NCERT Exemplar Class 12 Maths Solutions Chapter 3 Matrices

NCERT Exemplar Class 12 Maths Solutions Chapter 3, Matrix is one of the most interesting chapters to study. Matrices are much faster and more efficient than the usual direct-solving method. In our daily lives, we see tables, spreadsheets, seating charts, and cinema bookings arranged in rows and columns resembling matrices. So, what is a matrix? A matrix is a rectangular arrangement of numbers, symbols, or expressions in rows and columns . The numbers inside a matrix are called its elements . It is usually written as A=[a_ij], where a_ij denotes the element in the i -th row and j -th column. Matrices are used in various fields like computer graphics, economics, and engineering to solve systems of equations, data representation, and transformations. They are a key concept in linear algebra.

NCERT Exemplar Class 12 Math chapter 3 solutions cover various matrix-related topics like the types, the operations on two or matrices, invertible matrices, etc. It is a highly scoring chapter of NCERT Class 12 Maths Solutions that a student can utilize to gain higher scores in their exams.

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 $mn$ elements, then the dimension of the order can be given by $m\ast n$, where $m$ and $n$ are natural numbers.

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

$\because mn=28$

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

So, let $\mathrm{m}=1$ 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, $mn=13$.

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 $a_{2\times 2}$ matrix where
(i) $a_{ij}=\frac{(\hat{i}-2\hat{j})}{2}$
(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 ${\mathrm{a}}_{\mathrm{ij}}=\frac{(\mathrm{i}-2\mathrm{j}{)}^2}{2}$

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

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

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

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:

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

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

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

Question:5

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

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

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

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

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

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

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

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

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

Equating the corresponding elements, we get

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

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

Answer:

The given equation can be written as

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

Equating the corresponding elements we get

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

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 $(A+B)(A-B)\ne A^2-B^2.$

Answer:

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

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

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

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

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

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

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

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

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

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

Hence, the values of $x$ are -2 and -14.

Question:11

Show that $\left[\begin{array}{cc}5& 3\\ -1& -2\end{array}\right]$ satisfies the equation $A^2-3A-7I=0$ and hence find $A^{-1}$.

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Comparing elements of both sides

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

And 4z = 4

⇒ z = 1

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

Question:14

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

Answer:

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

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

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

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

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

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