<s>[INST] Find the inverse a matrix with trigonometic entries 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.
 [/INST] Implement the formula $\def\adj{\operatorname{adj}}A^{-1}=\frac{1}{\det A}\cdot \adj A$
Find $\det A$
$\det A=\begin{vmatrix} 1&0&0\\0&\cos x &\sin x\\ 0 &\sin x &-\cos x \end{vmatrix}=-1$
Find $\adj A$
$A_{11}=(-1)^{1+1}\left\lvert \begin{array}{cc} \cos x & \sin x \\ \sin x & -\cos x \end{array} \right\rvert=-1$
$A_{12}=(-1)^{1+2}\left\lvert \begin{array}{cc} 0 & \sin x \\ 0 & -\cos x \end{array} \right\rvert=0$ 
$A_{13}=(-1)^{1+3}\left\lvert \begin{array}{cc} 0 & \cos x \\ 0 & \sin x \end{array} \right\rvert=0$
$A_{21}=(-1)^{2+1}\left\lvert \begin{array}{cc} 0 & 0 \\ \sin x & -\cos x \end{array} \right\rvert=0$
$A_{22}=(-1)^{2+2}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & -\cos x \end{array} \right\rvert=-\cos x$
$A_{23}=(-1)^{2+3}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \sin x \end{array} \right\rvert=-\sin x$
$A_{31}=(-1)^{3+1}\left\lvert \begin{array}{cc} 0 & 0 \\ \cos x & \sin x \end{array} \right\rvert=0$
$A_{32}=(-1)^{3+2}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \sin x \end{array} \right\rvert=-\sin x$
$A_{33}=(-1)^{3+3}\left\lvert \begin{array}{cc} 1 & 0 \\ 0 & \cos x \end{array} \right\rvert=\cos x$
$A^{-1}=\left\lvert \begin{array}{ccc} 1&0&0\\ 0 &\cos x & \sin x \\ 0 & \sin x & -\cos x \end{array} \right\rvert$
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