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.
 A: It is the rotation matrix, so you can simply plug in $-\theta$.
On the other hand you could also use the $2\times2$ inverse identity:
$$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}
=\frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$$
A: \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}
A: Several people have posted ways to do this, but none did it the way the original posted proposed, i.e. row operations.  Nor any with trigonometric identities.  So here are those two methods:
\begin{align}
& \begin{bmatrix} \cos\theta & -\sin\theta & 1 & 0 \\ \sin\theta & \cos\theta & 0 & 1
\end{bmatrix} \\
\text{Now put }\\
\text{$(\sin\theta)\cdot$1st row${}-(\cos\theta)\cdot$2nd row } \\
\text{in the 2nd row: } & \begin{bmatrix} \cos\theta & -\sin\theta & 1 & 0 \\ 0 & -1 & \sin\theta & -\cos\theta \end{bmatrix} \\
\text{Then put }\\
(-\sin\theta)\cdot\text{2nd row}+\text{1st row } \\
\text{in the 1st row: } &
\begin{bmatrix} \cos\theta & 0 & 1-\sin^2\theta & \sin\theta\cos\theta \\ 0 & -1 & \sin\theta & -\cos\theta \end{bmatrix} \\[8pt]
= {} & \begin{bmatrix} \cos\theta & 0 & \cos^2\theta & \sin\theta\cos\theta \\ 0 & -1 & \sin\theta & -\cos\theta \end{bmatrix} \\[6pt]
\text{Then multiply the 1st row by $1/\cos\theta$: } \\
\text{and the 2nd by $-1$: } &
\begin{bmatrix} 1 & 0 & \cos\theta & \sin\theta \\ 0 & 1 & -\sin\theta & \cos\theta \end{bmatrix}
\end{align}
That does it via row operations.
Now let's try trigonometric identities.  First prove that
$$
\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}
\begin{bmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{bmatrix}
=\begin{bmatrix} \cos(\alpha+\beta) & -\sin(\alpha+\beta) \\ \sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix}.
$$
The look at what happens when $\beta=-\alpha$ and you'll see the answer.
