Question

Find the inverse of the following matrix:

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

Answer 1

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