Use the classical adjoint formula to compute the inverses of each of the following \(3 \times 3\) real matrices. \[ \begin{bmatrix} -2 & 3 & 2 \\ 6 & 0 & 3 \\ 4 & 1 & -1 \end{bmatrix}, \qquad \begin{bmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{bmatrix} \]
Solution
For \[ A=\begin{bmatrix}-2&3&2\\6&0&3\\4&1&-1\end{bmatrix} \] one computes \(\det A=72\) and the cofactor matrix \[ C=\begin{bmatrix}-3&18&6\\5&-6&14\\9&18&-18\end{bmatrix}. \] Hence \[ A^{-1}=\frac1{72}C^t =\begin{bmatrix}-1/24&5/72&1/8\\1/4&-1/12&1/4\\1/12&7/36&-1/4\end{bmatrix}. \]
For \[ R=\begin{bmatrix}\cos\theta&0&-\sin\theta\\0&1&0\\\sin\theta&0&\cos\theta\end{bmatrix}, \] \(\det R=\cos^2\theta+\sin^2\theta=1\). Its classical adjoint is \[ \operatorname{adj}R= \begin{bmatrix}\cos\theta&0&\sin\theta\\0&1&0\\-\sin\theta&0&\cos\theta\end{bmatrix}, \] so \[ R^{-1}=\operatorname{adj}R=R^t. \]