$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} A & =\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right], B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \text { and } x^2=-1 \\ \therefore\quad (A+B) & =\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { and } \quad (A+B)^2 & =\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x+1 \\ x+1 & 0 \end{array}\right] \\ & =\left[\begin{array}{cc} 1-x^2 & 0 \\ 0 & 1-x^2 \end{array}\right]\quad\text{.... (i)} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Also, }\quad & A^2=A \cdot A=\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -x \\ x & 0 \end{array}\right]=\left[\begin{array}{cc} -x^2 & 0 \\ 0 & -x^2 \end{array}\right] \\ \text{and}\quad & B^2=B \cdot B=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned} \end{aligned}$$

$$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} A^2+B^2 & =\left[\begin{array}{cc} -x^2+1 & 0 \\ 0 & -x^2+1 \end{array}\right]=\left[\begin{array}{cc} 1-x^2 & 0 \\ 0 & 1-x^2 \end{array}\right] \quad\text{using Eq. (i)]}\\ & =(A+B)^2\quad\text{Hence proved.} \end{aligned} \end{aligned}$$

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

$\therefore\quad A^2=\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right] \cdot\left[\begin{array}{ccc}0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4\end{array}\right] \quad\left[\because A^2=A \cdot A\right]$

$$=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I$$ Hence proved.

Let $$P(n):\left(A^n\right)^n=\left(A^n\right)^{\prime}$$

$\therefore \quad P(1):(A)^1=(A)^{\prime}$

$\Rightarrow \quad A^{\prime}=A^{\prime} \Rightarrow P(1)$ is true.

Now, $$\quad P(k):(A)^k=\left(A^k\right)^{\prime}$$

where $k \in N$

and $$\quad P(k+1):\left(A^{\prime}\right)^{k+1}=\left(A^{k+1}\right)^{\prime}$$

$$\begin{aligned} &\text { where } P(k+1) \text { is true whenever } P(k) \text { is true. }\\ &\begin{aligned} \therefore \quad P(k+1):\left(A^{\prime}\right)^k \cdot\left(A^{\prime}\right)^{\prime} & =\left[A^{k+1}\right]^{\prime} \\ \left(A^k\right)^{\prime} \cdot(A)^{\prime} & =\left[A^{k+1}\right]^{\prime} \\ \left(A \cdot A^k\right)^{\prime} & =\left[A^{k+1}\right]^{\prime} \quad \left[\because(A)^{\prime k}=\left(A^k\right)^{\prime} \text { and }(A B)=B^{\prime} A^{\prime}\right]\\ \left(A^{k+1}\right)^{\prime} & =\left[A^{k+1}\right]^{\prime}\text{Hence proved.} \end{aligned} \end{aligned}$$

(i) Let $A=\left[\begin{array}{cc}1 & 3 \\ -5 & 7\end{array}\right]$

In order to use elementary row operations we may write $A=I A$.

$$\begin{array}{ll} \therefore\quad{\left[\begin{array}{cc} 1 & 3 \\ -5 & 7 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A} \\ \Rightarrow\quad{\left[\begin{array}{cc} 1 & 3 \\ 0 & 22 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 5 & 1 \end{array}\right] A} & {\left[\because R_2 \rightarrow R_2+5 R_1\right]} \\ \Rightarrow\quad{\left[\begin{array}{ll} 1 & 3 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 5 / 22 & 1 / 22 \end{array}\right] A} & {\left[\because R_2 \rightarrow \frac{1}{22} R_2\right]} \\ \Rightarrow\quad{\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 7 / 22 & -3 / 22 \\ 5 / 22 & 1 / 22 \end{array}\right] A} & {\left[\because R_1 \rightarrow R_1-3 R_2\right]} \\ \Rightarrow\quad{\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\frac{1}{22}\left[\begin{array}{cc} 7 & -3 \\ 5 & 1 \end{array}\right] A} \end{array}$$

$$\begin{aligned} &\Rightarrow I=B A \text {, where } B \text { is the inverse of } A \text {. }\\ & \therefore\quad B=\frac{1}{22}\left[\begin{array}{ll} 7 & -3 \\ 5 & -1 \end{array}\right] \end{aligned}$$

(ii) Let $A=\left[\begin{array}{cc}1 & -3 \\ -2 & 6\end{array}\right]$

In order to use elementary row operations, we write $\mathrm{A}=I \mathrm{~A}$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{cc} 1 & -3 \\ -2 & 6 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A} & \\ \Rightarrow & {\left[\begin{array}{cc} 1 & -3 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] A} & {\left[\because R_2 \rightarrow R_2+2 R_1\right]} \end{array}$$

Since, we obtain all zeroes in a row of the matrix $A$ on LHS, so $A^{-1}$ does not exist.

We have,

$$\left[\begin{array}{cc} x y & 4 \\ z+6 & x+y \end{array}\right]=\left[\begin{array}{cc} 8 & w \\ 0 & 6 \end{array}\right]$$

$$ \begin{aligned} & \text { By equality of matrix, } \\ & \Rightarrow \quad x=6-y \text { and }(6-y) \cdot y=8 \\ & \Rightarrow \quad y^2-6 y+8=0 \\ & \Rightarrow \quad y^2-4 y-2 y+8=0 \\ & \Rightarrow \quad(y-2)(y-4)=0 \\ & \Rightarrow \quad y=2 \text { or } y=4 \\ & \therefore \quad x=6-2=4 \\ & \text { or } \\ & x=6-4=2 \quad [\because x=6-y]\\ & \text { Also, } \\ & z+6=0 \\ & \Rightarrow \quad z=-6 \text { and } w=4 \\ & \therefore \quad x=2, y=4 \text { or } x=4, y=2, z=-6 \text { and } w=4 \end{aligned}$$