Exercise 7.1 - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Matrices and Determinants
Ex 7.1
Question 1.
Construct an $\mathrm{m} \times \mathrm{n}$ matrix $\mathrm{A}=\left[\mathrm{a}_{\mathrm{j}}\right]$, where $\mathrm{a}_{\mathrm{ij}}$ is given by
(i) $a_{i j}=\frac{(i-2 j)^2}{2}$ with $m=2, n=3$
(ii) $a_{i j}=\frac{|3 i-4 j|}{4}$ with $m=3, n=4$
Solution:
(i) $\mathrm{a}_{\mathrm{ij}}=\frac{(i-2 j)^2}{2}$
Here $\mathrm{m}=2, \mathrm{n}=3$
So we have to construct a matrix of order $2 \times 3$
Now A matrix of order $2 \times 3$ will be of the form $A=\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\end{array}\right)$
Here $\quad a_{i j}=\frac{(i-2 j)^2}{2}$
$\therefore \quad a_{11}=\frac{[1-2(1)]^2}{2}=\frac{(1-2)^2}{2}=\frac{1}{2}=\frac{1}{2}$
$a_{12}=\frac{[1-2(2)]^2}{2}=\frac{(1-4)^2}{2}=\frac{9}{2}$
$a_{13}=\frac{[1-2(3)]^2}{2}=\frac{(1-6)^2}{2}=\frac{25}{2}$
$a_{21}=\frac{[2-2(1)]^2}{2}=\frac{(2-2)^2}{2}=0$
$a_{22}=\frac{[2-2(2)]^2}{2}=\frac{(2-4)^2}{2}=\frac{4}{2}=2$
$a_{23}=\frac{[2-2(3)]^2}{2}=\frac{(2-6)^2}{2}=\frac{16}{2}=8$
$A=\left(\begin{array}{lll}\frac{1}{2} & \frac{9}{2} & \frac{25}{2} \\ 0 & 2 & 8\end{array}\right)=\frac{1}{2}\left(\begin{array}{lll}1 & 9 & 25 \\ 0 & 4 & 16\end{array}\right)$

(ii) Here $\mathrm{m}=3$ and $\mathrm{n}=4$
So we have to construct a matrix order $3 \times 4$ The general form of a matrix of order $3 \times 4$ will be
Here
$
\begin{aligned}
& \mathrm{A}=\left(\begin{array}{llll}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{21} & a_{22} & a_{23} & a_{24} \\
a_{31} & a_{32} & a_{33} & a_{34}
\end{array}\right) \\
& a_{i j}=\frac{|3 i-4 j|}{4} \\
& a_{12}=\frac{|3-4(2)|}{4}=\frac{|3-8|}{4}=\frac{5}{4} \\
& a_{13}=\frac{|3-4(3)|}{4}=\frac{|3-12|}{4}=\frac{9}{4} \\
& a_{14}=\frac{|3-4(4)|}{4}=\frac{|3-16|}{4}=\frac{13}{4} \\
& a_{21}=\frac{|3(2)-4(1)|}{4}=\frac{|6-4|}{4}=\frac{2}{4}=\frac{1}{2} \\
& a_{22}=\frac{|3(2)-4(2)|}{4}=\frac{|6-8|}{4}=\frac{2}{4}=\frac{1}{2} \\
& a_{23}=\frac{|3(2)-4(3)|}{4}=\frac{|6-12|}{4}=\frac{6}{4}=\frac{3}{2} \\
& a_{24}=\frac{|3(2)-4(4)|}{4}=\frac{|6-16|}{4}=\frac{10}{4}=\frac{5}{2} \\
&
\end{aligned}
$
$
\therefore \quad a_{11}=\frac{|3-4|}{4}=\frac{1}{4}
$

$\begin{aligned}
& a_{31}=\frac{|3(3)-4(1)|}{4}=\frac{|9-4|}{4}=\frac{5}{4} \\
& a_{32}=\frac{|3(3)-4(2)|}{4}=\frac{|9-8|}{4}=\frac{1}{4} \\
& a_{33}=\frac{|3(3)-4(3)|}{4}=\frac{|9-12|}{4}=\frac{3}{4} \\
& a_{34}=\frac{|3(3)-4(4)|}{4}=\frac{|9-16|}{4}=\frac{7}{4} \\
& \mathrm{~A}=\left(\begin{array}{cccc}
\frac{1}{4} & \frac{5}{4} & \frac{9}{4} & \frac{13}{4} \\
\frac{1}{2} & \frac{1}{2} & \frac{3}{2} & \frac{5}{2} \\
\frac{5}{4} & \frac{1}{4} & \frac{3}{4} & \frac{7}{4}
\end{array}\right)=\frac{1}{4}\left(\begin{array}{cccc}
1 & 5 & 9 & 13 \\
2 & 2 & 6 & 10 \\
5 & 1 & 3 & 7
\end{array}\right) \\
&
\end{aligned}$

