Question

Find the inverse of the following matrix.

\[\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & - \cos \alpha\end{bmatrix}\]

Answer 1

\[G = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & - \cos\alpha\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & - \cos\alpha\end{vmatrix} = - 1, C_{12} = - \begin{vmatrix}0 & \sin\alpha \\ 0 & - \cos\alpha\end{vmatrix} = 0\text{ and }C_{13} = \begin{vmatrix}0 & \cos\alpha \\ 0 & \sin\alpha\end{vmatrix} = 0\]
\[ C_{21} = - \begin{vmatrix}0 & 0 \\ \sin\alpha & - \cos\alpha\end{vmatrix} = 0, C_{22} = \begin{vmatrix}1 & 0 \\ 0 & - \cos\alpha\end{vmatrix} = - \cos\alpha\text{ and }C_{23} = - \begin{vmatrix}1 & 0 \\ 0 & \sin\alpha\end{vmatrix} = - \sin\alpha\]
\[ C_{31} = \begin{vmatrix}0 & 0 \\ \cos\alpha & \sin\alpha\end{vmatrix} = 0, C_{32} = - \begin{vmatrix}1 & 0 \\ 0 & \sin\alpha\end{vmatrix} = - \sin\alpha\text{ and }C_{33} = \begin{vmatrix}1 & 0 \\ 0 & \cos\alpha\end{vmatrix} = \cos\alpha\]
\[adjF = \begin{bmatrix}- 1 & 0 & 0 \\ 0 & - \cos\alpha & - \sin\alpha \\ 0 & - \sin\alpha & \cos\alpha\end{bmatrix}^T = \begin{bmatrix}- 1 & 0 & 0 \\ 0 & - \cos\alpha & - \sin\alpha \\ 0 & - \sin\alpha & \cos\alpha\end{bmatrix}\]
\[and \left| F \right| = - 1\]
\[ \therefore F^{- 1} = - 1\begin{bmatrix}- 1 & 0 & 0 \\ 0 & - \cos\alpha & - \sin\alpha \\ 0 & - \sin\alpha & \cos\alpha\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & - \cos\alpha\end{bmatrix}\]