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प्रश्न
Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]
उत्तर
\[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\]
\[ \therefore \left| A \right| = \cos^2 \theta + \sin^2 \theta = 1 \neq 0\]
A is a singular matrix . Therefore, it is invertible .
\[\text{ Let }C_{ij}\text{ be a cofactor of }a_{ij}\text{ in A .} \]
The cofactors of element A are given by
\[ C_{11} = \cos\theta \]
\[ C_{12} = \sin\theta\]
\[ C_{21} = - \sin\theta\]
\[ C_{22} = \cos\theta\]
Now,
\[adj A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}^T = \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\]
\[ \therefore A^{- 1} = \frac{1}{\left| A \right|}adj A = \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\]
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