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If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$, then the matrix $\mathrm{A}^{-50}$
when $\theta=\frac{\pi}{12}$, is equal to :
Correct Option: 1
Here, $\mathrm{AA}^{\mathrm{T}}=\mathrm{I}$
$\Rightarrow A^{-1}=A^{\mathrm{T}}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$
Also, $\mathrm{A}^{-\mathrm{n}}=\left[\begin{array}{cc}\cos (\mathrm{n} \theta) & \sin (\mathrm{n} \theta) \\ -\sin (\mathrm{n} \theta) & \cos (\mathrm{n} \theta)\end{array}\right]$
$\therefore \mathrm{A}^{-50}=\left[\begin{array}{cc}\cos (50) \theta & \sin (50) \theta \\ -\sin (50) \theta & \cos (50) \theta\end{array}\right]$
$=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]$
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