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Question
Find the inverse of the following matrix:
Solution
\[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\]
\[\left| A \right| = \cos^2 \theta + \sin^2 \theta = 1 \neq 0\]
A is a singular matrix; therefore, it is invertible .
\[\text{ Let }C_{ij}\text{ be a cofactor of }a_{ij}\text{ in A. }\]
Now,
\[ C_{11} = \cos\theta \]
\[ C_{12} = \sin\theta\]
\[ C_{21} = - \sin\theta\]
\[ C_{22} = \cos\theta\]
\[adjA = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}^T = \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\]
\[ \therefore A^{- 1} = \frac{1}{\left| A \right|}adjA = \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\]
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