i) If $A=\left(\begin{array}{l}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\end{array}\right)$ then for any integer $n\ge 1$ Show that $A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$

ii) If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ then show that for all the positive integers n, $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

using mathematical indication 

$\text{ Let }n=1\text{ then }{A}^1=\left[\begin{array}{cc}1+2\left(1\right)& -4\left(1\right)\\ 1& 1-2\left(1\right)\end{array}\right]=\left[\begin{array}{l}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\end{array}\right]=A$

$\therefore s\left(1\right)$ is true

Assume that s(k) is true $\text{ i.e., }{A}^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$

Now, he have to show that s$(k + 1)$is true $A^{k+1}=A^k\cdot A=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$=\left[\begin{array}{l}3(1+2k)-4k\left(1\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4(1+2k)-4k(-1)\\ k\left(3\right)+(1-2k)\left(1\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4\left(k\right)+(1-2k)(-1)\end{array}\right]$

$=\left[\begin{array}{l}3+6k-4k\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4-8k+4k\\ 3k+1-2k\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4k-1+2k\end{array}\right]$

$=\left[\begin{array}{cc}3+2k& -4-4k\\ 1+k& -1-2k\end{array}\right]=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ (k+1)& 1-2(k+1)\end{array}\right]$

$\therefore s(k+1)$ is true. The given statement is true $\mathrm{\forall }n\in N$

We solve this problem by using the principle of mathematical induction

Consider statement $S\left(n\right):A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

Since $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

$p\left(n\right)\text{ is true for }n=1$

Assume that S(k) is true for $k\in N$

then $A^k=\left[\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right]$

Now, we show that $s(k+1)$ is true

$A^{k+1}=A^k\times A=\left[\begin{array}{cc}\cos k\theta & \sin k\theta \\ -\sin k\theta & \cos k\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

$=\left[\begin{array}{cc}\cos k\theta \cos \theta -\sin k\theta \sin \theta & \cos k\theta \sin \theta +\sin k\theta \cos \theta \\ -\sin k\theta \cos \theta -\cos k\theta \sin \theta & -\sin k\theta \sin \theta +\cos k\theta \cos \theta \end{array}\right]$

$=\left[\begin{array}{cc}\cos (k\theta +\theta )& \sin (k\theta +\theta )\\ -\sin (k\theta +\theta )& \cos (k\theta +\theta )\end{array}\right]$

$=\left[\begin{array}{cc}\cos (k+1)\theta & \sin (k+1)\theta \\ -\sin (k+1)\theta & \cos (k+1)\theta \end{array}\right]$

Therefore S(n) is true for$n = k + 1$ Hence, by the principle of methematical induction$S\left(n\right)$ is true for all positive integeral values of n.