We have, \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right] \)
\({ A+A }^{ \prime }=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A+A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] =0\)
and \({ A-A }^{ \prime }=\left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow \frac { 1 }{ 2 } { A-A }^{ \prime }=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{matrix} \right] \)
\(\Rightarrow A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)

If A is symmetric matrix then
\(A={ A }^{ \prime }\)
\(\Rightarrow \left[ \begin{matrix} 0 & b & -2 \\ 3 & 1 & 3 \\ 2a & 3 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 & 2a \\ b & 1 & 3 \\ -2 & 3 & -1 \end{matrix} \right] \)
\(\therefore\)  By equality of matrices, 
b = 3 and a = - 1

Given \(A=\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix},\) then 
\({ A }^{ 2 }=\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}=O \)
\(\Rightarrow { A }^{ 16 }=({ A }^{ 2 })^{ 8 }=O.\)

Given \({ a }_{ ij }=\frac { 1 }{ 3 } \left| -3i+j \right| \)
\(\therefore \ { a }_{ 23 }=\frac { 1 }{ 3 } \left| -6+3 \right| =1\)
 

\(\begin{bmatrix} 3x\quad + & 4 \\ 2x\quad + & x \end{bmatrix}=\left[ \begin{matrix} 19 \\ 15 \end{matrix} \right] \Rightarrow 3x=15\Rightarrow x=5\)

\(\left[ \begin{matrix} x & +3y & y \\ 7 & -x & 4 \end{matrix} \right] =\begin{bmatrix} 4 & -1 \\ 0 & 4 \end{bmatrix}\)
\(\Rightarrow x+3y=4;y=-1;7-x=0 \)
\(\Rightarrow x=7,y=-1\)

For getting A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \)
For getting A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
Simplifying A³ - 6A² + 7A + kI³ as
\(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
Equating  \(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)
\(\Rightarrow\)  k - 2 = 0
\(\Rightarrow\)  k = 2
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)
A³ = A² . A = \(\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] \)
Now A³ - 6A² + 7A + kP₃ = 0
\(\Rightarrow \left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] -\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right] +\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow\)   21 - 30 + 7 + k = 0
\(\Rightarrow\)  k = 2

\(\begin{bmatrix} \cos { 2\alpha } & \sin { 2\alpha } \\ -\sin { 2\alpha } & \cos { 2\alpha } \end{bmatrix}\)

