm{k} & -4 \mathrm{k}\end{array}\right] \\ & (\mathrm{kA})^{\prime}=\left[\begin{array}{cc}0 & 4 \mathrm{k} \\ -\mathrm{k} & 3 \mathrm{k} \\ 2 \mathrm{k} & -\mathrm{k}\end{array}\right]\end{aligned}$

R.H.S. $\mathrm{kA}^{\prime}=\mathrm{k}\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]^{\prime}$

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

Hence, L.H.S. = R.H.S.

$(\mathrm{kA})^{\prime}=\left(\mathrm{kA}^{\prime}\right)$ is verified.

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:

i) Given that: $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$

L.H.S. $(2 \mathrm{~A}+\mathrm{B})^{\prime}=\left[2\left(\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right)+\left(\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right)\right]^{\prime}$

$\begin{aligned} & =\left[\left(\begin{array}{cc}2 & 4 \\ 8 & 2 \\ 10 & 12\end{array}\right)+\left(\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right)\right]^{\prime} \\ & =\left[\begin{array}{cc}2+1 & 4+2 \\ 8+6 & 2+4 \\ 10+7 & 12+3\end{array}\right]^{\prime} \\ & =\left[\begin{array}{cc}3 & 6 \\ 14 & 6 \\ 17 & 15\end{array}\right]^{\prime} \\ & =\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]\end{aligned}$

R.H.S. $2 A^{\prime}+\mathrm{B}^{\prime}=2\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]^{\prime}+\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]^{\prime}$

$\begin{aligned} & =2\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right] \\ & =\left[\begin{array}{lll}2 & 8 & 10 \\ 4 & 2 & 12\end{array}\right]+\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}2+1 & 8+6 & 10+7 \\ 4+2 & 2+4 & 12+3\end{array}\right] \\ & =\left[\begin{array}{ccc}3 & 14 & 17 \\ 6 & 6 & 15\end{array}\right]\end{aligned}$

Hence, L.H.S. = R.H.S.

$(2 \mathrm{~A}+\mathrm{B})^{\prime}=2 \mathrm{~A}^{\prime}+\mathrm{B}^{\prime}$ is verified.

ii) Given that: $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]$

L.H.S. $(\mathrm{A}-\mathrm{B})^{\prime}=\left[\left(\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right)-\left(\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right)\right]^{\prime}$

$\begin{aligned} & =\left[\begin{array}{cc}1-1 & 2-2 \\ 4-6 & 1-4 \\ 5-7 & 6-3\end{array}\right]^{\prime} \\ & =\left[\begin{array}{cc}0 & 0 \\ -2 & -3 \\ -2 & 3\end{array}\right]^{\prime} \\ & =\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]\end{aligned}$

R.H.S. $\mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ll}1 & 2 \\ 4 & 1 \\ 5 & 6\end{array}\right]^{\prime}-\left[\begin{array}{ll}1 & 2 \\ 6 & 4 \\ 7 & 3\end{array}\right]^{\prime}$

$\begin{aligned} & =\left[\begin{array}{lll}1 & 4 & 5 \\ 2 & 1 & 6\end{array}\right]-\left[\begin{array}{lll}1 & 6 & 7 \\ 2 & 4 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}1-1 & 4-6 & 5-7 \\ 2-2 & 1-4 & 6-3\end{array}\right] \\ & =\left[\begin{array}{ccc}0 & -2 & -2 \\ 0 & -3 & 3\end{array}\right]\end{aligned}$

Hence, L.H.S. = R.H.S. $(\mathrm{A}-\mathrm{B})^{\prime}=\mathrm{A}^{\prime}-\mathrm{B}^{\prime}$ is verified.
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.
We know that the transposition of AB is given by
(AB)’ = B’A’
Using this result, and by taking the 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 that A’A is the symmetric matrix for any matrix A.
Now, take the transpose of AA’.
Transpose of AA’ = (AA’)’
⇒ (AA’)’ = (A’)’A’ [ (AB)’ = B’A’]
⇒ (AA’)’ = AA’ [(A’)’ = A]
Since (AA’)’ = AA’
This means, AA’ is the symmetric matrix for any matrix A.
Thus, A’A and AA’ are symmetric matrices 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:

As A and B are square matrices of order $3 \times 3$.

