If `A=[(cosalpha+sinalpha,sqrt(2)sinalpha),(-sqrt(2)sinalpha,cosalpha-sinalpha)]` , prove that `A^n=[(cosn\ alpha+sinn\ alpha,sqrt(2)sinn\ alpha),(-sqrt(2)sinn\ alpha,cosn\ alpha-sinn\ alpha)]` for all `n in Ndot`

Explanation

To prove that \( A^n = \begin{pmatrix} \cos(n\alpha) + \sin(n\alpha) & \sqrt{2}\sin(n\alpha) \\ -\sqrt{2}\sin(n\alpha) & \cos(n\alpha) - \sin(n\alpha) \end{pmatrix} \) for all \( n \in \mathbb{N} \), we will use the principle of mathematical induction.

### Step 1: Base Case (n = 1)

We first check the base case when \( n = 1 \).

\[
A^1 = A = \begin{pmatrix} \cos(\alpha) + \sin(\alpha) & \sqrt{2}\sin(\alpha) \\ -\sqrt{2}\sin(\alpha) & \cos(\alpha) - \sin(\alpha) \end{pmatrix}
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