(i) \(A=\frac { 1 }{ 2 } (A+A')+\frac { 1 }{ 2 } (A-A');\) 
where \(A=\frac { 1 }{ 2 } (A+A')\) is symmetric
and \(\frac { 1 }{ 2 } (A-A')\) is skew symmetric Proceed.
\(\text { Let } A=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right] \text { , then } A^{\prime}=\left[\begin{array}{ll} 3 & 1 \\ 5 & -1 \end{array}\right]\)
\(\text { Now, } A+A^{\prime}=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]+\left[\begin{array}{rr} 3 & 1 \\ 5 & -1 \end{array}\right]=\left[\begin{array}{rr} 6 & 6 \\ 6 & -2 \end{array}\right]\)
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr} 6 & 6 \\ 6 & -2 \end{array}\right]=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]\)
\(\text { Now, } P^{\prime}=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]=P\)
\(\text { Thus, }P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]-\left[\begin{array}{cc} 3 & 1 \\ 5 & -1 \end{array}\right]=\left[\begin{array}{rr} 0 & 4 \\ -4 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rr} 0 & 4 \\ -4 & 0 \end{array}\right]=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]=-Q\)
Thus \(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{rr} 3 & 3 \\ 3 & -1 \end{array}\right]+\left[\begin{array}{rr} 0 & 2 \\ -2 & 0 \end{array}\right]=\left[\begin{array}{rr} 3 & 5 \\ 1 & -1 \end{array}\right]=A\)
(ii) \(\text { Let } A=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right], \text { then } A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]\)
\(\text { Now, } A+A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{rrr} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{array}\right]\)
\(\text { Now, } P^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=P\)
Thus,\(P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=-Q\)
Thus \(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]+\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]=\left[\begin{array}{rrr} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{array}\right]=A\)
(iii) Let \(A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]\), then \(A^{\prime}=\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]\)
Now, \(A+A=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]+\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]\)
\(P=\frac{1}{2}\left(A+A^{\prime}\right)=\frac{1}{2}\left[\begin{array}{ccc}6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{-5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]\)
Now, \(P^{\prime}=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=\left[\begin{array}{ccc}3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2\end{array}\right]=P\)
Thus, \(P=\frac{1}{2}\left(A+A^{\prime}\right)\) is a symmetric matrix. 
Now, \(A-A^{\prime}=\left[\begin{array}{ccc}3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2\end{array}\right]-\left[\begin{array}{ccc}3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2\end{array}\right]=\left[\begin{array}{ccc}0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0\end{array}\right]\)
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, \(Q=\left[\begin{array}{ccc}0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0\end{array}\right]=\left[\begin{array}{ccc}0 & -\frac{5}{2} & -\frac{3}{2} \\ \frac{5}{2} & 0 & -3 \\ \frac{3}{2} & 3 & 0\end{array}\right]=-Q\)
Thus, \(Q=\frac{1}{2}\left(A-A^{\prime}\right)\) is a skew -symmetric matrix.
Representing A as the sum of P and Q
\(P+Q=\left[\begin{array}{ccc} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -22 & ] \end{array}\right]+\left[\begin{array}{ccc} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{array}\right]=\left[\begin{array}{ccc} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{array}\right]=A\)
(iv)
\(\text { Let } A=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right] \text { , then } A^{\prime}=\left[\begin{array}{lr} 1 & -1 \\ 5 & 2 \end{array}\right]\)
\(\text { Now } A+A^{\prime}=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]+\left[\begin{array}{rr} 1 & -1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 4 & 4 \end{array}\right]\)
\(\text { Let } P=\frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ll} 1 & 2 \\ 2 & 2 \end{array}\right]\)
\(P=\frac{1}{2}\left(A+A^{\prime}\right) \text { is a symmetric matrix. }\)
\(\text { Now, } A-A^{\prime}=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]-\left[\begin{array}{rr} 1 & -1 \\ 5 & 2 \end{array}\right]=\left[\begin{array}{rr} 0 & 6 \\ -6 & 0 \end{array}\right]\)
\(\text { Let } Q=\frac{1}{2}\left(A-A^{\prime}\right)=\left[\begin{array}{rr} 0 & 3 \\ -3 & 0 \end{array}\right]\)
\(\text { Now, } Q^{\prime}=\left[\begin{array}{rr} 0 & -3 \\ 3 & 0 \end{array}\right]=-Q\)
\(Q=\frac{1}{2}\left(A-A^{\prime}\right) \text { is a skew-symmetric matrix. }\)
\(\text { Representing } A \text { as the sum of } P \text { and } Q \text { : }\)
\(P+Q=\left[\begin{array}{ll} 1 & 2 \\ 2 & 2 \end{array}\right]+\left[\begin{array}{rr} 0 & 3 \\ -3 & 0 \end{array}\right]=\left[\begin{array}{rr} 1 & 5 \\ -1 & 2 \end{array}\right]=A\)

\(Chem.\quad \quad \quad Phy.\quad \quad \quad Eco.\\ A=\left[ 10\times 12\quad \quad 8\times 12\quad \quad 10\times 12 \right] \)
\(Selling\ Price\ matrix,\ B=\left[ \begin{matrix} 80 \\ 60 \\ 40 \end{matrix} \right] .\)
\(\therefore \ Reqd.\ amount=AB\)
\(=\left[ 120\quad 96\quad 120 \right] \left[ \begin{matrix} 80 \\ 60 \\ 40 \end{matrix} \right] \)
\(=\left[ 120\times 80+96\times 60+120\times 40 \right] \)
\(=\left[ 9600+5760+4800 \right] =\left[ 20160 \right] .\)
Hence, the bookshop will receive Rs. 20, 160 by selling all the books.