<s>[INST] Seriously: What is the inverse of this 2-by-2 matrix? Okay, I have a hangover and it must be a stupid error, but I just don't get it:
The inverse of a 2-by-2 matrix $A=\left( \matrix{a & b \\ c& d} \right )$ is  $\frac{1}{det A}\left( \matrix{d & -b \\ -c& a} \right ) = \frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right )$. Now when I apply this recipe to the matrix
$$G:=\frac{1}{1+x^2+y^2}\left( \matrix{1+y^2 & -xy \\ -xy & 1+x^2} \right ) = \left( \matrix{\frac{1+y^2}{1+x^2+y^2} & \frac{-xy}{1+x^2+y^2} \\ \frac{-xy}{1+x^2+y^2} & \frac{1+x^2}{1+x^2+y^2}} \right )$$ I get 
$$det G = \frac{1+x^2+y^2+x^2y^2 -x^2y^2}{(1+x^2+y^2)^2} = \frac{1}{1+x^2+y^2}$$
and hence
$$G^{-1} = (1+x^2+y^2)\left( \matrix{1+x^2 & xy \\ xy & 1+y^2} \right )$$
But direct calculation shows that the inverse of $G$ is $\left( \matrix{1+x^2 & xy \\ xy & 1+y^2} \right )$, without the factor in front. Am I mad or still drunk? Can someone help my addled brain? Thanks - seriously!
 [/INST] You just made a mistake in "... and hence $G^{-1}=$...". Your $\det G$ is correct.  
In your problem $\frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right )=(1+x^2+y^2)\left( \matrix{d & -b \\ -c& a} \right )$. You should know what $a,b,c,d$ are now.
</s>