Solve this

We shall prove the result by the principle of mathematical induction on n.

Step 1: If $n=1$, by definition of integral power of a matrix, we have

$A^{1}=\left[\begin{array}{cc}\cos 1 \alpha+\sin 1 \alpha & \sqrt{2} \sin 1 \alpha \\ -\sqrt{2} \sin 1 \alpha & \cos 1 \alpha-\sin 1 \alpha\end{array}\right]=\left[\begin{array}{cc}\cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \alpha & \cos \alpha-\sin \alpha\end{array}\right]=A$

So, the result is true for $n=1$.

Step 2: Let the result be true for $n=m$. Then,

$A^{m}=\left[\begin{array}{cc}\cos m \alpha+\sin m \alpha & \sqrt{2} \sin m \alpha \\ -\sqrt{2} \sin m \alpha & \cos m \alpha-\sin m \alpha\end{array}\right]$  ...(1)

Now we shall show that the result is true for n=m+1">n=m+1n=m+1.
Here,

$A^{m+1}=\left[\begin{array}{ccc}\cos (m+1) \alpha+\sin (m+1) \alpha & \sqrt{2} \sin (m+1) \alpha \\ -\sqrt{2} \sin (m+1) \alpha & \cos (m+1) \alpha-\sin (m+1) \alpha\end{array}\right]$

By definition of integral power of matrix, we have

$A^{m+1}=A^{m} \cdot A$

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}\cos m \alpha+\sin m \alpha & \sqrt{2} \sin m \alpha \\ -\sqrt{2} \sin m \alpha & \cos m \alpha-\sin m \alpha\end{array}\right]\left[\begin{array}{cc}\cos \alpha+\sin \alpha & \sqrt{2} \sin \alpha \\ -\sqrt{2} \sin \alpha & \cos \alpha-\sin \alpha\end{array}\right]$        $[$ From eq. $(1)]$

$\Rightarrow A^{m+1}=\left[\begin{array}{c}(\cos m \alpha+\sin m \alpha)(\cos \alpha+\sin \alpha)-\sqrt{2} \sin m \alpha(\sqrt{2} \sin \alpha) \quad(\cos m \alpha+\sin m \alpha)(\sqrt{2} \sin \alpha)+\sqrt{2} \sin m \alpha(\cos \alpha-\sin \alpha) \\ -\sqrt{2} \sin m \alpha(\cos \alpha+\sin \alpha)-(\cos m \alpha-\sin m \alpha)(\sqrt{2} \sin \alpha)-\sqrt{2} \sin m \alpha(\sqrt{2} \sin \alpha)+(\cos m \alpha-\sin m \alpha)(\cos \alpha-\sin \alpha)\end{array}\right]$

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}\cos m \alpha \cos \alpha+\sin m \alpha \cos \alpha+\cos m \alpha \sin \alpha+\sin m \alpha \sin \alpha-2 \sin m \alpha \sin \alpha & \sqrt{2} \sin \alpha \cos m \alpha+\sqrt{2} \sin \alpha \sin m \alpha+\sqrt{2} \sin m \alpha \cos \alpha-\sqrt{2} \sin m a \sin \alpha \\ -\sqrt{2} \sin m a \cos \alpha-\sqrt{2} \sin m a \sin \alpha-\sqrt{2} \sin \alpha \cos m \alpha+\sqrt{2} \sin \alpha \sin m \alpha & -2 \sin \alpha \sin m \alpha+\cos m \alpha \cos \alpha-\sin m \alpha \cos \alpha-\cos m \alpha \sin \alpha+\sin m \alpha \sin \alpha\end{array}\right]$

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}\cos (m \alpha-\alpha)+\sin (m \alpha+\alpha)-\cos (m \alpha-\alpha)+\cos (m \alpha+\alpha) & \sqrt{2} \sin (m \alpha+\alpha) \\ -\sqrt{2} \sin (m \alpha+\alpha) & \cos (m \alpha+\alpha)-\sin (m \alpha+\alpha)\end{array}\right]$

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}\cos (m \alpha+\alpha)+\sin (m \alpha+\alpha) & \sqrt{2} \sin (m \alpha+a) \\ -\sqrt{2} \sin (m \alpha+\alpha) & \cos (m \alpha+\alpha)-\sin (m \alpha+\alpha)\end{array}\right]$

$\Rightarrow A^{m+1}=\left[\begin{array}{cc}\cos (m+1) \alpha+\sin (m+1) \alpha & \sqrt{2} \sin (m+1) \alpha \\ -\sqrt{2} \sin (m+1) \alpha & \cos (m+1) \alpha-\sin (m+1) \alpha\end{array}\right]$

This show that when the result is true for $n=m$, it is also true for $n=m+1$.

Hence, by the principle of mathematical induction, the result is valid for all $\mathrm{n} \in N$.