Exercise 3.3 (Revised) - Chapter 3 - Matrices - Ncert Solutions class 12 - Maths

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NCERT Solutions Class 12 Maths: Chapter 3 - Matrices | Comprehensive Guide

Ex 3.3 Question 1:
Find the transpose of each of the following matrices:
$
\text { (i) }\left[\begin{array}{c}
5 \\
\frac{1}{2} \\
-1
\end{array}\right]_{(\mathrm{ii})}\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]_{(\mathrm{iii})}\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right]
$

Answer
(i) Let $A=\left[\begin{array}{c}5 \\ \frac{1}{2} \\ -1\end{array}\right]$, then $A^{\top}=\left[\begin{array}{lll}5 & \frac{1}{2} & -1\end{array}\right]$

(ii)

$\text { Let } A=\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right] \text {, then } A^{\mathrm{T}}=\left[\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right]$

$\text { (iii) Let } A=\left[\begin{array}{ccc}
-1 & 5 & 6 \\
\sqrt{3} & 5 & 6 \\
2 & 3 & -1
\end{array}\right] \text {, then } A^{\mathrm{T}}=\left[\begin{array}{ccc}
-1 & \sqrt{3} & 2 \\
5 & 5 & 3 \\
6 & 6 & -1
\end{array}\right]$

Ex 3.3 Question 2:
$
A=\left[\begin{array}{rrr}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right] \text { and } B=\left[\begin{array}{rrr}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right] \text {, then verify that }
$
(i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}$
(ii) $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

Answer
We have:

$
A^{\prime}=\left[\begin{array}{rrr}
-1 & 5 & -2 \\
2 & 7 & 1 \\
3 & 9 & 1
\end{array}\right], B^{\prime}=\left[\begin{array}{rrr}
-4 & 1 & 1 \\
1 & 2 & 3 \\
-5 & 0 & 1
\end{array}\right]
$
(i)
$
A+B=\left[\begin{array}{rrr}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]+\left[\begin{array}{rrr}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 3 & -2 \\
6 & 9 & 9 \\
-1 & 4 & 2
\end{array}\right]
$

$\begin{aligned}
& \therefore(A+B)^{\prime}=\left[\begin{array}{rrr}
-5 & 6 & -1 \\
3 & 9 & 4 \\
-2 & 9 & 2
\end{array}\right] \\
& A^{\prime}+B^{\prime}=\left[\begin{array}{rrr}
-1 & 5 & -2 \\
2 & 7 & 1 \\
3 & 9 & 1
\end{array}\right]+\left[\begin{array}{rrr}
-4 & 1 & 1 \\
1 & 2 & 3 \\
-5 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 6 & -1 \\
3 & 9 & 4 \\
-2 & 9 & 2
\end{array}\right]
\end{aligned}$

Hence, we have verified that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$
(ii)
$
\begin{aligned}
& A-B=\left[\begin{array}{rrr}
-1 & 2 & 3 \\
5 & 7 & 9 \\
-2 & 1 & 1
\end{array}\right]-\left[\begin{array}{rrr}
-4 & 1 & -5 \\
1 & 2 & 0 \\
1 & 3 & 1
\end{array}\right]=\left[\begin{array}{rrr}
3 & 1 & 8 \\
4 & 5 & 9 \\
-3 & -2 & 0
\end{array}\right] \\
& \therefore(A-B)^{\prime}=\left[\begin{array}{rrr}
3 & 4 & -3 \\
1 & 5 & -2 \\
8 & 9 & 0
\end{array}\right] \\
& A^{\prime}-B^{\prime}=\left[\begin{array}{rrr}
-1 & 5 & -2 \\
2 & 7 & 1 \\
3 & 9 & 1
\end{array}\right]-\left[\begin{array}{rrr}
-4 & 1 & 1 \\
1 & 2 & 3 \\
-5 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
3 & 4 & -3 \\
1 & 5 & -2 \\
8 & 9 & 0
\end{array}\right] \\
&
\end{aligned}
$

Hence, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.

