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

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

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

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

$A^{m}=\left[\begin{array}{cc}\cos m \theta & i \sin m \theta \\ i \sin m \theta & \cos m \theta\end{array}\right]$

$A^{m}=\left[\begin{array}{cc}\cos m \theta & i \sin m \theta \\ i \sin m \theta & \cos m \theta\end{array}\right]$

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

$A^{m+1}=\left[\begin{array}{ccc}\cos (m+1) \theta & i \sin (m+1) \theta \\ i \sin (m+1) \theta & \cos (m+1) \theta\end{array}\right]$              ...(1)

By definition of integral power of matrix, we have

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

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

$\Rightarrow A^{m+1}=\left[\begin{array}{ccc}\cos m \theta \cdot \cos \theta+i \sin m \theta \cdot i \sin \theta & \cos m \theta \cdot i \sin \theta+i \sin m \theta \cdot \cos \theta \\ i \sin m \theta \cdot \cos \theta+\cos m \theta \cdot i \sin \theta & i \sin m \theta \cdot i \sin \theta+\cos m \theta \cdot \cos \theta\end{array}\right]$

$\Rightarrow A^{m+1}=\left[\begin{array}{ccc}\cos m \theta \cdot \cos \theta-\sin m \theta \cdot \sin \theta & i(\cos m \theta \cdot \sin \theta+\sin m \theta \cdot \cos \theta) \\ i(\sin m \theta \cdot \cos \theta+\cos m \theta \sin \theta) & -\sin m \theta \cdot \sin \theta+\cos m \theta \cdot \cos \theta\end{array}\right]$

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

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

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

This shows that when the result is true for $n=m$, it is true for $n=m+1$. Hence, by the principle of mathematical induction, the result is valid for all $\mathrm{n} \in N$.

Disclaimer: $n$ is missing before $\theta$ in $a_{12}$ in $A^{n}$.