Question 2.
Find the values of $p, q, r$ and $s$ if
$
\left[\begin{array}{ccc}
p^2-1 & 0 & -31-q^3 \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right]
$
Solution:
When two matrices (of same order) are equal then their corresponding entries are equal.
$
\begin{aligned}
& \text { Here }\left(\begin{array}{ccc}
p^2-1 & 0 & -31-q^3 \\
7 & r+1 & 9 \\
-2 & 8 & s-1
\end{array}\right)=\left(\begin{array}{ccc}
1 & 0 & -4 \\
7 & \frac{3}{2} & 9 \\
-2 & 8 & -\pi
\end{array}\right) \\
& \Rightarrow \mathrm{p}^2-1=1 \quad \Rightarrow \mathrm{p}^2=1+1=2 \\
& \mathrm{p}= \pm \sqrt{2} \\
& -31-q^3=-4 \\
& -\mathrm{q}^3=-4+31=27 \\
& \mathrm{q}^3=-27=(-3)^3 \\
& \Rightarrow \mathrm{q}=-3 \\
& \mathrm{r}+1=\frac{3}{2} \\
& \Rightarrow \mathrm{r}=\frac{3}{2}-1=\frac{3-2}{2}=\frac{1}{2} \\
& \mathrm{~s}-1=\pi \\
& \Rightarrow \mathrm{s}=-\pi+1 \text { (i.e.,) } \mathrm{s}=1-\pi \\
& \text { So, } \mathrm{p}= \pm \sqrt{2}, \mathrm{q}=-3, \mathrm{r}=1 / 2 \text { and } \mathrm{s}=1-\pi \\
&
\end{aligned}
$

Question 3.
Determine the value of $\mathrm{x}+\mathrm{y}$ if $\left[\begin{array}{cc}2 x+y & 4 x \\ 5 x-7 & 4 x\end{array}\right]=\left[\begin{array}{cc}7 & 7 y-13 \\ y & x+6\end{array}\right]$. Solution:
$
\begin{aligned}
& {\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]} \\
& \Rightarrow 2 \mathrm{x}+\mathrm{y}=7 \ldots \ldots \ldots \ldots(1) \\
& 4 \mathrm{x}=7 \mathrm{y}-13 \ldots \ldots \ldots \ldots(2) \\
& 5 \mathrm{x}-7=\mathrm{y} \ldots \ldots \ldots \ldots \ldots(3) \\
& 4 \mathrm{x}=\mathrm{x}+6 \ldots \ldots \ldots \ldots(4) \\
& \text { from (4) } 4 \mathrm{x}-\mathrm{x}=6 \\
& 3 \mathrm{x}=6 \Rightarrow \mathrm{x}=\frac{6}{3}=2
\end{aligned}
$
Substituting $x=2$ in (1), we get
$
2(2)+y=7 \Rightarrow 4+y=7 \Rightarrow y=7-4=3
$
So $x=2$ and $y=3$
$
\therefore \mathrm{x}+\mathrm{y}=2+3=5
$
Question 4.
Determine the matrices A and B if they satisfy
$
2 A-B+\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]=0 \text { and } A-2 B=\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]
$
Solution:

$
\begin{aligned}
& 2 \mathrm{~A}-\mathrm{B}+\left(\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right)=0 \\
& \Rightarrow \quad 2 \mathrm{~A}-\mathrm{B}=-\left(\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right)=\left(\begin{array}{ccc}
-6 & 6 & 0 \\
4 & -2 & -1
\end{array}\right) \\
& A-2 B=\left(\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right) \\
&
\end{aligned}
$
Solving (1) and (2)
(1) $\Rightarrow \quad 2 \mathrm{~A}-\mathrm{B}=\left(\begin{array}{ccc}-6 & 6 & 0 \\ 4 & -2 & -1\end{array}\right)$
$(2) \times 2 \Rightarrow$ $2 A-4 B=2\left(\begin{array}{ccc}3 & 2 & 8 \\ -2 & 1 & -7\end{array}\right)=\left(\begin{array}{ccc}6 & 4 & 16 \\ -4 & 2 & -14\end{array}\right)$
(1)-(3) $\Rightarrow$ $3 B=\left(\begin{array}{ccc}-6 & 6 & 0 \\ 4 & -2 & -1\end{array}\right)-\left(\begin{array}{ccc}6 & 4 & 16 \\ -4 & 2 & -14\end{array}\right)$
(i.e.)
$
\begin{aligned}
3 \mathrm{~B} & =\left(\begin{array}{ccc}
-6 & 6 & 0 \\
4 & -2 & -1
\end{array}\right)+\left(\begin{array}{ccc}
-6 & -4 & -16 \\
4 & -2 & 14
\end{array}\right) \\
3 \mathrm{~B} & =\left(\begin{array}{ccc}
-12 & 2 & -16 \\
8 & -4 & 13
\end{array}\right) \\
\mathrm{B} & =\frac{1}{3}\left(\begin{array}{ccc}
-12 & 2 & -16 \\
8 & -4 & 13
\end{array}\right)
\end{aligned}
$
Substituting B value in (2) we get
$
\begin{aligned}
A-2\left(\frac{1}{3}\right)\left(\begin{array}{rrr}
-12 & 2 & -16 \\
8 & -4 & 13
\end{array}\right) & =\left(\begin{array}{rrr}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right) \\
\mathrm{A} & =\left(\begin{array}{rrr}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right)+\frac{2}{3}\left(\begin{array}{rrr}
-12 & 2 & -16 \\
8 & -4 & 13
\end{array}\right) \\
& =\left(\begin{array}{rrr}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right)+\frac{1}{3}\left(\begin{array}{rrr}
-24 & 4 & -32 \\
16 & -8 & 26
\end{array}\right)
\end{aligned}
$

$=\frac{1}{3} \times 3\left(\begin{array}{rrr}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right)+\frac{1}{3}\left(\begin{array}{rrr}
-24 & 4 & -32 \\
16 & -8 & 26
\end{array}\right)$

$\begin{aligned}
& =\frac{1}{3}\left(\begin{array}{rrr}
9 & 6 & 24 \\
-6 & 3 & -21
\end{array}\right)+\frac{1}{3}\left(\begin{array}{rrr}
-24 & 4 & -32 \\
16 & -8 & 26
\end{array}\right) \\
& =\frac{1}{3}\left(\begin{array}{rrr}
-15 & 10 & -8 \\
10 & -5 & 5
\end{array}\right)
\end{aligned}$

Question 5.

