Since, $A$ and $B$ are square matrices of same order and $B$ is a skew-symmetric matrix i.e., $B^{\prime}=-B$

$$ \begin{aligned} &\text { Now, we have to prove that } A^{\prime} B A \text { is a skew-symmetric matrix. }\\ &\begin{array}{rlrl} \therefore \quad A^{\prime} B A^{\prime} & =A^{\prime} B A^{\prime}=B A^{\prime} A^{\prime} \quad [\because AB'=B'A']\\ & =A^{\prime} B^{\prime} A=A^{\prime}-B A=-A^{\prime} B A \end{array} \end{aligned}$$

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

$$\begin{aligned} &\begin{aligned} \text { Let } \quad & P(n):(A B)^n=A^n B^n \\ \therefore\quad & P(1):(A B)^1=A^1 B^1 \Rightarrow A B=A B \end{aligned} \end{aligned}$$

So, $P(1)$ is true.

Now, $$P(k):(A B)^k=A^k B^k, k \in N$$

So, $P(K)$ is true, whenever $P(k+1)$ is true.

$$\therefore \quad P(K+1: A B)^{K+1}=A^{k+1} B^{k+1}\quad\text{.... (i)}$$

$$\begin{array}{ll} \Rightarrow & A B^k \cdot A B^1 \quad [\because AB=BA]\\ \Rightarrow & A^k B^k \cdot B A \Rightarrow A^k B^{k+1} A \\ \Rightarrow & A^k \cdot A \cdot B^{k+1} \Rightarrow A^{k+1} B^{k+1} \\ \Rightarrow & (A \cdot B)^{k+1}=A^{k+1} B^{k+1} \end{array}$$

So, $P(k+1)$ is true for all $n \in N$, whenever $P(k)$ is true.

By mathematical induction $(A B)=A^n B^n$ is true for all $n \in N$.

We have, $A=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]$ and $A^{\prime}=\left[\begin{array}{ccc}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right]$

By using elementary row transformations, we get

$$\mathrm{A}=I \mathrm{~A}$$

$$\begin{aligned} & \Rightarrow\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A \\ & \Rightarrow \quad\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ 0 & -2 y & 2 z \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right] A \quad\left[\because R_3 \rightarrow R_3-R_2\right] \end{aligned}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} 0 & 2 y & z \\ x & 3 y & 0 \\ 0 & 0 & 3 z \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & -1 & 1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_3 \rightarrow R_3+R_1 \\ \text { and } R_2 \rightarrow R_2+R_1 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} -x & -y & z \\ x & 3 y & 0 \\ 0 & 0 & z \end{array}\right]=\left[\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 1 & 0 \\ \frac{1}{3} & \frac{-1}{3} & \frac{1}{3} \end{array}\right] A} & {\left[\begin{array}{l} \because \rightarrow R_1 \rightarrow R_2 \\ \text { and } R_3 \rightarrow \frac{1}{3} R_3 \end{array}\right]} \end{array}$$

$\Rightarrow \quad\left[\begin{array}{ccc}-x & -y & 0 \\ x & 3 y & 0 \\ 0 & 0 & z\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{3} & \frac{-2}{3} & \frac{-1}{3} \\ 1 & 1 & 0 \\ \frac{1}{3} & \frac{-1}{3} & \frac{1}{3}\end{array}\right] A \quad\left[\because R_1 \rightarrow R_1-R_3\right]$

$\Rightarrow \quad\left[\begin{array}{ccc}-x & -y & 0 \\ 0 & 2 y & 0 \\ 0 & 0 & z\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{3} & \frac{-2}{3} & \frac{-1}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{3} & \frac{1}{3}\end{array}\right] A \quad\left[\because R_2 \rightarrow R_2+R_1\right]$

$\Rightarrow \quad\left[\begin{array}{ccc}-x & 0 & 0 \\ 0 & 2 y & 0 \\ 0 & 0 & z\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{-1}{2} & \frac{-1}{2} \\ \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{3} & \frac{1}{3}\end{array}\right] A \quad\left[\because R_1 \rightarrow R_1+\frac{1}{2} R_2\right]$