Ex 3.3 Question 3:
$A^{\prime}=\left[\begin{array}{rr}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{rrr}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right]$, then verify that
(i) $(A+B)^{\prime}=A^{\prime}+B^{\prime}$
(ii) $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

Answer

(i) It is known that $A=\left(A^{\prime}\right)^{\prime}$

Therefore, we have:
$
\begin{aligned}
& A=\left[\begin{array}{rrr}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right] \\
& B^{\prime}=\left[\begin{array}{rr}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right] \\
& A+B=\left[\begin{array}{rrr}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]+\left[\begin{array}{rrr}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 1 \\
5 & 4 & 4
\end{array}\right]
\end{aligned}
$

$
\begin{aligned}
& \therefore(A+B)^{\prime}=\left[\begin{array}{ll}
2 & 5 \\
1 & 4 \\
1 & 4
\end{array}\right] \\
& A^{\prime}+B^{\prime}=\left[\begin{array}{rr}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]+\left[\begin{array}{rr}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]=\left[\begin{array}{ll}
2 & 5 \\
1 & 4 \\
1 & 4
\end{array}\right]
\end{aligned}
$

Thus, we have verified that $(A+B)^{\prime}=A^{\prime}+B^{\prime}$.

(ii)
$
\begin{aligned}
& A-B=\left[\begin{array}{rrr}
3 & -1 & 0 \\
4 & 2 & 1
\end{array}\right]-\left[\begin{array}{rrr}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
4 & -3 & -1 \\
3 & 0 & -2
\end{array}\right] \\
& \therefore(A-B)^{\prime}=\left[\begin{array}{rr}
4 & 3 \\
-3 & 0 \\
-1 & -2
\end{array}\right] \\
& A^{\prime}-B^{\prime}=\left[\begin{array}{rr}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right]-\left[\begin{array}{rr}
-1 & 1 \\
2 & 2 \\
1 & 3
\end{array}\right]=\left[\begin{array}{rr}
4 & 3 \\
-3 & 0 \\
-1 & -2
\end{array}\right]
\end{aligned}
$

Thus, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$.

Ex 3.3 Question 4:
$
\text { If } A^{\prime}=\left[\begin{array}{rr}
-2 & 3 \\
1 & 2
\end{array}\right] \text { and } B=\left[\begin{array}{rr}
-1 & 0 \\
1 & 2
\end{array}\right] \text {, then find }(A+2 B)^{\prime}
$

Answer
We know that $A=\left(A^{\prime}\right)^{\prime}$
$
\therefore A=\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]
$

$\begin{aligned}
& \therefore A+2 B=\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]+2\left[\begin{array}{rr}
-1 & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
-2 & 1 \\
3 & 2
\end{array}\right]+\left[\begin{array}{rr}
-2 & 0 \\
2 & 4
\end{array}\right]=\left[\begin{array}{rr}
-4 & 1 \\
5 & 6
\end{array}\right] \\
& \therefore(A+2 B)^{\prime}=\left[\begin{array}{rr}
-4 & 5 \\
1 & 6
\end{array}\right]
\end{aligned}$

Ex 3.3 Question 5:
For the matrices $A$ and $B$, verify that $(A B)^{\prime}=B^{\prime} A^{\prime}$ where
$
\text { (i) } A=\left[\begin{array}{r}
1 \\
-4 \\
3
\end{array}\right], B=\left[\begin{array}{lll}
-1 & 2 & 1
\end{array}\right]
$

(ii)

$
A=\left[\begin{array}{l}
0 \\
1 \\
2
\end{array}\right], B=\left[\begin{array}{lll}
1 & 5 & 7
\end{array}\right]
$

Answer
(i)
$
\begin{aligned}
& A B=\left[\begin{array}{r}
1 \\
-4 \\
3
\end{array}\right]\left[\begin{array}{lll}
-1 & 2 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-1 & 2 & 1 \\
4 & -8 & -4 \\
-3 & 6 & 3
\end{array}\right] \\
& \therefore(A B)^{\prime}=\left[\begin{array}{rrr}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]
\end{aligned}
$

$
\begin{aligned}
& \text { Now, } A^{\prime}=\left[\begin{array}{lll}
1 & -4 & 3
\end{array}\right], B^{\prime}=\left[\begin{array}{r}
-1 \\
2 \\
1
\end{array}\right] \\
& \therefore B^{\prime} A^{\prime}=\left[\begin{array}{r}
-1 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
1 & -4 & 3
\end{array}\right]=\left[\begin{array}{rrr}
-1 & 4 & -3 \\
2 & -8 & 6 \\
1 & -4 & 3
\end{array}\right]
\end{aligned}
$

Hence, we have verified that $(A B)^{\prime}=B^{\prime} A^{\prime}$.