If $\mathrm{A}=$, then compute $\mathrm{A}$
$
\left[\begin{array}{ll}
1 & a \\
0 & 1
\end{array}\right]
$
Solution:
$
A=\left(\begin{array}{ll}
1 & a \\
0 & 1
\end{array}\right)
$

Question 6.
Consider the matrix $A_\alpha=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
(i) Show that $\mathrm{A}_\alpha \mathrm{A}_\beta=\mathrm{A}_{\alpha+\beta}$.
(ii) Find all possible real values of satisfying the condition $A_\alpha+A^T{ }_\alpha=1$.
Solution:

$\begin{aligned}
A_\alpha & =\left(\begin{array}{lr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right) \\
\mathrm{A}_\beta & =\left(\begin{array}{lr}
\cos \beta & -\sin \beta \\
\sin \beta & \cos \beta
\end{array}\right) \\
\mathrm{A}_{\alpha+\beta} & =\left(\begin{array}{rr}
\cos (\alpha+\beta) & -\sin (\alpha+\beta) \\
\sin (\alpha+\beta) & \cos (\alpha+\beta)
\end{array}\right)
\end{aligned}$

(i) To Prove:
$
\begin{aligned}
\mathrm{A}_\alpha & \cdot \mathrm{A}_\beta=\mathrm{A}_{\alpha+\beta} \\
\mathrm{LHS} & =A_\alpha \cdot A_\beta=\left(\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right)\left(\begin{array}{cc}
\cos \beta & -\sin \beta \\
\sin \beta & \cos \beta
\end{array}\right) \\
& =\left(\begin{array}{lr}
\cos \alpha \cos \beta-\sin \alpha \sin \beta & -\cos \alpha \sin \beta-\sin \alpha \cos \beta \\
\sin \alpha \cos \beta+\cos \alpha \sin \beta & -\sin \alpha \sin \beta+\cos \alpha \cos \beta
\end{array}\right) \\
& =\left(\begin{array}{lr}
\cos (\alpha+\beta) & \sin (\alpha+\beta) \\
\sin (\alpha+\beta) & \cos (\alpha+\beta)
\end{array}\right)=\mathrm{A}_{\alpha+\beta}=\mathrm{RHS}
\end{aligned}
$

(ii) $\mathrm{A}_\alpha+\mathrm{A}_\alpha^{\mathrm{T}}=\mathrm{I}$
$
\begin{aligned}
& \Rightarrow\left(\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right)+\left(\begin{array}{rr}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right)=I \\
& \left(\begin{array}{ll}
2 \cos \alpha & 0 \\
0 & 2 \cos \alpha
\end{array}\right)=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& \Rightarrow \quad 2 \cos \alpha=1 \\
& \Rightarrow \quad \cos \alpha=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3} \\
&
\end{aligned}
$
General solution is $\alpha=2 \mathrm{n} \pi+\frac{\pi}{3}, \mathrm{n} \in \mathrm{Z}$
Question 7.
If $A=\left[\begin{array}{rr}4 & 2 \\ -1 & x\end{array}\right]$ such that $(A-2 I)(A-3 I)=0$, find the value of $\mathrm{x}$.

Solution:
Given
$
A=\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)
$
$
\begin{aligned}
& \therefore \quad A-2 I=\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)-2\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& =\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)+\left(\begin{array}{rr}
-2 & 0 \\
0 & -2
\end{array}\right)=\left(\begin{array}{rr}
2 & 2 \\
-1 & x-2
\end{array}\right) \\
& A-3 I=\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)-3\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{rr}
4 & 2 \\
-1 & x
\end{array}\right)+\left(\begin{array}{rr}
-3 & 0 \\
0 & -3
\end{array}\right) \\
& =\left(\begin{array}{rr}
1 & 2 \\
-1 & x-3
\end{array}\right) \\
& (A-2 I)(A-3 I)=0 \\
&
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \quad\left(\begin{array}{rr}
0 & 4+2(x-3) \\
-1-(x-2) & -2+(x-2)(x-3
\end{array}\right)=\left(\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right) \\
& \Rightarrow \quad-1-x+2=0 \text { (i.e. })-x+1=0 \\
& \Rightarrow \quad-x=-1 \\
& \Rightarrow \quad x=1 \\
&
\end{aligned}
$
Question 8.
If $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1\end{array}\right]$, show that $\mathrm{A}^2$ is a unit matrix.