$\Rightarrow \quad\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{1}{2 x} & \frac{1}{2 x} \\ \frac{1}{3 y} & \frac{1}{6 y} & \frac{-1}{6 y} \\ \frac{1}{3 z} & \frac{-1}{3 z} & \frac{1}{3 z}\end{array}\right] A \quad\left[\begin{array}{l}\because R_1 \rightarrow \frac{-1}{x} R_1 \\ R_2 \rightarrow \frac{1}{2 y} R_2 \\ \text { and } R_3 \rightarrow \frac{1}{z} R_3\end{array}\right]$

$$\begin{array}{ll} \therefore & A^{-1}=\left[\begin{array}{ccc} 0 & \frac{1}{2 x} & \frac{1}{2 x} \\ \frac{1}{3 y} & \frac{1}{6 y} & \frac{-1}{6 y} \\ \frac{1}{3 z} & \frac{-1}{3 z} & \frac{1}{3 z} \end{array}\right]=\left[\begin{array}{ccc} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{array}\right] \\ \Rightarrow & \frac{1}{2 x}=x \Rightarrow= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow & \frac{1}{6 y}=y \Rightarrow y= \pm \frac{1}{\sqrt{6}} \\ \text { and } & \frac{1}{3 z}=z \Rightarrow z= \pm \frac{1}{\sqrt{3}} \end{array}$$

We have, $A=\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]$ and $A^{\prime}=\left[\begin{array}{ccc}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right]$

$$\begin{aligned} &\begin{array}{ll} \text { Also, } & A^{\prime}=A^{-1} \\ \Rightarrow & A A^{\prime}=A A^{-1} \quad \left[\because \mathrm{AA}^{-1}=I\right]\\ \Rightarrow & A A^{\prime}=I \end{array}\\ \end{aligned}$$

$\Rightarrow \quad\left[\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right]\left[\begin{array}{ccc}0 & x & x \\ 2 y & y & -y \\ z & -z & z\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}4 y^2+z^2 & 2 y^2-z^2 & -2 y^2+z^2 \\ 2 y^2-z^2 & x^2+y^2+z^2 & x^2-y^2-z^2 \\ -2 y^2+z^2 & x^2-y^2-z^2 & x^2+y^2+z^2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$$\begin{array}{lr} \Rightarrow & 2 y^2-z^2=0 \Rightarrow 2 y^2=z^2 \\ \Rightarrow & 4 y^2+z^2=1 \\ \Rightarrow & 2 \cdot z^2+z^2=1 \end{array}$$

$$\begin{aligned} z & = \pm \frac{1}{\sqrt{3}} \\ \therefore\quad y^2=\frac{z^2}{2} \Rightarrow y & = \pm \frac{1}{\sqrt{6}} \end{aligned}$$

$$\begin{aligned} &\text { Also, }\\ &x^2+y^2+z^2=1\\ &\begin{array}{lrl} \Rightarrow \quad x^2=1-y^2-z^2 =1-\frac{1}{6}-\frac{1}{3} \\ =1-\frac{3}{6}=\frac{1}{2} \\ \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}} \\ \therefore \quad x = \pm, \frac{1}{\sqrt{2}}, y= \pm \frac{1}{\sqrt{6}} \\ \text { and } \quad z = \pm \frac{1}{\sqrt{3}} \end{array} \end{aligned}$$

For getting the inverse of the given matrix $A$ by row elementary operations we may write the given matrix as

$$A=I A$$

(i) $\because\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

$$\begin{array}{lll} \Rightarrow \quad {\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} & {\left[\because R_2 \rightarrow R_2+R_1\right]} \\ \Rightarrow \quad\left[\begin{array}{ccc} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] A & {\left[\because R_3 \rightarrow R_3-R_2\right]} \end{array}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{array}\right] A} & {\left[\because R_1 \rightarrow R_1+R_2\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] A} & {\left[\because R_2 \rightarrow R_2-3 R_1\right]} \end{array}$$

$$\begin{aligned} & \Rightarrow \quad\left[\begin{array}{ccc} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{array}\right]=\left[\begin{array}{ccc} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{array}\right] A \quad\left[\because R_2 \rightarrow R_2-3 R_1\right] \\ & \Rightarrow \quad\left[\begin{array}{ccc} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{array}\right] A \quad\left[\begin{array}{l} \because R_1 \rightarrow R_1+R_2 \\ \text { and } R_3 \rightarrow-1 \cdot R_3 \end{array}\right] \end{aligned}$$

$$\begin{array}{llc} \Rightarrow & {\left[\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_1 \rightarrow R_1+10 R_3 \\ \text { and } R_2 \rightarrow R_2+17 R_3 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_1 \rightarrow-1 R_1 \\ \text { and } R_2 \rightarrow-1 R_2 \end{array}\right]} \end{array}$$

