If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then $A^2=$
Answer & explanation
Correct answer: option 3
$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \Rightarrow A^2=A \times A= \left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
So $A^2=\left[\begin{array}{cc}\cos ^2 \theta-\sin ^2 \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos ^2 \theta-\sin ^2 \theta\end{array}\right]$
since,
$\cos ^2 \theta - \sin ^2 \theta=\cos 2 \theta$
$2 \sin 0 \cos \theta=\sin 2 \theta$
$A^2=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$