Solution:
Given
Now
$
\begin{aligned}
& A=\left(\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right) \\
& \mathrm{A}^2=\mathrm{A} \times \mathrm{A}=\overline{\left(\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right)} \downarrow\left(\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
a & b & -1
\end{array}\right) \\
& =\left[\begin{array}{lll}
1+0+0 & 0+0+0 & 0+0+0 \\
0+0+0 & 0+1+0 & 0+0+0 \\
a+0-a & 0+b-b & 0+0+1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
&
\end{aligned}
$
Question 9.
$
\text { If } A=\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]
$
and $A^3-6 A^2+7 A+K I=0$, find the value of $k$.
Solution:

$
\begin{aligned}
& \text { Given } \quad A^3-6 A^2+7 A+k I=0 \\
& \Rightarrow\left(\begin{array}{rrr}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right)-6\left(\begin{array}{rrr}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right)+7\left(\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right)+k\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)=\left(\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) \\
& \Rightarrow\left(\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right)+\left(\begin{array}{rrr}
-30 & 0 & -48 \\
-12 & -24 & -30 \\
-48 & 0 & -78
\end{array}\right)+\left(\begin{array}{rrr}
7 & 0 & 14 \\
0 & 14 & 7 \\
14 & 0 & 21
\end{array}\right)+\left(\begin{array}{lll}
k & 0 & 0 \\
0 & k & 0 \\
0 & 0 & k
\end{array}\right)=\left(\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) \\
& \Rightarrow\left(\begin{array}{rrr}
-2 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array}\right)+\left(\begin{array}{lll}
k & 0 & 0 \\
0 & k & 0 \\
0 & 0 & k
\end{array}\right)=\left(\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) \\
& \Rightarrow\left(\begin{array}{rrr}
k-2 & 0 & 0 \\
0 & k-2 & 0 \\
0 & 0 & k-2
\end{array}\right)=\left(\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) \\
& \Rightarrow k-2=0 \Rightarrow k=2 \\
&
\end{aligned}
$
Question 10.
Give your own examples of matrices satisfying the following conditions in each case:
(i) $\mathrm{A}$ and $\mathrm{B}$ such that $\mathrm{AB} \neq \mathrm{BA}$.
(ii) $\mathrm{A}$ and $\mathrm{B}$ such that $\mathrm{AB}=0=\mathrm{BA}, \mathrm{A} \neq 0$ and $\mathrm{B} \neq 0$.
(iii) $A$ and $B$ such that $A B=0$ and $B A \neq 0$.
Solution:

(i)

(ii) Let $\quad \mathrm{A}=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0\end{array}\right)$ and $\mathrm{B}=\left(\begin{array}{rrr}0 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right)$ Now $\quad A B=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0\end{array}\right)\left(\begin{array}{rrr}0 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right)=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$ and $\quad \mathrm{BA}=\left(\begin{array}{rrr}0 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 0\end{array}\right)\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0\end{array}\right)=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$ $\Rightarrow \quad \mathrm{AB}=\mathrm{BA}=0$

(iii) Let $\quad A=\left(\begin{array}{rrr}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 0\end{array}\right)$ and $B=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 4 & 0 & 0\end{array}\right)$ Now $\quad \mathrm{AB}=\left(\begin{array}{rrr}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 0\end{array}\right)\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 4 & 0 & 0\end{array}\right)=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$ and $\quad \mathrm{BA}=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 4 & 0 & 0\end{array}\right)\left(\begin{array}{rrr}2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 0\end{array}\right)=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 8 & 0 & 0\end{array}\right)$ $\Rightarrow \quad \mathrm{AB}=0$ But $\mathrm{BA} \neq 0$

Question 11.
Show that $f(x) f(y)=f(x+y)$, where $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$.
Solution:
Given
$
f(x)=\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]
$
So, $\quad f(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$ and $f(x+y)=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$
To Prove: $\quad f(x) f(y)=f(x+y)$

LHS

$
\begin{aligned}
& =\left[\begin{array}{ccc}
\cos x \cos y-\sin x \sin y & -\cos x \sin y-\sin x \cos y & 0 \\
\sin x \cos y+\cos x \sin y & -\sin x \sin y+\cos x \cos y & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
0 & 0 & 1
\end{array}\right]=f(x+y)=\text { RHS }
\end{aligned}
$
Question 12.
If $A$ is a square matrix such that $A^2=A$, find the value of $7 A-(I+A)^3$.
Solution:
Given $\mathrm{A}^2=\mathrm{A}$
So $7 \mathrm{~A}-(\mathrm{I}+\mathrm{A})^3=7 \mathrm{~A}-\left(\mathrm{I}+3 \mathrm{~A}+3 \mathrm{~A}^2+\mathrm{A}^3\right.$ ]
$
=7 \mathrm{~A}-\mathrm{I}-3 \mathrm{~A}-3 \mathrm{~A}^2-\mathrm{A}^3
$
Given $A^2=A$
$
\begin{aligned}
& 7 A-I-3 A-3 A-A^3=-I+A-A^3 \\
& =-I+A-\left(A^2 \times A\right) \\
& =-I+A-(A \times A)=-I+A-A^2 \\
& =-I+A-A=-I
\end{aligned}
$
So the value of $7 A-(I+A)^3=-I$.
Question 13.
Verify the property $\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC}$, when the matrices $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are given by

$A=\left[\begin{array}{ccc}2 & 0 & -3 \\ 1 & 4 & 5\end{array}\right], B=\left[\begin{array}{cc}3 & 1 \\ -1 & 0 \\ 4 & 2\end{array}\right]$, and $C=\left[\begin{array}{cc}4 & 7 \\ 2 & 1 \\ 1 & -1\end{array}\right]$
Solution:
Given
$
A=\left(\begin{array}{rrr}
2 & 0 & -3 \\
1 & 4 & 5
\end{array}\right) ; B=\left(\begin{array}{rr}
3 & 1 \\
-1 & 0 \\
4 & 2
\end{array}\right)
$
and
$
C=\left(\begin{array}{rr}
4 & 7 \\
2 & 1 \\
1 & -1
\end{array}\right)
$
Now $B+C=\left(\begin{array}{rr}3 & 1 \\ -1 & 0 \\ 4 & 2\end{array}\right)+\left(\begin{array}{rr}4 & -1 \\ 2 & 1 \\ 1 & -1\end{array}\right)=\left(\begin{array}{ll}7 & 8 \\ 1 & 1 \\ 5 & 1\end{array}\right)$