We have, $(\mathrm{AB})^2=\mathrm{AB} \cdot \mathrm{AB}$

$\begin{aligned} & =\mathrm{A}(\mathrm{BA}) \mathrm{B} \\ & =\mathrm{A}(\mathrm{AB}) \mathrm{B} \ldots \ldots .[\mathrm{If} \mathrm{AB}=\mathrm{BA}] \\ & =\mathrm{AABB} \\ & =\mathrm{A}^2 \mathrm{~B}^2\end{aligned}$

Thus, $(\mathrm{AB})^2=\mathrm{A}^2 \mathrm{~B}^2$ is true only if $\mathrm{AB}=\mathrm{BA}$.

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:

Given that $A$ and $B$ are square matrices such that $A B=B A$.

So, $(\mathrm{A}+\mathrm{B})^2=(\mathrm{A}+\mathrm{B}) \cdot(\mathrm{A}+\mathrm{B})$

$\begin{aligned} & =A^2+A B+B A+B^2 \\ & =A^2+A B+A B+B^2 \ldots \ldots .[\text { Since, } A B=B A] \\ & =A^2+2 A B+B^2\end{aligned}$

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

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

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

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

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

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

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

Answer:

We have,

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

And $a=4, b=-2$

And $a(C-A)=4(C-A)$

$=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$

Also, $\mathrm{aC}-\mathrm{aA}=4 \mathrm{C}-4 \mathrm{~A}$

$\begin{aligned} & =\left[\begin{array}{cc}8 & 0 \\ 4 & -8\end{array}\right]-\left[\begin{array}{cc}4 & 8 \\ -4 & 12\end{array}\right] \\ & =\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right] \\ & =\mathrm{a}(\mathrm{C}-\mathrm{A})\end{aligned}$

Clearly $L H S=\mathrm{RHS}=\left[\begin{array}{cc}4 & -8 \\ 8 & -20\end{array}\right]$

Hence proved.

Question 32.5

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

Answer:

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

Question 32.6

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

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

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

Answer:

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

Question 32.8

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

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

Answer:

We have,

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

And $a=4, b=-2$

$\begin{aligned} & (\mathrm{A}-\mathrm{B})=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]-\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right] \\ & =\left[\begin{array}{cc}1-4 & 2-0 \\ -1-1 & 3-5\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\end{aligned}$

$\begin{aligned} & \therefore(\mathrm{A}-\mathrm{B}) \mathrm{C}=\left[\begin{array}{cc}-3 & 2 \\ -2 & -2\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right]\end{aligned}$

Now, $A C=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$

$=\left[\begin{array}{ll}4 & -4 \\ 1 & -6\end{array}\right]$

And $B C=\left[\begin{array}{ll}4 & 0 \\ 1 & 5\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ 1 & -2\end{array}\right]$

$=\left[\begin{array}{cc}8 & 0 \\ 7 & -10\end{array}\right]$

$\begin{aligned} & \therefore \mathrm{AC}-\mathrm{BC}=\left[\begin{array}{ll}4-8 & -4-0 \\ 1-7 & -6+10\end{array}\right] \\ & =\left[\begin{array}{cc}-4 & -4 \\ -6 & 4\end{array}\right] \\ & =(\mathrm{A}-\mathrm{B}) \mathrm{C}\end{aligned}$

Hence proved.

Question 32.9

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

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

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 of 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:

Let $\mathrm{P}(\mathrm{n}):\left(\mathrm{A}^{\prime}\right)^{\mathrm{n}}=\left(\mathrm{A}^{\mathrm{n}}\right)^{\prime}$

$\therefore \mathrm{P}(1):\left(\mathrm{A}^{\prime}\right)=(\mathrm{A})^{\prime}$

$\Rightarrow \mathrm{A}^{\prime}=\mathrm{A}^{\prime}$

$\Rightarrow \mathrm{P}(1)$ is true.

