Question

If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ then prove that $A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$, $n\in N$.

Answer 1

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

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

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

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

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

$\therefore A^3=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

$\therefore A^3=\left[\begin{array}{cc}\cos 2\theta \cos \theta -\sin 2\theta \sin \theta & \sin \theta \cos 2\theta +\sin 2\theta \cos \theta \\ -\sin 2\theta \cos \theta -\sin \theta \cos 2\theta & -\sin \theta \sin \theta +\cos 2\theta \cos \theta \end{array}\right]$

$\therefore A^3=\left[\begin{array}{cc}\cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{array}\right]$

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

$n\in R$.