Question 14.
Find the matrix A which satisfies the matrix relation
$
A\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]=\left[\begin{array}{rrr}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right]
$
Solution:
Let $\mathrm{B}=\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right)$ and $\mathrm{C}=\left(\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right)$
Here $B$ is of order $2 \times 3$ and $C$ is of order $2 \times 3$
Given $A B=C$
$\Rightarrow$ A should be a matrix of order $2 \times 2$
Let $\mathrm{A}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$
Now $\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right)=\left(\begin{array}{rrr}-7 & -8 & -9 \\ 2 & 4 & 6\end{array}\right)$
$
\Rightarrow\left(\begin{array}{lll}
a+4 b & 2 a+5 b & 3 a+6 b \\
c+4 d & 2 c+5 d & 3 c+6 d
\end{array}\right)=\left(\begin{array}{rrr}
-7 & -8 & -9 \\
2 & 4 & 6
\end{array}\right)
$

Question 15.
If $A^T=\left[\begin{array}{rr}4 & 5 \\ -1 & 0 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & -1 & 1 \\ 7 & 5 & -2\end{array}\right]$, verify the following
(i) $(\mathrm{A}+\mathrm{B})^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}+\mathrm{B}^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}}+\mathrm{A}^{\mathrm{T}}$
(ii) $(A-B)^{\mathrm{T}}=A^{\mathrm{T}}-B^{\mathrm{T}}$
(iii) $\left(B^T\right)^{\mathrm{T}}=\mathrm{B}$.
Solution:
$
\text { Given } \begin{aligned}
A^{\mathrm{T}} & =\left(\begin{array}{rr}
4 & 5 \\
-1 & 0 \\
2 & 3
\end{array}\right) \Rightarrow A=\left(\begin{array}{rrr}
4 & -1 & 2 \\
5 & 0 & 3
\end{array}\right) \\
B & =\left(\begin{array}{rrr}
2 & -1 & 1 \\
7 & 5 & -2
\end{array}\right) \Rightarrow B^{\mathrm{T}}=\left(\begin{array}{rr}
2 & 7 \\
-1 & 5 \\
1 & -2
\end{array}\right)
\end{aligned}
$
(i) To verify $(\mathrm{A}+\mathrm{B})^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}+\mathrm{B}^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}}+\mathrm{A}^{\mathrm{T}}$
$
\begin{aligned}
A+B & =\left(\begin{array}{rrr}
4 & -1 & 2 \\
5 & 0 & 3
\end{array}\right)+\left(\begin{array}{rrr}
2 & -1 & 1 \\
7 & 5 & -2
\end{array}\right) \\
& =\left(\begin{array}{rrr}
6 & -2 & 3 \\
12 & 5 & 1
\end{array}\right) \\
\therefore(A+B)^{\mathrm{T}} & =\left(\begin{array}{rr}
6 & 12 \\
-2 & 5 \\
3 & 1
\end{array}\right)
\end{aligned}
$

$\begin{aligned}
A^{\mathrm{T}} & =\left(\begin{array}{rr}
4 & 5 \\
-1 & 0 \\
2 & 3
\end{array}\right) \text { and } B^{\mathrm{T}}=\left(\begin{array}{rr}
2 & 7 \\
-1 & 5 \\
1 & -2
\end{array}\right) \\
\therefore \quad A^{\mathrm{T}}+\mathrm{B}^{\mathrm{T}} & =\left(\begin{array}{rr}
4 & 5 \\
-1 & 0 \\
2 & 3
\end{array}\right)+\left(\begin{array}{rr}
2 & 7 \\
-1 & 5 \\
1 & -2
\end{array}\right)=\left(\begin{array}{rr}
6 & 12 \\
-2 & 5 \\
3 & 1
\end{array}\right)
\end{aligned}$

Also $B^T+A^T=\left(\begin{array}{rr}2 & 7 \\ -1 & 5 \\ 1 & -2\end{array}\right)+\left(\begin{array}{rr}4 & 5 \\ -1 & 0 \\ 2 & 3\end{array}\right)=\left(\begin{array}{rr}6 & 12 \\ -2 & 5 \\ 3 & 1\end{array}\right)$
Here
$
(1)=(2)=(3)
$

