Find the inverse of \(\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\) by elementary row transformations.

Answer: \(A^{-1} = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

\(|A| = \cos^2\theta + \sin^2\theta = 1 \ne 0\), so \(A^{-1}\) exists. Write \(AA^{-1} = I\), i.e. apply row transformations to \(A\) and \(I\) together: \([A \mid I]\). (Assume \(\cos\theta \ne 0\).)

\(\left[\begin{array}{ccc|ccc} \cos\theta & -\sin\theta & 0 & 1 & 0 & 0 \\ \sin\theta & \cos\theta & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]\)

\(R_1 \to \dfrac{1}{\cos\theta}R_1\): \(\left[\begin{array}{ccc|ccc} 1 & -\tan\theta & 0 & \sec\theta & 0 & 0 \\ \sin\theta & \cos\theta & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]\)

\(R_2 \to R_2 - \sin\theta\,R_1\): the (2,2) entry becomes \(\cos\theta + \sin\theta\tan\theta = \dfrac{\cos^2\theta + \sin^2\theta}{\cos\theta} = \sec\theta\) and the right side of row 2 becomes \(-\sin\theta\sec\theta = -\tan\theta,\ 1,\ 0\):

\(\left[\begin{array}{ccc|ccc} 1 & -\tan\theta & 0 & \sec\theta & 0 & 0 \\ 0 & \sec\theta & 0 & -\tan\theta & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]\)

\(R_2 \to \cos\theta\,R_2\): \(\left[\begin{array}{ccc|ccc} 1 & -\tan\theta & 0 & \sec\theta & 0 & 0 \\ 0 & 1 & 0 & -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]\)

\(R_1 \to R_1 + \tan\theta\,R_2\): the right side of row 1 becomes \(\sec\theta - \tan\theta\sin\theta = \dfrac{1 - \sin^2\theta}{\cos\theta} = \cos\theta\), and \(\tan\theta\cos\theta = \sin\theta\):

\(\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \cos\theta & \sin\theta & 0 \\ 0 & 1 & 0 & -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right]\)

Hence \(A^{-1} = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\). (Check: \(AA^{-1} = I\).)