So, the inverse of $A$ is $\left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]$.

$$\begin{aligned} &\text { (ii) } \therefore\\ &\begin{aligned} & {\left[\begin{array}{ccc} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} \\ \Rightarrow\quad & {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right] A} \quad \left[\begin{array}{l} \because R_2 \rightarrow R_2+R_3 \\ \text { and } R_1 \rightarrow R_1-2 R_3 \end{array}\right]\\ \Rightarrow\quad & {\left[\begin{array}{ccc} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 2 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right] A}\quad \left[\because R_2 \rightarrow R_2+R_1\right] \end{aligned} \end{aligned}$$

Since, second row of the matrix $A$ on LHS is containing all zeroes, so we can say that inverse of matrix $A$ does not exist.

$$\begin{aligned} &\begin{aligned} \text { (iii) } \therefore \quad & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} \\ \Rightarrow\quad & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 3 & 1 & 1 \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} \quad \left[\because R_2 \rightarrow R_2-R_1\right] \end{aligned}\\ \end{aligned}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 2 & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_2 \rightarrow R_2-R_1 \\ \text { and } R_3 \rightarrow R_3+R_1 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 4 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 2 & 0 & 1 \end{array}\right] A} & {\left[\begin{array}{l} \because R_3 \rightarrow R_3+R_1 \\ \text { and } R_2 \rightarrow R_2-\frac{1}{2} R_1 \end{array}\right]} \end{array}$$

$$\begin{array}{lll} \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] A} & {\left[\because R_3 \rightarrow R_3-2 R_1\right]} \\ \Rightarrow & {\left[\begin{array}{ccc} 2 & 0 & -1 \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & \frac{1}{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ \frac{5}{2} & -1 & 1 \end{array}\right] A} & {\left[\because R_3 \rightarrow R_3-R_2\right]} \end{array}$$

$$ \begin{array}{lll} \Rightarrow & {\left[\begin{array}{lll} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ \frac{-5}{2} & 1 & 0 \\ 5 & -2 & 2 \end{array}\right] A} & {\left[\begin{array}{l} \because R_1 \rightarrow \frac{1}{2} R_1 \\ \text { and } R_3 \rightarrow 2 R_3 \end{array}\right]} \\ \Rightarrow & {\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right] A} \end{array} \quad\left[\begin{array}{l} \because \rightarrow R_1+\frac{1}{2} R_3 \\ \text { and } R_2 \rightarrow R_2-\frac{5}{2} R_3 \end{array}\right] .$$

Hence, $\left[\begin{array}{rrr}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ is the inverse of given matrix $A$.

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} & A=\left[\begin{array}{ccc} 2 & 3 & 1 \\ 1 & -1 & 2 \\ 4 & 1 & 2 \end{array}\right] \\ \therefore\quad & A^{\prime}=\left[\begin{array}{ccc} 2 & 1 & 4 \\ 3 & -1 & 1 \\ 1 & 2 & 2 \end{array}\right] \end{aligned} \end{aligned}$$

$$\begin{aligned} \text{Now,}\quad & \frac{A+A^{\prime}}{2}=\frac{1}{2}\left[\begin{array}{ccc} 4 & 4 & 5 \\ 4 & -2 & 3 \\ 5 & 3 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right] \\ \text{and}\quad & \frac{A-A^{\prime}}{2}=\frac{1}{2}\left[\begin{array}{ccc} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right] \end{aligned}$$

$$\begin{aligned} &\therefore \quad \frac{A+A^{\prime}}{2}+\frac{A-A^{\prime}}{2}=\left[\begin{array}{ccc} 2 & 2 & \frac{5}{2} \\ 2 & -1 & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} & 2 \end{array}\right]+\left[\begin{array}{ccc} 0 & 1 & \frac{-3}{2} \\ -1 & 0 & \frac{1}{2} \\ \frac{3}{2} & \frac{-1}{2} & 0 \end{array}\right]\\ &\text { which is the required expression. } \end{aligned}$$