$
\Rightarrow(A+B)^T=A^T+B^T=B^T+A^T
$
(ii) To verify
$
\begin{aligned}
(A-B)^T & =A^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}} \\
A-B & =\left(\begin{array}{lrr}
4 & -1 & 2 \\
5 & 0 & 3
\end{array}\right)-\left(\begin{array}{rrr}
2 & -1 & 1 \\
7 & 5 & -2
\end{array}\right) \\
& =\left(\begin{array}{rrr}
4 & -1 & 2 \\
5 & 0 & 3
\end{array}\right)+\left(\begin{array}{rrr}
-2 & 1 & -1 \\
-7 & -5 & 2
\end{array}\right)=\left(\begin{array}{rrr}
2 & 0 & 1 \\
-2 & -5 & 5
\end{array}\right)
\end{aligned}
$
$
\therefore \quad(A-B)^{\mathrm{T}}=\left(\begin{array}{rr}
2 & -2 \\
0 & -5 \\
1 & 5
\end{array}\right)
$
Also $A^{\mathrm{T}}=\left(\begin{array}{rr}4 & 5 \\ -1 & 0 \\ 2 & 3\end{array}\right)$ and $B^T=\left(\begin{array}{rr}2 & 7 \\ -1 & 5 \\ 1 & -2\end{array}\right)$
$
\begin{aligned}
\therefore \quad A^{\mathrm{T}}-\mathrm{B}^{\mathrm{T}} & =\left(\begin{array}{rr}
4 & 5 \\
-1 & 0 \\
2 & 3
\end{array}\right)-\left(\begin{array}{rr}
2 & 7 \\
-1 & 5 \\
1 & -2
\end{array}\right)=\left(\begin{array}{rr}
4 & 5 \\
-1 & 0 \\
2 & 3
\end{array}\right)+\left(\begin{array}{rr}
-2 & -7 \\
1 & -5 \\
-1 & 2
\end{array}\right) \\
& =\left(\begin{array}{rr}
2 & -2 \\
0 & -5 \\
1 & 5
\end{array}\right)
\end{aligned}
$
Here $\quad(1)=(2) \Rightarrow(A-B)^T=A^T-B^T$

(iii) To verify
Here
$
\begin{aligned}
\left(B^{\mathrm{T}}\right)^{\mathrm{T}} & =\mathrm{B} \\
\mathrm{B} & =\left(\begin{array}{rrr}
2 & -1 & 1 \\
7 & 5 & -2
\end{array}\right) \\
\therefore \quad B^{\mathrm{T}} & =\left(\begin{array}{rr}
2 & 7 \\
-1 & 5 \\
1 & -2
\end{array}\right) \text { and }\left(B^{\mathrm{T}}\right)^{\mathrm{T}}=\left\{\left(\begin{array}{rr}
2 & 7 \\
-1 & 5 \\
1 & -2
\end{array}\right)\right]^{\mathrm{T}} \\
& =\left(\begin{array}{rrr}
2 & -1 & 1 \\
7 & 5 & -2
\end{array}\right) \\
\Rightarrow\left(B^{\mathrm{T}}\right)^{\mathrm{T}} & =\mathrm{B}
\end{aligned}
$
$
\text { (1) }=(2) \Rightarrow\left(B^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{B}
$
Question 16.
If $A$ is a $3 \times 4$ matrix and $B$ is a matrix such that both ATB and BAT are defined, what is the order of the matrix B?
Sol.
$A$ is a matrix of order $3 \times 4$
So AT will be a matrix of order $4 \times 3$
AT B will be defined when B is a matrix of order $3 \times n$
$\mathrm{BA}^{\mathrm{T}}$ will be defirted when $\mathrm{B}$ is of order $m \times 4$
from (1) and (2) we see that B should be a matrix of order $3 \times 4$
Question 17.
Express the following matrices as the sum of a symmetric matrix and a skew-symmetric matrix:
(i) $\left[\begin{array}{rr}4 & -2 \\ 3 & -5\end{array}\right]$ and
(ii) $\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$
Solution:
(i)
Let $A=\left(\begin{array}{ll}4 & -2 \\ 3 & -5\end{array}\right) \Rightarrow A^T=\left(\begin{array}{rr}4 & 3 \\ -2 & -5\end{array}\right)$
Let $\quad P=\frac{1}{2}\left(A+A^{\mathrm{T}}\right)=\frac{1}{2}\left\{\left(\begin{array}{rr}4 & -2 \\ 3 & -5\end{array}\right)+\left(\begin{array}{rr}4 & 3 \\ -2 & -5\end{array}\right)\right\}=\frac{1}{2}\left(\begin{array}{rr}8 & 1 \\ 1 & -10\end{array}\right)$
Now $P^{\mathrm{T}}=\frac{1}{2}\left(\begin{array}{rr}8 & 1 \\ 1 & -10\end{array}\right)=\mathrm{P} \Rightarrow \mathrm{P}$ is a symmetric matrix
Let $\mathrm{Q}=\frac{1}{2}\left(A-A^T\right)=\frac{1}{2}\left[\left(\begin{array}{ll}4 & -2 \\ 3 & -5\end{array}\right)-\left(\begin{array}{rr}4 & 3 \\ -2 & -5\end{array}\right)\right]=\frac{1}{2}\left(\begin{array}{rr}0 & -5 \\ 5 & 0\end{array}\right)$
Here $Q^T=\left(\begin{array}{rr}0 & 5 \\ -5 & 0\end{array}\right)=-Q \Rightarrow Q$ is a skew symmetric matrix

$
A=P+Q=\frac{1}{2}\left[\left(\begin{array}{rr}
8 & 1 \\
1 & -10
\end{array}\right)+\left(\begin{array}{rr}
0 & -5 \\
5 & 0
\end{array}\right)\right]
$
(ii) Let
$
A=\left(\begin{array}{rrr}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right) \Rightarrow A^{\mathrm{T}}=\left(\begin{array}{rrr}
3 & -2 & -4 \\
3 & -2 & -5 \\
-1 & 1 & 2
\end{array}\right)
$
Let
$
\mathrm{P}=\frac{1}{2}\left(A+A^T\right)=\frac{1}{2}\left\{\left(\begin{array}{rrr}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right)+\left(\begin{array}{rrr}
3 & -2 & -4 \\
3 & -2 & -5 \\
-1 & 1 & 2
\end{array}\right)\right\}=\frac{1}{2}\left(\begin{array}{rrr}
6 & 1 & -5 \\
1 & -4 & -4 \\
-5 & -4 & 4
\end{array}\right)
$

