$$\begin{aligned} &\text { We have, }\\ &\begin{array}{ll} & P(x)=\left[\begin{array}{lll} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right] \\ \therefore & P(y)=\left[\begin{array}{ll} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \end{array} \end{aligned}$$

$$\begin{aligned} \text { Now, } \quad P(x) \cdot P(y) & =\left[\begin{array}{ll} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]\left[\begin{array}{cc} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \\ & =\left[\begin{array}{ll} \cos x \cdot \cos y-\sin x \cdot \sin y & \cos x \cdot \sin y+\sin x \cdot \cos y \\ -\sin x \cdot \cos y-\cos x \cdot \sin y & -\sin x \cdot \sin y+\cos x \cdot \cos y \end{array}\right]\quad\text{.... (i)} \end{aligned}$$

$$\begin{aligned} & \quad\left[\begin{array}{l} \because \cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y \\ \text { and } \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y \end{array}\right] \\ & \text { and } \quad P(x+y)=\left[\begin{array}{lr} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$

Also, $\quad P(y) \cdot P(x)=\left[\begin{array}{ll}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]\left[\begin{array}{lc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$

$$\begin{aligned} & =\left[\begin{array}{lr} \cos y \cdot \cos x-\sin y \cdot \sin x & \cos y \cdot \sin x+\sin y \cdot \cos x \\ -\sin y \cdot \cos x-\sin x \cdot \cos y & -\sin y \cdot \sin x+\cos y \cdot \cos x \end{array}\right] \\ & =\left[\begin{array}{ll} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{array}\right]\quad\text{.... (iii)} \end{aligned}$$

Thus, we see from the Eqs. (i), (ii) and (iii) that,

$$P(x) \cdot P(y)=P(x+y)=P(y) \cdot P(x)$$ Hence proved.

$$\begin{aligned} &\begin{aligned} \text { Since, } A^2=A \text { and }(I+A) \cdot(I+A)=I^2 & +I A+A I+A^2 \\ & =I^2+2 A I+A^2 \\ & =I+2 A+A=I+3 A \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { and } \quad (I+A) \cdot(I+A)(I+A) & =(I+A)(I+3 A) \\ & =I^2+3 A I+A I+3 A^2 \\ & =I+4 A I+3 A \\ & =I+7 A=7 A+I\quad\text{Hence proved.} \end{aligned} \end{aligned}$$

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

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

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

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

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

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

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