(ii)
$
\begin{aligned}
& A B=\left[\begin{array}{l}
0 \\
1 \\
2
\end{array}\right]\left[\begin{array}{lll}
1 & 5 & 7
\end{array}\right]=\left[\begin{array}{rrr}
0 & 0 & 0 \\
1 & 5 & 7 \\
2 & 10 & 14
\end{array}\right] \\
& \therefore(A B)^{\prime}=\left[\begin{array}{rrr}
0 & 1 & 2 \\
0 & 5 & 10 \\
0 & 7 & 14
\end{array}\right] \\
& \text { Now, } A^{\prime}=\left[\begin{array}{lll}
0 & 1 & 2
\end{array}\right], B^{\prime}=\left[\begin{array}{l}
1 \\
5 \\
7
\end{array}\right] \\
& \therefore B^{\prime} A^{\prime}=\left[\begin{array}{l}
1 \\
5 \\
7
\end{array}\right]\left[\begin{array}{lll}
0 & 1 & 2
\end{array}\right]=\left[\begin{array}{lll}
0 & 1 & 2 \\
0 & 5 & 10 \\
0 & 7 & 14
\end{array}\right] \\
&
\end{aligned}
$

Hence, we have verified that $(A B)^{\prime}=B^{\prime} A^{\prime}$.

Ex 3.3 Question 6:
If (i) $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$, then verify that $A^{\prime} A=I$
(ii) $A=\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]$, then verify that $A^{\prime} A=I$

Answer

(i)

$\begin{aligned}
& A=\left[\begin{array}{rr}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\
& \therefore A^{\prime}=\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
& A^{\prime} A=\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right] \\
& =\left[\begin{array}{ll}
(\cos \alpha)(\cos \alpha)+(-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha)+(-\sin \alpha)(\cos \alpha) \\
(\sin \alpha)(\cos \alpha)+(\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha)+(\cos \alpha)(\cos \alpha)
\end{array}\right]
\end{aligned}$

$
\begin{aligned}
& =\left[\begin{array}{lc}
\cos ^2 \alpha+\sin ^2 \alpha & \sin \alpha \cos \alpha-\sin \alpha \cos \alpha \\
\sin \alpha \cos \alpha-\sin \alpha \cos \alpha & \sin ^2 \alpha+\cos ^2 \alpha
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I
\end{aligned}
$

Hence, we have verified that $A^{\prime} A=I$.

Ex 3.3 Question 7:
(i) Show that the matrix $A=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 5 & 1 & 3\end{array}\right]_{\text {is a symmetric matrix }}$

(ii) Show that the matrix $A=\left[\begin{array}{ccc}0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0\end{array}\right]_{\text {is a skew symmetric matrix }}$

Answer
(i) We have:

$
\begin{aligned}
& A^{\prime}=\left[\begin{array}{ccc}
1 & -1 & 5 \\
-1 & 2 & 1 \\
5 & 1 & 3
\end{array}\right]=A \\
& \therefore A^{\prime}=A
\end{aligned}
$

Hence, $A$ is a symmetric matrix.
(ii) We have:
$
\begin{aligned}
& A^{\prime}=\left[\begin{array}{ccc}
0 & -1 & 1 \\
1 & 0 & -1 \\
-1 & 1 & 0
\end{array}\right]=-\left[\begin{array}{ccc}
0 & 1 & -1 \\
-1 & 0 & 1 \\
1 & -1 & 0
\end{array}\right]=-A \\
& \therefore A^{\prime}=-A
\end{aligned}
$

Hence, $A$ is a skew-symmetric matrix.

