# Solve this

Question:

If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then for any natural number, find the value of $\operatorname{Det}\left(A^{n}\right)$.

Solution:

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

Then, $A^{2}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

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

$=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$

Similarly, $A^{n}=\left[\begin{array}{cc}\cos (n \theta) & \sin (n \theta) \\ -\sin (n \theta) & \cos (n \theta)\end{array}\right]$

Therefore,

$\left|A^{n}\right|=\left|\begin{array}{cc}\cos (n \theta) & \sin (n \theta) \\ -\sin (n \theta) & \cos (n \theta)\end{array}\right|$

$=\cos ^{2}(n \theta)+\sin ^{2}(n \theta)$

$=1$

Hence, $\operatorname{Det}\left(A^{n}\right)=1$