Now, let $\mathrm{P}(\mathrm{k})=\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}}=\left(\mathrm{A}^{\mathrm{k}}\right)^{\prime}$

Where $k \in N$

And $\mathrm{P}(\mathrm{k}+1):\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}+1}=\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}} \mathrm{A}^{\prime}$

$=\left(\mathrm{A}^{\prime}\right)^{\mathrm{k}^{\prime}} \mathrm{A}^{\prime}$

$=\left(\mathrm{AA}^{\mathrm{k}}\right)^{\prime} \quad . . . . .\left(\mathrm{As}(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right)$

$=\left(\mathrm{A}^{\mathrm{k}+1}\right)^{\prime}$

Thus $\mathrm{P}(1)$ is true and whenever $\mathrm{P}(\mathrm{k})$ is true $\mathrm{P}(\mathrm{k}+1)$ is true, So, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$

Question 37.1

Find the 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 the 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 the 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 all zeroes in one of the rows of the matrix in the LHS.
So by any means, we can't make an identity matrix in LHS.
∴ The 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:

Given that: $\left[\begin{array}{cc}x y & 4 \\ z+6 & x+y\end{array}\right]=\left[\begin{array}{cc}8 & w \\ 0 & 6\end{array}\right]$

Equating the corresponding elements,

$\begin{aligned} & x y=8 \\ & w=4 \\ & z+6=0 \\ & \Rightarrow z=-6, x+y=6\end{aligned}$

Now, solving $x+y=6$ $\qquad$

And $x y=8$ $\qquad$ (ii)

From equation (i), $y=6-x$ $\qquad$

Putting the value of $y$ in equation (ii), we get,

$\begin{aligned} & x(6-x)=8 \\ & \Rightarrow 6 x-x^2=8 \\ & \Rightarrow x^2-6 x+8=0 \\ & \Rightarrow x^2-4 x-2 x+8=0 \\ & \Rightarrow x(x-4)-2(x-4)=0 \\ & \Rightarrow(x-4)(x-2)=0 \\ & \therefore x=4,2\end{aligned}$

From equation (iii)

$y=2,4$

Hence, $x=4$ or $2, y=2$ or $4, z=-6$ 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 that: $\mathrm{A}=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$

$\mathrm{A}^2=\mathrm{A} \cdot \mathrm{A}$

$\begin{aligned} & =\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right] \\ & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]\end{aligned}$

$\therefore \mathrm{A}^2-5 \mathrm{~A}-14 \mathrm{I}=\left[\begin{array}{cc}29 & -25 \\ -20 & -24\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\ & =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right] \\ & =\left[\begin{array}{cc}29-29 & -25+25 \\ -20+20 & 24-24\end{array}\right] \\ & =\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}$

Hence, $\mathrm{A}^2-5 \mathrm{~A}-14 \mathrm{I}=0$

Now, multiplying both sides by A, we get,

$A^2 \cdot A-5 A \cdot A-14 I A=0 A$

$\begin{aligned} & \Rightarrow A^3-5 A^2-14 A=0 \\ & \Rightarrow A^3=5 A^2+14 A \\ & \Rightarrow A^3=5\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]+14\left[\begin{array}{cc}3 & -5 \\ -4 & -2\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}145 & -125 \\ -100 & 120\end{array}\right]+\left[\begin{array}{cc}42 & -70 \\ -56 & 28\end{array}\right] \\ & =\left[\begin{array}{cc}145+42 & -125-70 \\ -100-56 & 120+28\end{array}\right] \\ & =\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]\end{aligned}$

Hence, $A^3=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]$

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:

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

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

Equating the corresponding elements, we get,

$\begin{aligned} & 3 \mathrm{a}=\mathrm{a}+4 \\ & \Rightarrow 3 \mathrm{a}-\mathrm{a}=4 \\ & \Rightarrow 2 \mathrm{a}=4 \\ & \Rightarrow \mathrm{a}=2 \\ & 3 \mathrm{~b}=6+\mathrm{a}+\mathrm{b} \\ & \Rightarrow 3 \mathrm{~b}-\mathrm{b}-\mathrm{a}=6 \\ & \Rightarrow 2 \mathrm{~b}-\mathrm{a}=6 \\ & \Rightarrow 2 \mathrm{~b}-2=6 \\ & \Rightarrow 2 \mathrm{~b}=8 \\ & \Rightarrow \mathrm{~b}=4\end{aligned}$