Ex 3.3 Question 8:
For the matrix $A=\left[\begin{array}{ll}1 & 5 \\ 6 & 7\end{array}\right]$, verify that
(i) $\left(A+A^{\prime}\right)$ is a symmetric matrix
(ii) $\left(A-A^{\prime}\right)$ is a skew symmetric matrix

Answer

$
A^{\prime}=\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]
$
(i)
$
A+A^{\prime}=\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]+\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]=\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]
$
$
\therefore\left(A+A^{\prime}\right)^{\prime}=\left[\begin{array}{ll}
2 & 11 \\
11 & 14
\end{array}\right]=A+A^{\prime}
$

Hence, $\left(A+A^{\prime}\right)$ is a symmetric matrix.
$
\begin{aligned}
& \text { (ii) } A-A^{\prime}=\left[\begin{array}{ll}
1 & 5 \\
6 & 7
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
5 & 7
\end{array}\right]=\left[\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right] \\
& \left(A-A^{\prime}\right)^{\prime}=\left[\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}\right]=-\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]=-\left(A-A^{\prime}\right)
\end{aligned}
$

Hence, $\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.

Ex 3.3 Question 9:
Find $\frac{1}{2}\left(A+A^{\prime}\right)$ and $\frac{1}{2}\left(A-A^{\prime}\right)$, when $A=\left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right]$
Answer

$\begin{aligned}
&A=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]\\
&\begin{aligned}
& \therefore A^{\prime}=\left[\begin{array}{ccc}
0 & -a & -b \\
a & 0 & -c \\
b & c & 0
\end{array}\right] \\
& A+A^{\prime}=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]+\left[\begin{array}{ccc}
0 & -a & -b \\
a & 0 & -c \\
b & c & 0
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& \therefore \frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{aligned}
\end{aligned}$

$\begin{aligned}
& \text { Now, } A-A^{\prime}=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]-\left[\begin{array}{ccc}
0 & -a & -b \\
a & 0 & -c \\
b & c & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & 2 a & 2 b \\
-2 a & 0 & 2 c \\
-2 b & -2 c & 0
\end{array}\right] \\
& \therefore \frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]
\end{aligned}$

$\begin{aligned}
& \text { Now, } A-A^{\prime}=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]-\left[\begin{array}{ccc}
0 & -a & -b \\
a & 0 & -c \\
b & c & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & 2 a & 2 b \\
-2 a & 0 & 2 c \\
-2 b & -2 c & 0
\end{array}\right] \\
& \therefore \frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right]
\end{aligned}$

Ex 3.3 Question 10:
Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
(i) $\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]$
(ii) $\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$
(iii) $\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$
(iv) $\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]$

Answer

(i)

Let $A=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rr}3 & 1 \\ 5 & -1\end{array}\right]$

Now, $A+A^{\prime}=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]+\left[\begin{array}{rr}3 & 1 \\ 5 & -1\end{array}\right]=\left[\begin{array}{rr}6 & 6 \\ 6 & -2\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr}6 & 6 \\ 6 & -2\end{array}\right]=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{rr}3 & 3 \\ 3 & -1\end{array}\right]=P$

Thus,
$
P=\frac{1}{2}\left(A+A^{\prime}\right)
$
is a symmetric matrix.

Now, $A-A^{\prime}=\left[\begin{array}{rr}3 & 5 \\ 1 & -1\end{array}\right]-\left[\begin{array}{rr}3 & 1 \\ 5 & -1\end{array}\right]=\left[\begin{array}{rr}0 & 4 \\ -4 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr}0 & 4 \\ -4 & 0\end{array}\right]=\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{rr}0 & 2 \\ -2 & 0\end{array}\right]=-Q$
Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$
Thus, 2 is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :

$
P+Q=\left[\begin{array}{rr}
3 & 3 \\
3 & -1
\end{array}\right]+\left[\begin{array}{rr}
0 & 2 \\
-2 & 0
\end{array}\right]=\left[\begin{array}{rr}
3 & 5 \\
1 & -1
\end{array}\right]=A
$
(ii)

Let $A=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$

Now, $A+A^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{rrr}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rrr}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=P$
Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix.

Now, $A-A^{\prime}=\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]+\left[\begin{array}{rrr}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
Now, $Q^{\prime}=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=-Q$

Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :
$
P+Q=\left[\begin{array}{rrr}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]+\left[\begin{array}{rrr}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{rrr}
6 & -2 & 2 \\
-2 & 3 & -1 \\
2 & -1 & 3
\end{array}\right]=A
$

(iii)

Let $A=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]$

Now, $A+A^{\prime}=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{rrr}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{rrr}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=P$

Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix. 

