If $A=\left[\begin{array}{rr}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$, find $A^{2}$
Given : $A=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
Now,
$A^{2}=A A$
$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos ^{2}(2 \theta)-\sin ^{2}(2 \theta) & \cos (2 \theta) \sin 2 \theta+\cos (2 \theta) \sin 2 \theta \\ -\cos (2 \theta) \sin 2 \theta-\sin 2 \theta \cos 2 \theta & -\sin ^{2}(2 \theta)+\cos ^{2}(2 \theta)\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos (2 \times 2 \theta) & 2 \sin 2 \theta \cos 2 \theta \\ -2 \sin 2 \theta \cos (2 \theta) & \cos (2 \times 2 \theta)\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos 4 \theta & \sin (2 \times 2 \theta) \\ -\sin (2 \times 2 \theta) & \cos 4 \theta\end{array}\right]\left[\begin{array}{c}\cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta\end{array}\right]$ $\left[\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}\cos 4 \theta & \sin 4 \theta \\ -\sin 4 \theta & \cos 4 \theta\end{array}\right]$
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