Question

Find the inverse of the rotation matrix in (7.13); you should get $\mathit{C}$ in (11.14). Replace role="math" localid="1664340540940" $\mathit{\theta }$ by $\mathbf{-}\mathit{\theta }$ in (7.13) to see that the matrix C corresponds to a rotation through $\mathbf{-}\mathit{\theta }$.

Short answer

Inverse is$\left(\begin{array}{c}\cos \theta \\ \sin \theta \end{array}\begin{array}{c}-\sin \theta \\ \cos \theta \end{array}\right)$

Step-by-step solution

Given information

The rotation matrix is $\left(\begin{array}{c}\cos \theta \\ -\sin \theta \end{array}\begin{array}{c}\sin \theta \\ \cos \theta \end{array}\right)$

Rotation matrix

A rotation matrix is a transformation matrix that operates on a vector and outputs a rotated vector while keeping the coordinate axes constant. These matrices rotate a vector by an angle in a counter clockwise direction. A square matrix with real entities is always a rotation matrix.

Calculate the inverse of the rotation matrix

Replace $\theta$by $-\theta$in the given rotation matrix.

$$\begin{gathered} {\left(\begin{array}{c}\cos \theta \\ -\sin \theta \end{array}\begin{array}{c}\sin \theta \\ \cos \theta \end{array}\right)}^{-1}=\left(\begin{array}{c}\cos (-\theta )\\ -\sin (-\theta )\end{array}\begin{array}{c}\sin (-\theta )\\ \cos (-\theta )\end{array}\right) \\ =\left(\begin{array}{c}\cos \theta \\ \sin \theta \end{array}\begin{array}{c}-\sin \theta \\ \cos \theta \end{array}\right) \end{gathered}$$