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 Det(Aⁿ).

Answer 1

$$\begin{gathered} \mathrm{Let}A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]. \\ \mathrm{Then},A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ =\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & \cos \theta \sin \theta +\sin \theta \cos \theta \\ -\sin \theta \cos \theta -\cos \theta \sin \theta & -{\sin }^2\theta +{\cos }^2\theta \end{array}\right] \\ =\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right] \\ \mathrm{Similarly},A^n=\left[\begin{array}{cc}\cos \left(n\theta \right)& \sin \left(n\theta \right)\\ -\sin \left(n\theta \right)& \cos \left(n\theta \right)\end{array}\right] \\ \mathrm{Therefore}, \\ \left|A^n\right|=\left|\begin{array}{cc}\cos \left(n\theta \right)& \sin \left(n\theta \right)\\ -\sin \left(n\theta \right)& \cos \left(n\theta \right)\end{array}\right| \\ ={\cos }^2\left(n\theta \right)+{\sin }^2\left(n\theta \right) \\ =1 \\ \mathrm{Hence},\mathrm{Det}\left(A^n\right)=1. \end{gathered}$$