Term of rotation matrix entries equals 1 - proof concept?! I derived some stuff and it is happening that i come to the following expression:
$\frac{r_{13}^2 + r_{23}^2}{(r_{11}r_{23} - r_{13}r_{21})^2 + (r_{12}r_{23} - r_{13}r_{22})^2}$ 
that must equal 1 for all first two rows of a rotation matrix (orthogonal matrix).
$$  \begin{matrix}
    r_{11} & r_{12} & r_{13} & \\
    \end{matrix}   $$
$$  \begin{matrix}
 r_{21} & r_{22} & r_{23} & \\
    \end{matrix}   $$
I am confused how to proof this! I see a bit of a cross product in the denominator but can't identify the relation combined with the numerator.

I did this just by using the facts that the rows and columns have length one and are mutually orthogonal.
$$\begin{align}
&(r_{11}r_{23}-r_{13}r_{21})^2+(r_{12}r_{23}-r_{13}r_{22})^2\\
&=r_{11}^2r_{23}^2+r_{13}^2r_{21}^2-2r_{11}r_{21}r_{13}r_{23}+
r_{12}^2r_{23}^2+r_{13}^2r_{22}^2-
2r_{12}r_{22}r_{13}r_{23}\\
&=r_{23}^2(r_{11}^2+r_{12}^2)+r_{13}^2(r_{21}^2+r_{22}^2)
-2r_{13}r_{23}(r_{11}r_{21}+r_{12}r_{22})\\
&=r_{23}^2(1-r_{13}^2)+r_{13}^2(1-r_{23}^2)-2r_{13}r_{23}(-r_{13}r_{23})\\
&=r_{13}^2+r_{23}^2-2r_{13}^2r_{23}^2+2r_{13}^2r_{23}^2\\
&=r_{23}^2+r_{13}^2
\end{align}$$
I see that an answer has been posted since I started typing this (I'm awfully slow), but I'm going to post it anyway, since it's a bit different.
