Invert a $2\times 2$ Matrix containing trig functions Invert the $2\times 2$ matrix: 
\begin{bmatrix}
\cos\theta & -\sin\theta  \\
\sin\theta &  \cos\theta
\end{bmatrix}  
My thought was to append the  $2\times 2$ identity matrix to the right of the trig matrix and use row operations to get the answer.  
I have to show all steps, so I cannot just flip it and call it a day.

\begin{align}
\left[\begin{matrix}
\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right]^{-1} & = \frac{1}{{\left|\begin{matrix}
\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right|}}\cdot\left(\begin{matrix}\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)\end{matrix}\right)\tag{1} \\[2em]
& = 1\cdot \left(\begin{matrix}\cos\left(\theta\right) & \sin\left(\theta\right) \\
-\sin\left(\theta\right) & \cos\left(\theta\right)\end{matrix}\right),\tag{2}
\end{align}
because
\begin{align}\left|\begin{matrix}\cos\left(\theta\right) & -\sin\left(\theta\right) \\
\sin\left(\theta\right) & \cos\left(\theta\right)
\end{matrix}\right|&=\cos^2\left(\theta\right)-\left\{-\sin^2\left(\theta\right)\right\}=\sin^2\left(\theta\right)+\cos^2\left(\theta\right)=1.\tag{3}
\end{align}
