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प्रश्न
उत्तर
\[Given: A = \begin{bmatrix}\sin\alpha & \cos\alpha \\ - \cos\alpha & \sin\alpha\end{bmatrix} \]
\[ A^T = \begin{bmatrix}\sin\alpha & - \cos\alpha \\ \cos\alpha & \sin\alpha\end{bmatrix}\]
\[Now, \]
\[ A^T A = \begin{bmatrix}\sin\alpha & - \cos\alpha \\ \cos\alpha & \sin\alpha\end{bmatrix}\begin{bmatrix}\sin\alpha & \cos\alpha \\ - \cos\alpha & \sin\alpha\end{bmatrix} \]
\[ \Rightarrow A^T A = \begin{bmatrix}\left( \sin\alpha \right)\left( \sin\alpha \right) + \left( - \cos\alpha \right)\left( - \cos\alpha \right) & \left( \sin\alpha \right)\left( \cos\alpha \right) + \left( - \cos\alpha \right)\left( \sin\alpha \right) \\ \left( \cos\alpha \right)\left( \sin\alpha \right) + \left( \sin\alpha \right)\left( - \cos\alpha \right) & \left( \cos\alpha \right)\left( \cos\alpha \right) + \left( \sin\alpha \right)\left( \sin\alpha \right)\end{bmatrix}\]
\[ \Rightarrow A^T A = \begin{bmatrix}\sin^2 \alpha + \cos^2 \alpha & \sin\alpha \cos\alpha - \sin\alpha \cos\alpha \\ \sin\alpha \cos\alpha - \sin\alpha \cos\alpha & \cos^2 \alpha + \sin^2 \alpha\end{bmatrix}\]
\[ \Rightarrow A^T A = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = I\]
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