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If $\mathrm{A}=\left[\begin{array}{ll}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$, then what is $\mathbf{A}^{3}$ equal to?
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The correct answer is:
$\left[\begin{array}{ll}\cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta\end{array}\right]$
$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
We know, $A^{n}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$
$\therefore A^{3}=\left[\begin{array}{cc}\cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta\end{array}\right]$
We know, $A^{n}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$
$\therefore A^{3}=\left[\begin{array}{cc}\cos 3 \theta & \sin 3 \theta \\ -\sin 3 \theta & \cos 3 \theta\end{array}\right]$
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