We know that, the notation, namely $A=\left[a_{i j}\right]_{m \times n}$ indicates that $A$ is a matrix of order $m \times n$, also $1 \leq i \leq m, 1 \leq j \leq n ; i, j \in N$.

(i) Here, $A=\left[a_{i j}\right]_{2 \times 2}$

$$\Rightarrow \quad A=\frac{(i-2 j)^2}{2}, 1 \leq i \leq 2 ; 1 \leq j \leq 2\quad\text{..... (i)}$$

$$\begin{aligned} \therefore\quad& a_{11}=\frac{(1-2)^2}{2}=\frac{1}{2} \\ & a_{12}=\frac{(1-2 \times 2)^2}{2}=\frac{9}{2} \\ & a_{21}=\frac{(2-2 \times 1)^2}{2}=0 \\ & a_{22}=\frac{(2-2 \times 2)^2}{2}=2 \\ \text{Thus,}\quad & A=\left[\begin{array}{ll} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{array}\right]_{2 \times 2} \end{aligned}$$

(ii) Here, $A=\left[a_{\mathrm{ij}}\right]_{2 \times 2}=|-2 i+3 j|, 1 \leq i \leq 2 ; 1 \leq j \leq 2$

$$\begin{aligned} \therefore\quad a_{11} & =|-2 \times 1+3 \times 1|=1 \\ a_{12} & =|-2 \times 1+3 \times 2|=4 \quad[\because|-1|=1]\\ a_{21} & =|-2 \times 2+3 \times 1|=1 \\ a_{22} & =|-2 \times 2+3 \times 2|=2 \\ \therefore\quad A & =\left|\begin{array}{ll} 1 & 4 \\ 1 & 2 \end{array}\right|_{2 \times 2} \end{aligned}$$

$$\begin{aligned} & \text { Since, } A=\left[a_{i j}\right]_{m \times n} \quad 1 \leq i \leq m \text { and } 1 \leq j \leq n, i, j \in N \\ & \therefore \quad A=\left[e^{i \cdot x} \sin j x\right]_{3 \times 2} ; 1 \leq i \leq 3 ; 1 \leq j \leq 2 \\ & \Rightarrow \quad a_{11}=e^{1 \cdot x} \cdot \sin 1 \cdot x=e^x \sin x \\ & a_{12}=e^{1 \cdot x} \cdot \sin 2 \cdot x=e^x \sin 2 x \\ & a_{21}=e^{2 \cdot x} \cdot \sin 1 \cdot x=e^{2 x} \sin x \\ & a_{22}=e^{2 \cdot x} \cdot \sin 2 \cdot x=e^{2 x} \sin 2 x \\ & a_{31}=e^{3 \cdot x} \cdot \sin 1 \cdot x=e^{3 x} \sin x \\ & a_{32}=e^{3 \cdot x} \cdot \sin 2 \cdot x=e^{3 x} \sin 2 x \\ \therefore\quad& A=\left[\begin{array}{ll} e^x \sin x & e^x \sin 2 x \\ e^{2 x} \sin x & e^{2 x} \sin 2 x \\ e^{3 x} \sin x & e^{3 x} \sin 2 x \end{array}\right]_{3 \times 2} \end{aligned}$$

We have, $$A=\left[\begin{array}{cc} a+4 & 3 b \\ 8 & -6 \end{array}\right]_{2 \times 2} \text { and } B=\left[\begin{array}{cc} 2 a+2 & b^2+2 \\ 8 & b^2-5 b \end{array}\right]_{2 \times 2}$$

Also, $$A=B$$

By equality of matrices we know that each element of $A$ is equal to the corresponding element of $B$, that is $a_{i j}=b_{i j}$ for all $i$ and $j$. $$ \begin{array}{lll} \therefore & a_{11}=b_{11} \Rightarrow a+4=2 a+2 \Rightarrow a=2 & \\ & a_{12}=b_{12} \Rightarrow 3 b=b^2+2 \Rightarrow b^2=3 b-2 & \\ \text { and } & a_{22}=b_{22} \Rightarrow-6=b^2-5 b & \\ \Rightarrow & -6=3 b-2-5 b \quad \left[\because b^2=3 b-2\right]\\ \Rightarrow & 2 b=4 \Rightarrow b=2 & \\ \therefore & a =2 \text { and } b=2 \end{array}$$

We have, $$A=\left[\begin{array}{cc} \sqrt{3} & 1 \\ 2 & 3 \end{array}\right]_{2 \times 2} \text { and } B=\left[\begin{array}{lll} x & y & z \\ a & b & 6 \end{array}\right]_{2 \times 3}$$

Here, $A$ and $B$ are of different orders. Also, we know that the addition of two matrices $A$ and $B$ is possible only if order of both the matrices $A$ and $B$ should be same. Hence, the sum of matrices $A$ and $B$ is not possible.

We have, $X=\left[\begin{array}{ccc}3 & 1 & -1 \\ 5 & -2 & -3\end{array}\right]_{2 \times 3}$ and $Y=\left[\begin{array}{ccc}2 & 1 & -1 \\ 7 & 2 & 4\end{array}\right]_{2 \times 3}$

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

$$\begin{aligned} \text{(ii) }\quad & \because 2 X=2\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]=\left[\begin{array}{rrr} 6 & 2 & -2 \\ 10 & -4 & -6 \end{array}\right] \\ & \text { and } \\ & 3 Y=3\left[\begin{array}{ccc} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 6 & 3 & -3 \\ 21 & 6 & 12 \end{array}\right] \\ & \therefore \quad 2 X-3 Y=\left[\begin{array}{ccc} 6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12 \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -11 & -10 & -18 \end{array}\right] \end{aligned}$$

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

Also, $$X+Y+Z=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$$

We see that $Z$ is the additive inverse of $(X+Y)$ or negative of $(X+Y)$.

$$\therefore \quad Z=\left[\begin{array}{ccc} -5 & -2 & 2 \\ -12 & 0 & -1 \end{array}\right] \quad [\because Z=-(X+Y)]$$