Question

Find the inverse of the matrix \[\begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{bmatrix}\]

Answer 1

\[A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}\]

\[ \therefore \left| A \right| = \cos^2 \theta + \sin^2 \theta = 1 \neq 0\]

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, 

\[adj A = \begin{bmatrix}\cos\theta & \sin\theta \\ - \sin\theta & \cos\theta\end{bmatrix}^T = \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\]

\[ \therefore A^{- 1} = \frac{1}{\left| A \right|}adj A = \begin{bmatrix}\cos\theta & - \sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}\]