$\begin{aligned} & 3 \mathrm{c}=-1+\mathrm{c}+\mathrm{d} \\ & \Rightarrow 3 \mathrm{c}-\mathrm{c}-\mathrm{d}=-1 \\ & \Rightarrow 2 \mathrm{c}-\mathrm{d}=-1\end{aligned}$

And $3 \mathrm{~d}=2 \mathrm{~d}+3$

$\begin{aligned} & \Rightarrow 3 d-2 d=3 \\ & \Rightarrow d=3\end{aligned}$

Now $2 \mathrm{c}-\mathrm{d}=-1$

$\begin{aligned} & \Rightarrow 2 \mathrm{c}-3=-1 \\ & \Rightarrow 2 \mathrm{c}=3-1 \\ & \Rightarrow 2 \mathrm{c}=2 \\ & \therefore \mathrm{c}=1 \\ & \therefore \mathrm{a}=2, \mathrm{~b}=4, \mathrm{c}=1 \text { and } \mathrm{d}=3 .\end{aligned}$

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:

Order of matrix $\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]$ is $3 \times 2$ and the matrix

$\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$ is $3 \times 3$

$\therefore$ Order of matrix A must be $2 \times 3$

Let $\mathrm{A}=\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{d} & \mathrm{e} & \mathrm{f}\end{array}\right]_{2 \times 3}$

So, $\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{d} & \mathrm{e} & \mathrm{f}\end{array}\right]=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$

$\left[\begin{array}{ccc}2 \mathrm{a}-\mathrm{d} & 2 \mathrm{~b}-\mathrm{e} & 2 \mathrm{c}-\mathrm{f} \\ \mathrm{a}+0 & \mathrm{~b}+0 & \mathrm{c}+0 \\ -3 \mathrm{a}+4 \mathrm{~d} & -3 \mathrm{~b}+4 \mathrm{e} & -3 \mathrm{c}+4 \mathrm{f}\end{array}\right]=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 5\end{array}\right]$

Equating the corresponding elements, we get,

$\begin{aligned} & 2 a-d=-1 \text { and } a=1 \\ & \Rightarrow 2 \times 1-d=-1 \\ & \Rightarrow d=2+1 \\ & \Rightarrow d=3\end{aligned}$

$\begin{aligned} & 2 \mathrm{~b}-\mathrm{e}=-8 \text { and } \mathrm{b}=-2 \\ & \Rightarrow 2(-2)-\mathrm{e} \\ & \Rightarrow-8 \\ & \Rightarrow-4-\mathrm{e}=-8 \\ & \Rightarrow \mathrm{e}=4 \\ & 2 \mathrm{c}-\mathrm{f}=-10 \text { and } \mathrm{c}=-5 \\ & \Rightarrow 2(-5)-\mathrm{f}=-10 \\ & \Rightarrow-10-\mathrm{f}=-10 \\ & \Rightarrow \mathrm{f}=0 \\ & \mathrm{a}=1, \mathrm{~b}=-2, \mathrm{c}=-5, \mathrm{~d}=3, \mathrm{e}=4 \text { and } \mathrm{f}=0\end{aligned}$

Hence, $A=\left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\end{array}\right]$.

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:

Here, $\mathrm{A}=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

Given that: $\mathrm{A}^{-1}=\mathrm{A}^{\prime}$

Pre-multiplying both sides by A

$\mathrm{AA}^{-1}=\mathrm{AA}^{\prime}$

$\begin{aligned} & \Rightarrow \mathrm{I}=\mathrm{AA}^{\prime} \quad \ldots \ldots .\left[\because \mathrm{AA}^{-1}=\mathrm{I}\right] \\ & \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\cos ^2 \alpha+\sin ^2 \alpha & -\sin \alpha \cos \alpha+\sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha+\cos \alpha \sin \alpha & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$

Hence, it is true for all values of a.