Now $P^{\mathrm{T}}=\left(\begin{array}{rrr}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right)=\mathrm{P} \Rightarrow \mathrm{P}$ is a symmetric matrix. et $Q=\frac{1}{2}\left(A-A^{\mathrm{T}}\right)=\frac{1}{2}\left\{\left(\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right)-\left(\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right)\right\}$
Let $=\frac{1}{2}\left[\begin{array}{rrr}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]$
Now $\mathrm{Q}^{\mathrm{T}}=\frac{1}{2}\left(\begin{array}{rrr}0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0\end{array}\right)=-Q \Rightarrow \mathrm{Q}$ is a skew symmetric matrics $A=P+Q=\frac{1}{2}\left\{\left(\begin{array}{rrr}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right)+\left(\begin{array}{rrr}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right)\right\}$

Question 18.
$
\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right] A^T=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & 2 & -5 \\
9 & 22 & 15
\end{array}\right]
$
Find the matrix A such that

Solution:

Let $B=\left(\begin{array}{rr}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right)$ and $C=\left(\begin{array}{rrr}-1 & -8 & -10 \\ 1 & 2 & -5 \\ 9 & 22 & 15\end{array}\right)$
$B$ is of order $3 \times 2$ and $C$ is of order $3 \times 3$
So
$
\left\{\begin{array}{c}
\text { Now }(3 \times 2) \\
\times \\
(2 \times 3) \Rightarrow(3 \times 3)
\end{array}\right\}
$
$\Rightarrow \quad \mathrm{A}^{\mathrm{T}}$ should be of order $2 \times 3$
$\Rightarrow \quad$ A should be of order $3 \times 2$
Let $\mathrm{A}=\left(\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right)$ so $\mathrm{A}^{\mathrm{T}}=\left(\begin{array}{lll}a & c & c \\ b & d & f\end{array}\right)$
Here $\mathrm{BA}^{\mathrm{T}}=\mathrm{C}$

Question 19.
$
\text { If } A=\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
x & 2 & y
\end{array}\right]
$
is a matrix such that $\mathrm{AA}^T=9 I$, find the values of $x$ and $y$. Solution:
$
\text { Given } \mathrm{A}=\left(\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
x & 2 & y
\end{array}\right)
$

$\begin{gathered}
=\left(\begin{array}{ccc}
9 & 0 & x+4+2 y \\
0 & 9 & 2 x+2-2 y \\
x+4+2 y & 2 x+2-2 y & x^2+4+y^2
\end{array}\right) \\
\Rightarrow\left(\begin{array}{ccc}
9 & 0 & x+\text { (given) } \\
0 & 9 & 2 x+2-2 y \\
x+4+2 y & 2 x+2-2 y & x^2+4+y^2
\end{array}\right)=9\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)=\left(\begin{array}{lll}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right)
\end{gathered}$

Question 20.
(i) For what value of $x$, the matrix $A=\left[\begin{array}{rrr}0 & 1 & -2 \\ -1 & 0 & x^3 \\ 2 & -3 & 0\end{array}\right]$ is skew-symmetric.
(ii) If $\left[\begin{array}{rrr}0 & p & 3 \\ 2 & q^2 & -1 \\ r & 1 & 0\end{array}\right]$ is skew-symmetric, find the values of $p, q$ and $r$.
Solution:

(i)
$
A=\left(\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 0 & x^3 \\
2 & -3 & 0
\end{array}\right)
$
Given $A$ is skew symmetric
$
\begin{aligned}
& \Rightarrow \quad A^{\mathrm{T}}=-\mathrm{A} \\
& \Rightarrow \quad\left(\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & -3 \\
-2 & x^3 & 0
\end{array}\right)=-A=-\left(\begin{array}{rrr}
0 & 1 & -2 \\
-1 & 0 & x^3 \\
2 & -3 & 0
\end{array}\right) \\
& =\left(\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & -x^3 \\
-2 & 3 & 0
\end{array}\right) \\
& \Rightarrow \quad-x^3=-3 \Rightarrow x^3=3 \Rightarrow x=3^{1 / 3} \\
&
\end{aligned}
$

(ii) Let
$
\mathrm{A}=\left(\begin{array}{rrr}
0 & p & 3 \\
2 & q^2 & -1 \\
r & 1 & 0
\end{array}\right)
$
Given $A$ is a skew symmetric matrix
$
\begin{aligned}
& \Rightarrow \quad A^{\mathrm{T}}=-\mathrm{A} \\
& \Rightarrow\left(\begin{array}{rrr}
0 & 2 & r \\
p & q^2 & 1 \\
3 & -1 & 0
\end{array}\right)=-\left(\begin{array}{rrr}
0 & p & 3 \\
2 & q^2 & -1 \\
r & 1 & 0
\end{array}\right) \\
& \Rightarrow \quad\left(\begin{array}{lll}
0 & 2 & r \\
p & q^2 & 1 \\
3 & -1 & 0
\end{array}\right)=\left(\begin{array}{ccc}
0 & -p & -3 \\
-2 & -q^2 & 1 \\
-r & -1 & 0
\end{array}\right) \\
& \text { 2. } \mid \begin{array}{rr}
r=-3 ; & q^2= \\
\Rightarrow & 2 q^2=0
\end{array} \\
& \Rightarrow \quad \begin{aligned}
-p & =2 \\
p & =-2
\end{aligned} \quad \Rightarrow \quad q^2=0 \\
& q=0 \\
& \therefore \quad p=-2 ; q=0 ; r=-3 \\
&
\end{aligned}
$
Question 21.
Construct the matrix $A=\left[a_j\right]_{3 \times 3}$, where $a_{i j}=i-j$. State whether $A$ is symmetric or skew- symmetric.

