$$ \begin{aligned} & P Q=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right] \quad\text{.... (i)}\\ & \text { and } Q P=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]\left[\begin{array}{ccc} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right]=\left[\begin{array}{ccc} a x & 0 & 0 \\ 0 & b y & 0 \\ 0 & 0 & z c \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$

Thus, we see that, $P Q=Q P$ [usings Eq. (i) and (ii)] Hence proved.

We have, $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=A$ $$ \begin{aligned} \therefore \quad\left[\begin{array}{lll} 2 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] & =\left[\begin{array}{lll} -2-1+0 & 0+1+3 & -2+0+3 \end{array}\right] \\ & =\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right] \end{aligned}$$

Now,

$$\begin{aligned} {\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] } & =A \\ \therefore\quad A & =\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \\ & =[-3+0-1]=[-4] \end{aligned}$$

We have to verify that, $A(B+C)=A B+A C$

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

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

$$\begin{aligned} \text { Also, } \quad A B & =\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right] \\ & =\left[\begin{array}{lll} 10+8 & 6+7 & 8+6 \end{array}\right]=\left[\begin{array}{lll} 18 & 13 & 14 \end{array}\right] \end{aligned}$$

$$\begin{aligned} \text{and}\quad A C & =\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right] \\ & =\left[\begin{array}{lll} -2+1 & 4+0 & 2+2 \end{array}\right]=\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right] \\ \therefore\quad A B+A C & =\left[\begin{array}{lll} 18 & 13 & 14 \end{array}\right]+\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right] \\ & =\left[\begin{array}{lll} 17 & 17 & 18 \end{array}\right] \quad\text{.... (ii)}\\ \therefore\quad A(B+C) & =(A B+A C)\quad\text{[using Eqs. (i) and (ii)]} \end{aligned}$$

Hence proved.

$$\begin{aligned} &\text { We have, }\\ &A=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\\ &\therefore \quad A^2=A \cdot A \end{aligned}$$

$$ \begin{aligned} & =\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{array}\right] \\ \therefore \quad A^2+A & =\left[\begin{array}{ccc} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{array}\right]+\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right] \\ & =\left[\begin{array}{ccc} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{array}\right]\quad\text{.... (i)} \end{aligned}$$

Now, and $$ \begin{aligned} A+I & =\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]+\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{array}\right] \\ \text{and}\quad A(A+I) & =\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right] \cdot\left[\begin{array}{ccc} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$

Thus, we see that $A^2+A=A(A+I)$ [using Eqs. (i) and (ii)]

We have, $A=\left[\begin{array}{ccc}0 & -1 & 2 \\ 4 & 3 & -4\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 0 \\ 1 & 3 \\ 2 & 6\end{array}\right]$

(i) We have to verify that, $A^{\prime}=A$

$$\begin{aligned} \therefore\quad & A^{\prime}=\left[\begin{array}{cc} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right] \\ \text{and}\quad & A^{\prime}=\left[\begin{array}{ccc} 0 & -1 & 2 \\ 4 & 3 & -4 \end{array}\right]=A\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\text { (ii) We have to verify that, } A B^{\prime}=B^{\prime} A^{\prime}\\ &\begin{array}{ll} \therefore & A B=\left[\begin{array}{cc} 3 & 9 \\ 11 & -15 \end{array}\right] \\ \Rightarrow & (A B)^{\prime}=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \end{array} \end{aligned}$$

$$\begin{array}{rlr} \text{and}\quad B^{\prime} A^{\prime} & =\left[\begin{array}{lll} 4 & 1 & 2 \\ 0 & 3 & 6 \end{array}\right]\left[\begin{array}{rr} 0 & 4 \\ -1 & 3 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{cc} 3 & 11 \\ 9 & -15 \end{array}\right] \\ & =(A B)^{\prime} \quad \text { Hence proved. } \end{array}$$

(iii) We have to verify that, $(k A)^{\prime}=\left(k A^{\prime}\right)$

$$\begin{aligned} \text{Now,}\quad (k A) & =\left[\begin{array}{ccc} 0 & -k & 2 k \\ 4 k & 3 k & -4 k \end{array}\right] \\ \text{and}\quad (k A)^{\prime} & =\left[\begin{array}{cc} 0 & 4 k \\ -k & 3 k \\ 2 k & -4 k \end{array}\right] \\ \text{Also,}\quad k A^{\prime} & =\left[\begin{array}{cc} 0 & 4 k \\ -k & 3 k \\ 2 k & -4 k \end{array}\right] \\ & =(k A)^{\prime}\quad\text{Hence proved.} \end{aligned}$$