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 a 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 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 have, $\mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos x & \operatorname{six} \\ -\sin x & \cos x\end{array}\right]$

$\therefore \mathrm{P}(\mathrm{y})=\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$

Now,

$\mathrm{P}(\mathrm{x}) \cdot \mathrm{P}(\mathrm{y})=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{cc}\cos x \cdot \cos y-\sin x \cdot \sin y & \cos x \cdot \sin y+\sin x \cdot \cos y \\ -\sin x \cdot \cos y-\cos x \cdot \sin y & -\sin x \cdot \sin y+\cos x \cdot \cos y\end{array}\right] \\ & =\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]\end{aligned}$

$=P(x+y) \ldots \ldots(i)$

Also,

$\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})=\left[\begin{array}{cc}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$

$=\left[\begin{array}{cc}\cos y \cdot \cos x-\sin y \cdot \sin x & \cos y \cdot \sin x+\sin y \cdot \cos x \\ -\sin y \cdot \cos x-\sin x \cdot \cos y & -\sin y \cdot \sin x+\cos y \cdot \cos x\end{array}\right]$

$=\left[\begin{array}{cc}\cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y)\end{array}\right]$

$\mathrm{P}(\mathrm{x}) \cdot(\mathrm{y})=\mathrm{P}(\mathrm{x}+\mathrm{y})=\mathrm{P}(\mathrm{y}) \cdot \mathrm{P}(\mathrm{x})$

Question 47

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

Answer:

We know that,

A. I = I. A

So, A and I are commutative.

Thus, we can expand $(\mathrm{I}+\mathrm{A})^3$ like real numbers expansion.

So, $(I+A)^3=I^3+3 I^2 A+3 \mathrm{IA}^2+A^3$

$\begin{aligned} & =\mathrm{I}+3 \mathrm{IA}+3 \mathrm{~A}^2+\mathrm{AA}^2 \ldots . .\left(\mathrm{As} \mathrm{I}^{\mathrm{n}}=\mathrm{I}, \mathrm{n} \in \mathrm{N}\right) \\ & =\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}+\mathrm{AA} \\ & =\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}+\mathrm{A}^2 \\ & =\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}+\mathrm{A} \\ & =\mathrm{I}+7 \mathrm{~A}\end{aligned}$

Hence proved,
$(I + A)^{3} = I + 7A$

Question 48

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

Answer:

Given that B is a skew-symmetric matrix

$\therefore \mathrm{B}^{\prime}=-\mathrm{B}$

Let $\mathrm{P}=\mathrm{A}^{\prime} \mathrm{BA}$

$\begin{aligned} & =\mathrm{A}^{\prime} \mathrm{B}^{\prime}\left(\mathrm{A}^{\prime}\right)^{\prime} \ldots \ldots .\left[(\mathrm{AB})^{\prime}=\mathrm{B}^{\prime} \mathrm{A}^{\prime}\right] \\ & =\mathrm{A}^{\prime}(-\mathrm{B}) \mathrm{A} \\ & =-\mathrm{A}^{\prime} \mathrm{BA} \\ & =-\mathrm{P}\end{aligned}$

So $\mathrm{P}^{\prime}=-\mathrm{P}$

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:

Let $\mathrm{P}(\mathrm{n}):(\mathrm{AB})^{\mathrm{n}}=\mathrm{A}^{\mathrm{n}} \mathrm{B}^{\mathrm{n}}$

So, $\mathrm{P}(1)$ : $(\mathrm{AB})^1=\mathrm{A}^1 \mathrm{~B}^1$

$\Rightarrow \mathrm{AB}=\mathrm{AB}$

So, $\mathrm{P}(1)$ is true.

Let $\mathrm{P}(\mathrm{n})$ is true for some $\mathrm{k} \in \mathrm{N}$

So, $P(k):(A B)^k=A^k B^k, k \in N$

Now $(A B)^{k+1}=(A B)^{\mathrm{k}}(\mathrm{AB}) \quad \ldots .($ Using $(\mathrm{i}))$

$\begin{aligned} & =A^k B^k(A B) \\ & =A^k B^{k-1}(B A) B \\ & =A^k B^{k-1}(A B) B \quad \ldots .(\text { As given } A B=B A) \\ & =A^k B^{k-1} A B^2 \\ & =A^k B^{k-2}(B A) B^2 \\ & =A^k B^{k-2} A B B^2 \\ & =A^k B^{k-2} A B^3\end{aligned}$
$=\mathrm{A}^{\mathrm{k}+1} \mathrm{~B}^{\mathrm{k}+1}$

Thus $\mathrm{P}(1)$ is true and whenever $\mathrm{P}(\mathrm{k})$ is true $\mathrm{P}(\mathrm{k}+1)$ is true.

