[Math] Find the inverse a matrix with trigonometic entries

Question

What is the inverse of
\[
\begin{pmatrix}
1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{pmatrix}
\]
Please help me to solve the above problem.

Answer 1

Let $A_x :=\left[ \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right]$, and let $R_x :=\left[ \begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array} \right]$ be the matrix of the rotation by angle $x$ in the plane (that is, for all ${\bf v}$ in $\mathbb R^2$, $\ R_x\cdot {\bf v}$ is the rotated version of $\bf v$), we have that $$A_x = R_x\cdot \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]\ \text{ and }\ (R_x)^{-1} = R_{-x} = \left[ \begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array} \right]\text{, so} $$ $$(A_x)^{-1} = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]\cdot R_{-x}$$ So, by easy matrix multiplication, one can verify that the additional $1$ in the additional dimension is not hurting much, ie. the requested inverse is: $$ \left[ \begin{array}{ccc} 1&0&0\\ 0 &\cos x & \sin x \\ 0 & \sin x & -\cos x \end{array} \right] $$