If $A=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right]$, then prove by principle of mathematical induction that
$A^{n}=\left[\begin{array}{cc}\cos n \theta & i \sin \theta \\ i \sin n \theta & \cos n \theta\end{array}\right]$ for all $n \in N$
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 $A^{n}$.