So, $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$.

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:

Matrix A is such that $\mathrm{A}^{\prime}=\mathrm{A}^{-1}$

$\Rightarrow \mathrm{AA}^{\prime}=\mathrm{I}$

$\Rightarrow\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]\left[\begin{array}{ccc}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}4 y^2+z^2 & 2 y^2-z^2 & -2 y^2+z^2 \\ 2 y^2-z^2 & x^2+y^2+z^2 & x^2-y^2-z^2 \\ -2^2+z^2 & x^2-y^2+z^2 & x^2+y^2+z^2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\begin{aligned} & \Rightarrow 4 y^2+z^2=1 \\ & 2 y^2-z^2=0 \\ & x^2+y^2+z^2=1 \\ & x^2-y^2-z^2=0\end{aligned}$

$\begin{aligned} & \Rightarrow y^2=\frac{1}{6}, z^2=\frac{1}{3}, x^2=\frac{1}{2} \\ & \Rightarrow \mathrm{x}= \pm \frac{1}{\sqrt{2}} \\ & \Rightarrow \mathrm{y}= \pm \frac{1}{\sqrt{6}}\end{aligned}$

And $z= \pm \frac{1}{\sqrt{3}}$

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.t
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 R 2 → R 2 - 3R 1
$\begin{aligned} &\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] \mathrm{A}\\ &\text { Applying } \mathrm{R}_{3} \rightarrow(-1) \mathrm{R}_{3}\\ &\Rightarrow\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \end{aligned}$
$\text { Applying } R_{1} \rightarrow R_{1}+R_{2}$
$\left[\begin{array}{ccc} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] A$
$\\\text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+10 \mathrm{R}_{3} \text { and } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+17 \mathrm{R}_{3}\\ \Rightarrow\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{array}\right] \mathrm{A}\\ \text { Applying } \mathrm{R}_{1} \rightarrow(-1) \mathrm{R}_{1} \text { and } \mathrm{R}_{2} \rightarrow(-1) \mathrm{R}_{2}\\$
$\Rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right] A$
$\text { As we have an Identity Matrix in LHS, }\\ \\\therefore \mathrm{A}^{-1}=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right]$

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.t
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:
$\begin{array}{l} {\left[\begin{array}{ccc} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 2 & 3 & -3 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}}\\ \\ \text { Applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-2 \mathrm{R}_{3} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \\\\ \text { Applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \\ {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ -1 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right] \mathrm{A}} \end{array}$
The second row of LHS contains all zeros, so we aren’t going to get any matrix in LHS.
∴ The 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 = $\begin{bmatrix} 2 &3 & 1\\1 &-1 &2 \\4 &1 &2 \end{bmatrix}$
The 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 an equal number of rows and columns and thus it matches the definition of a 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

The 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 of 3×3 = 9 elements, then
As each element can take 2 values (0 or 2)
∴ By simply counting principles, we can say that the total number of possible matrices = total number of ways in which 9 elements can take possible values = $2^9$ = 512
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:

Given that: $\left[\begin{array}{ll}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$

Equating the corresponding elements, we get,

$2 x+y=7 \ldots \ldots .(i)$

And $4 x=x+6$

From equations (ii)

$\begin{aligned} & 4 x-x=6 \\ & 3 x=6 \\ & \therefore x=2\end{aligned}$

From equations (i)

$\begin{aligned} & 2 \times 2+y=7 \\ & 4+y=7 \\ & \therefore y=7-4=3\end{aligned}$

$\therefore$ 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:

Given that: $\mathrm{A}=\frac{1}{\pi}\left[\begin{array}{ll}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]$

And $\mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}\left(\frac{x}{\pi}\right) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$

$\mathrm{A}-\mathrm{B}=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)\end{array}\right]-\frac{1}{\pi}\left[\begin{array}{cc}-\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x)\end{array}\right]$