Now, $A-A^{\prime}=\left[\begin{array}{rrr}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{rrr}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{ccc}0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=-Q$
Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is
is a skew-symmetric matrix.

Representing $A$ as the sum of $P$ and $Q$ :
$
P+Q=\left[\begin{array}{rrr}
3 & \frac{1}{2} & -\frac{5}{2} \\
\frac{1}{2} & -2 & -2 \\
-\frac{5}{2} & -2 & 2
\end{array}\right]+\left[\begin{array}{ccc}
0 & \frac{5}{2} & \frac{3}{2} \\
-\frac{5}{2} & 0 & 3 \\
-\frac{3}{2} & -3 & 0
\end{array}\right]=\left[\begin{array}{rrr}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]=A
$

(iv)

Let $A=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]$, then $A^{\prime}=\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]$

Now $A+A^{\prime}=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]+\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]=\left[\begin{array}{ll}2 & 4 \\ 4 & 4\end{array}\right]$

Let $P=\frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]$

Now, $P^{\prime}=\left[\begin{array}{ll}1 & 2 \\ 2 & 2\end{array}\right]=P$
Thus, $P=\frac{1}{2}\left(A+A^{\prime}\right)$ is a symmetric matrix.
Now, $A-A^{\prime}=\left[\begin{array}{rr}1 & 5 \\ -1 & 2\end{array}\right]-\left[\begin{array}{rr}1 & -1 \\ 5 & 2\end{array}\right]=\left[\begin{array}{rr}0 & 6 \\ -6 & 0\end{array}\right]$

Let $Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{rr}0 & 3 \\ -3 & 0\end{array}\right]$

Now, $Q^{\prime}=\left[\begin{array}{rr}0 & -3 \\ 3 & 0\end{array}\right]=-Q$
Thus, $Q=\frac{1}{2}\left(A-A^{\prime}\right)$ is a skew-symmetric matrix.
Representing $A$ as the sum of $P$ and $Q$ :
$
P+Q=\left[\begin{array}{ll}
1 & 2 \\
2 & 2
\end{array}\right]+\left[\begin{array}{rr}
0 & 3 \\
-3 & 0
\end{array}\right]=\left[\begin{array}{rr}
1 & 5 \\
-1 & 2
\end{array}\right]=A
$

Ex 3.3 Question 11:
If $A, B$ are symmetric matrices of same order, then $A B-B A$ is a
A. Skew symmetric matrix
B. Symmetric matrix
C. Zero matrix D. Identity matrix

Answer
The correct answer is $\mathrm{A}$.
$A$ and $B$ are symmetric matrices, therefore, we have:
$
A^{\prime}=A \text { and } B^{\prime}=B
$

Consider $(A B-B A)^{\prime}=(A B)^{\prime}-(B A)^{\prime}$
$
\left[(A-B)^{\prime}=A^{\prime}-B^{\prime}\right]
$

$\begin{aligned}
&\begin{aligned}
& =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} \quad\left[(A B)^{\prime}=B^{\prime} A^{\prime}\right] \\
& =B A-A B \quad[\text { by (1) }] \\
&
\end{aligned}\\
&=-(A B-B A)
\end{aligned}$

$
\therefore(A B-B A)^{\prime}=-(A B-B A)
$

Thus, $(A B-B A)$ is a skew-symmetric matrix.

Ex 3.3 Question 12:
$A=\left[\begin{array}{rr}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$, then $A+A^{\prime}=I$, if the value of $\mathrm{a}$ is
A. $\frac{\pi}{6}$ B. $\frac{\pi}{3}$
C. $ก$ D. $\frac{3 \pi}{2}$

Answer
The correct answer is B.
$
\begin{aligned}
& A=\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
& \Rightarrow A^{\prime}=\left[\begin{array}{lr}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]
\end{aligned}
$

Now, $A+A^{\prime}=I$
$
\begin{aligned}
& \therefore\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]+\left[\begin{array}{ll}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& \Rightarrow\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]
\end{aligned}
$

Comparing the corresponding elements of the two matrices, we have:
$
\begin{aligned}
& 2 \cos \alpha=1 \\
& \Rightarrow \cos \alpha=\frac{1 \pi}{2}=\cos \frac{\pi}{3} \\
& \therefore \alpha=\frac{\pi}{3}
\end{aligned}
$