Question

Find the inverse of the matrix \[ \begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \] by elementary row transformations.

Hint:

When finding the inverse using elementary row operations, always perform the same operations on the identity matrix.

Answer 1

We are given the matrix \( A = \begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \) and need to find its inverse using elementary row transformations. The augmented matrix for this operation is: \[ \left( \begin{array}{ccc|ccc} \cos \theta & -\sin \theta & 0 & 1 & 0 & 0 \\ \sin \theta & \cos \theta & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \]

Step 1: Make the first element in the first column equal to 1. \\ We divide the first row by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)): \[ R_1 \to \frac{1}{\cos \theta} R_1 \] The augmented matrix becomes: \[ \left( \begin{array}{ccc|ccc} 1 & -\tan \theta & 0 & \sec \theta & 0 & 0 \\ \sin \theta & \cos \theta & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \] Step 2: Make the first element in the second column equal to 0. \\ We subtract \( \sin \theta \) times the first row from the second row: \[ R_2 \to R_2 - \sin \theta \cdot R_1 \] The augmented matrix becomes: \[ \left( \begin{array}{ccc|ccc} 1 & -\tan \theta & 0 & \sec \theta & 0 & 0 \\ 0 & \cos \theta & 0 & -\sin \theta \sec \theta & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \] Step 3: Make the second element in the second column equal to 1. \\ We divide the second row by \( \cos \theta \): \[ R_2 \to \frac{1}{\cos \theta} R_2 \] The augmented matrix becomes: \[ \left( \begin{array}{ccc|ccc} 1 & -\tan \theta & 0 & \sec \theta & 0 & 0 \\ 0 & 1 & 0 & -\sin \theta \sec^2 \theta & \sec \theta & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \] Step 4: Make the second element in the first column equal to 0. \\ We add \( \tan \theta \) times the second row to the first row: \[ R_1 \to R_1 + \tan \theta \cdot R_2 \] The augmented matrix becomes: \[ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & \sec \theta & 0 & 0 \\ 0 & 1 & 0 & -\sin \theta \sec^2 \theta & \sec \theta & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right) \] Thus, the inverse of the matrix is: \[ A^{-1} = \begin{pmatrix} \sec \theta & 0 & 0 \\ -\sin \theta \sec^2 \theta & \sec \theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Final Answer: \[ \boxed{A^{-1} = \begin{pmatrix} \sec \theta & 0 & 0 \\ -\sin \theta \sec^2 \theta & \sec \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}} \]