$=\frac{1}{\pi}\left[\begin{array}{cc}\sin ^{-1}(x \pi)+\cos ^{-1}(x \pi) & \tan ^{-1}\left(\frac{x}{\pi}\right)-\tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right)-\sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)+\tan ^{-1}(\pi x)\end{array}\right]$

$=\frac{1}{\pi}\left[\begin{array}{ll}\frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2}\end{array}\right] \cdots \cdots\left[\begin{array}{l}\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\end{array}\right]$

$=\frac{1}{\pi} \times \frac{\pi}{2}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$=\frac{1}{2} \mathrm{I}$

∴ Option (D) is the only correct answer.

Question 57

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

Answer:

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

Question 58

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

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

Question 59

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

Answer:

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

Question 60

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

Answer:

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

Question 61

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

Answer:

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

Question 62

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

Answer:

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

Question 63

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

Answer:

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

Question 64

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

Answer:

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

Question 65

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

Answer:

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

Question 66

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

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

Answer:

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

Question 67

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

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

Answer:

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

Question 68

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

Answer:

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

Question 69

Fill in the blanks in each of the following:
The sum of two skew-symmetric matrices is always _______ matrix.

Answer:

A skew-symmetric matrix
Let A and B be any two matrices

$\therefore$ For skew-symmetric matrices

$\mathrm{A}=-\mathrm{A}^{\prime} \quad \ldots . . .(\mathrm{i})$

And B = - $\mathrm{B}^{\prime}$

Adding (i) and (ii), we get

$\begin{aligned} & \mathrm{A}+\mathrm{B}=-\mathrm{A}^{\prime}-\mathrm{B}^{\prime} \\ & \Rightarrow \mathrm{A}+\mathrm{B}=-\left(\mathrm{A}^{\prime}+\mathrm{B}^{\prime}\right)\end{aligned}$

So $\mathrm{A}+\mathrm{B}$ is skew-symmetric matrix.

Question 70

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

Answer:

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

Question 71

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

Answer:

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

Question 72

Fill in the blanks in each of the following:
A matrix that is not a square matrix is called a _____ matrix.

Answer:

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

Question 73

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

Answer:

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

Question 74

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

Answer:

Given A is a symmetric matrix

$\therefore \mathrm{A}^{\prime}=-\mathrm{A}$

$\operatorname{Now}\left(\mathrm{A}^3\right)^{\prime}=\left(\mathrm{A}^{\prime}\right)^3 \quad \ldots \ldots .\left[\because\left(\mathrm{A}^{\prime}\right)^{\mathrm{n}}=\left(\mathrm{A}^{\mathrm{n}}\right)^{\prime}\right]$

$=\mathrm{A}^3$

Question 75

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

Answer:

Given A is a skew-symmetric matrix.

$\begin{aligned} & \therefore \mathrm{A}^{\prime}=-\mathrm{A} \\ & \therefore\left(\mathrm{A}^2\right)^{\prime}=\left(\mathrm{A}^{\prime}\right)^2 \\ & =(-\mathrm{A})^2 \\ & =\mathrm{A}^2\end{aligned}$

So, $\mathrm{A}^2$ is a symmetric matrix.

Question 76

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

Answer:

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

Question 77

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

Answer:

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

Question 78

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

Answer:

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

Question 79

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

Answer:

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

Question 80

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

Answer:

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

Question 81

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

Answer:

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

Question 82

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

Answer:

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

Question 83

Which of the following statements are True or False

Matrices of any order can be added.

