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Question
If \[A = \begin{bmatrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\text{ and }A \left( adj A = \right)\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\], then find the value of k.
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Solution
\[A = \begin{bmatrix} \cos \theta & \sin \theta\\ - \sin \theta & \cos \theta \end{bmatrix}\]
\[ \therefore \left| A \right| = \begin{vmatrix} \cos \theta & \sin \theta\\ - \sin \theta & \cos \theta \end{vmatrix} = \cos^2 \theta + \sin^2 \theta = 1 \neq 0\]
\[\text{ Thus, }A^{- 1}\text{ exists . }\]
Now,
\[ A^{- 1} = \frac{adj A}{\left| A \right|} = \begin{bmatrix}[ \cos \theta & - \sin \theta\\\sin \theta & \cos \theta \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = adj A\]
\[ \Rightarrow A A^{- 1} = A adj A \]
\[ \Rightarrow A A^{- 1} = \begin{bmatrix} k & 0\\0 & k \end{bmatrix} \]
\[ \Rightarrow \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} = \begin{bmatrix} k & 0\\0 & k \end{bmatrix} [ \because A A^{- 1} = I] \]
\[ \Rightarrow k = 1\]
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