Rotation Matrix inverse using Gauss-Jordan elimination I'd like to calculate the inverse of a rotation matrix, let take the simplest case which is a $2$ x $2$ rotation matrix:
$R =\begin{bmatrix} \cos \theta & -\sin \theta \\[0.3em] \sin \theta & \cos \theta \end{bmatrix}$
I know that the inverse is the following
$R^{-1} =\begin{bmatrix} \cos \theta & \sin \theta \\[0.3em] -\sin \theta & \cos \theta \end{bmatrix}$ 
and I know that I can calculate it using the transpose method as such:
$R^{-1}=R^T$
but I fail to calculate the inverse using $Gauss-Jordan$ elimination, that is I don't know how to substract $\cos \theta$ from $\sin \theta$ in the second row. It all gets a bit complicated; I've looked around and nobody has a full step method using $G.-J.$ only the solution or the transpose method. Could someone provide me a full-step solution using $G.-J.$?
 A: $$
\begin{bmatrix}
\cos t&-\sin t&|&1&0\\ \sin t&\cos t&|&0&1
\end{bmatrix}
\xrightarrow{\frac1{\cos t}R1}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ \sin t&\cos t&|&0&1
\end{bmatrix}
\xrightarrow{R2-\sin t\,R1}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\cos t+\frac{\sin^2t}{\cos t}&|&-\frac{\sin t}{\cos t}&1
\end{bmatrix}=
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&\frac1{\cos t}&|&-\frac{\sin t}{\cos t}&1
\end{bmatrix}
\xrightarrow{\cos t\,R2}
\begin{bmatrix}
1 &-\frac{\sin t}{\cos t}&|&\frac1{\cos t}&0\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
\xrightarrow{R1+\frac{\sin t}{\cos t}R2}
\begin{bmatrix}
1 &0&|&\frac1{\cos t}-\frac{\sin^2t}{\cos t}&\sin t\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
=\begin{bmatrix}
1 &0&|&\cos t&\sin t\\ 0&1&|&-\sin t&\cos t
\end{bmatrix}
$$
This is a terrible method to calculate the inverse of any $2\times 2$ matrix. 
Edit: of course this does not work when $\cos t=0$; but this is a much easier case: you simply divide by $\sin t$ and permute the rows. 
A: Allow $\theta$ to remain symbolic until the operation is complete. I'm assuming if you had real numbers to plug in then taking the inverse would be trivial. 
\begin{bmatrix}
    c &-s  &| &1 &0\\
    s & c &| &0 & 1\\
\end{bmatrix}
Multiply top line by c
\begin{bmatrix}
    c^2  &-sc  &| &c &0\\
    s & c &| &0 & 1\\
\end{bmatrix}
Bottom line by s
\begin{bmatrix}
    c^2  &-sc  &| &c &0\\
    s^2 & sc &| &0 & s\\
\end{bmatrix}
Add bottom line to top
\begin{bmatrix}
    c^2+s^2  &sc-sc  &| &c &s\\
    s^2 & sc &| &0 & s\\
\end{bmatrix}
Reduce
\begin{bmatrix}
    1  & 0  &| &c &s\\
    s^2 & sc &| &0 & s\\
\end{bmatrix}
Multiply top line by $-s^2$ and add to bottom line
\begin{bmatrix}
    1  & 0  &| &c &s\\
    s^2-s^2 & sc &| &-cs^2 & s-s^3\\
\end{bmatrix}
Reduce
\begin{bmatrix}
    1  & 0  &| &c &s\\
    0 & sc &| &-cs^2 & s-s^3\\
\end{bmatrix}
Divide bottom line by $s$
\begin{bmatrix}
    1  & 0  &| &c &s\\
    0 & c &| &-cs & 1-s^2\\
\end{bmatrix}
Note that Pythagorean Theorem gives $c^2 = 1-s^2$
\begin{bmatrix}
    1  & 0  &| &c &s\\
    0 & c &| &-cs & c^2\\
\end{bmatrix}
Divide bottom line by $c$
\begin{bmatrix}
    1  & 0  &| &c &s\\
    0 & 1 &| &-s & c\\
\end{bmatrix}
Again, if you want to make specific cases for theta $(=\pi/2,0,etc.)$, then just plug those specific values of $\theta$ in. Notice I did not carry any fractions through; every time I divided by a trig term if fully canceled terms in the numerators. 
A: To simplify notation let $s = \sin (\theta)$, $c = \cos(\theta)$.
If $\theta \neq \pm\pi/2, \ $ ($c = 0$ here), then 
$R = $
$$
\begin{align*}
&\begin{bmatrix} 
c & -s \\
s & c 
\end{bmatrix} 
\sim \\
&\begin{bmatrix}
c^2 & -sc \\
s^2  & sc
\end{bmatrix}
\sim \\
(c^2 + s^2 = 1, right?)
&\begin{bmatrix}
1 & 0 \\
s^2 & sc 
\end{bmatrix}
\sim \\
&\begin{bmatrix}
1 & 0 \\
s/c & 1
\end{bmatrix}
\end{align*}
$$
then finish it off!
If $\theta = \pm \pi/2$, then $\theta \neq 0$ or $\pi$, so perform the similar set of operations first swapping the two rows and ending with a divide-by-$\sin$.
QED
