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प्रश्न
If \[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\] , then for any natural number, find the value of Det(An).
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उत्तर
\[\text{ Let }A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix} . \]
\[\text{ Then, } A^2 = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\]
\[ = \begin{bmatrix}\cos^2 \theta - \sin^2 \theta & \cos\theta\sin\theta + \sin\theta\cos\theta \\ - \sin\theta\cos\theta - \cos\theta\sin\theta & - \sin^2 \theta + \cos^2 \theta\end{bmatrix}\]
\[ = \begin{bmatrix}\cos2\theta & \sin2\theta \\ - \sin2\theta & \cos2\theta\end{bmatrix}\]
\[\text{ Similarly, }A^n = \begin{bmatrix}\cos\left( n\theta \right) & \sin\left( n\theta \right) \\ - \sin\left( n\theta \right) & \cos\left( n\theta \right)\end{bmatrix}\]
Therefore,
\[\left| A^n \right| = \begin{vmatrix}\cos\left( n\theta \right) & \sin\left( n\theta \right) \\ - \sin\left( n\theta \right) & \cos\left( n\theta \right)\end{vmatrix}\]
\[ = \cos^2 \left( n\theta \right) + \sin^2 \left( n\theta \right)\]
\[ = 1\]
\[\text{ Hence, Det}( A^n ) = 1 .\]
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