Answer:

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

Question 84

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

Answer:

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

Question 85

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

Answer:

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

Question 86

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

Answer:

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

Question 87

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

Answer:

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

Question 88

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

Answer:

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

Question 89

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

Answer:

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

Question 90

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

Answer:

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

Question 91

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

Answer:

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

Question 92

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

Answer:

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

Question 93

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

Answer:

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

Question 94

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

Answer:

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

Question 95

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

Answer:

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

Question 96

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

Answer:
Let $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}}$ and $\mathrm{B}=\left[\mathrm{b}_{\mathrm{ij}}\right]_{\mathrm{p} \times \mathrm{q}}$

$A B$ is defined when $n=P$

$\therefore$ Order of $\mathrm{AB}=\mathrm{m} \times \mathrm{q}$

$\Rightarrow \operatorname{Order}$ of $(\mathrm{AB})^{\prime}=\mathrm{q} \times \mathrm{m}$

Order of $\mathrm{B}^{\prime}$ is $\mathrm{q} \times \mathrm{p}$ and order of $\mathrm{A}^{\prime}$ is $\mathrm{n} \times \mathrm{m}$

$\therefore \mathrm{B}^{\prime} \mathrm{A}^{\prime}$ is defined when $\mathrm{P}=\mathrm{n}$

And the order of $\mathrm{B}^{\prime} \mathrm{A}^{\prime}$ is $\mathrm{q} \times \mathrm{m}$

Hence, order of $(A B)^{\prime}=$ Order of $B^{\prime} A^{\prime}$ i.e., $q \times m$
Hence, the given statement is true.

Question 97

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

Answer:

False
Let $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]$

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

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

$\therefore \mathrm{AB}=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{ll}0 & 0 \\ 2 & 0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

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

Here $\mathrm{AB}=\mathrm{AC}=0$ but $\mathrm{B} \neq \mathrm{C}$.

Question 98

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

Answer:

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

Question 99

Which of the following statements are True or False

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

Answer:

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

Question 100

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

Answer:

True
For skew-symmetric matrix A’=-A
$\begin{gathered}\Rightarrow\left(A^2\right)^{\prime}=(A A)^{\prime}=A^{\prime} A^{\prime} \\ \Rightarrow A^{\prime} A^{\prime}=(-A)(-A)=A^2 \\ \left.\Rightarrow\left(A^2\right)^{\prime}=A^2 \quad \text { (since } A \text { is symmetric }\right)\end{gathered}$

This equation shows that the transpose of $A^2$ is equal to $A^2$ itself when $A$ is a symmetric matrix (i.e., $\left.A^{\prime}=A\right)$.

Question 101:

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

Answer:

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

NCERT Maths Exemplar Solutions Chapter 3 Matrices: Topics

• Introduction
• Matrix
• Order of matrix
• Types of matrices
• Operations on matrices
• Addition of matrices
• Multiplication of a matrix by a scalar
• Properties of matrix addition
• Properties of scalar multiplication of a matrix
• Multiplication of matrices
• Properties of the multiplication of matrices
• Transpose of a matrix
• Properties of the transpose of matrices
• Symmetric and skew-symmetric matrices
• Elementary operation of a matrix
• Invertible matrices
• The inverse of matrices by elementary operations

NCERT Class 12 Maths Solutions Chapter-wise

Solutions are gathered together on Careers360 for quick access. Click the links below to view them.

Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 3

• NCERT Exemplar Solutions for Class 12 Math Chapter 3 will help the students understand the basics. It will help in developing the concepts.
• From this article, a student will be able to solve the problems based on the applications of matrices.
• NCERT Exemplar Class 12 Maths Solutions Chapter 3 covers all types of problems, which enable a student to solve higher-order problems.
• NCERT Exemplar Class 12 Maths Chapter 3 Solutions are prepared in a simple language only so that it is easier for the student to understand the solution and hence develop the right approach for a particular question.

NCERT Exemplar Class 12 Maths Solutions Chapterwise

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NCERT Solutions of Class 12 - Subject-wise

Here are the subject-wise links for the NCERT Solutions of Class 12:

• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Chemistry
• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 12 Biology

NCERT Notes of Class 12: Subject Wise

Given below are the subject-wise NCERT Notes of Class 12 :

• NCERT Notes for Class 12 Physics
• NCERT Notes for Class 12 Chemistry
• NCERT Notes for Class 12 Maths
• NCERT Notes for Class 12 Biology

NCERT Books and NCERT Syllabus

Students are advised to go through the current syllabus at the start of the academic year to understand the topics to be covered. The updated syllabus links and recommended books are provided below.

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12

NCERT Exemplar Class 12 Solutions: Subject Wise

Given below are the subject-wise Exemplar Solutions of Class 12 NCERT:

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