Solution:
Given $A$ is a matrix of order $3 \times 3$
$
\therefore \mathrm{A}=\left(\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right)
$
We are given $a_{i j}=i-j$
$
\begin{aligned}
& \begin{array}{l|l|l}
a_{11}=1-1=0 & a_{21}=2-1=1 & a_{31}=3-1=2 \\
a_{12}=1-2=-1 & a_{22}=2-2=0 & a_{32}=3-2=1 \\
a_{13}=1-3=-2 & a_{23}=2-3=-1 & a_{33}=3-3=0
\end{array} \\
& \therefore A=\left(\begin{array}{rrr}
0 & -1 & -2 \\
1 & 0 & -1 \\
2 & 1 & 0
\end{array}\right) \text { and } A^{\mathrm{T}}=\left(\begin{array}{rrr}
0 & 1 & 2 \\
-1 & 0 & 1 \\
-2 & -1 & 0
\end{array}\right) \\
&
\end{aligned}
$
Here $A^{\mathrm{T}}=-\mathrm{A}$
$\Rightarrow A$ is skew symmetric

Question 22.
Let $\mathrm{A}$ and $\mathrm{B}$ be two symmetric matrices. Prove that $\mathrm{AB}=\mathrm{BA}$ if and only if $\mathrm{AB}$ is a symmetric matrix.
Solution:
Let $A$ and $B$ be two symmetric matrices
$\Rightarrow A^{\mathrm{T}}=\mathrm{A}$ and $\mathrm{B}^{\mathrm{T}}=\mathrm{B}$
Given that $\mathrm{AB}=\mathrm{BA}$ (2)
To prove $A B$ is symmetric:
Now $(\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}=\mathrm{BA}$
(from(1)) But $(A B)^{\mathrm{T}}=\mathrm{AB}$ by
$\Rightarrow A B$ is symmetric.
Conversely let $\mathrm{AB}$ be a symmetric matrix.
$
\begin{aligned}
& \Rightarrow(A B)^T=A B \\
& \text { i.e. } B^T A^T=A B \\
& \text { i.e. } B A=A B \text { (from (1)) } \\
& \Rightarrow A B \text { is symmetric }
\end{aligned}
$
Question 23.
If $A$ and $\mathrm{B}$ are symmetric matrices of same order, prove that
(i) $\mathrm{AB}+\mathrm{BA}$ is a symmetric matrix.
(1i) $\mathrm{AB}-\mathrm{BA}$ is a skew-symmetric matrix.
Solution:
Given $A$ and $B$ are symmetric matrices
$
\Rightarrow-\mathrm{A}^{\mathrm{T}}=\mathrm{A} \text { and } \mathrm{B}^{\mathrm{T}}=\mathrm{B}
$
(i) To prove $A B+B A$ is a symmetric matrix.
Proof: Now $(A B+B A)^T=(A B)^T+(B A)^T=B^T A^T+A^T B^T$
$=\mathrm{BA}+\mathrm{AB}=\mathrm{AB}+\mathrm{BA}$
i.e. $(A B+B A)^T=A B+B A$
$\Rightarrow(A B+B A)$ is a symmetric matrix.

(ii) To prove $A B-B A$ is a skew symmetric matrix.
Proof: $(A B-B A)^T=(A B)^T-(B A)^T=B^T A^T-A^T B^T=B A-A B$
i.e. $(A B-B A)^T=-(A B-B A)$
$\Rightarrow A B-B A$ is a skew symmetric matrix.
Question 24.
A shopkeeper in a Nuts and Spices shop makes gift packs of cashew nuts, raisins and almonds. Pack I contains $100 \mathrm{gm}$ of cashew nuts, $100 \mathrm{gm}$ of raisins and $50 \mathrm{gm}$ of almonds. Pack-II contains 200 gm of cashew nuts, $100 \mathrm{gm}$ of raisins and $100 \mathrm{gm}$ of almonds. Pack-III contains $250 \mathrm{gm}$ of cashew nuts, $250 \mathrm{gm}$ of raisins and $150 \mathrm{gm}$ of almonds. The cost of $50 \mathrm{gm}$ of cashew nuts is ₹ $50,50 \mathrm{gm}$ of raisins is $₹ 10$, and $50 \mathrm{gm}$ of almonds is 60 . What is the cost of each gift pack?
Solution:

$
\begin{aligned}
& =\left(\begin{array}{c}
100+20+60 \\
200+20+120 \\
250+50+180
\end{array}\right)=\left(\begin{array}{l}
180 \\
340 \\
480
\end{array}\right) \\
\therefore \quad \text { Cost of pack I } & =₹ 180 \\
\text { Cost of pack II } & =₹ 340 \\
\text { Cost of pack III } & =₹ 480
\end{aligned}
$