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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!

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Look at that square in the denominator of $detG$... – leonbloy Jul 3 '11 at 15:12
up vote 4 down vote accepted

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.

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Oh my god you are right - thank you! – Embarassed Guy Jul 3 '11 at 15:15

Here is another way to solve the problem altogether. Notice that $$\det\left(\begin{array}{cc} 1+y^{2} & -xy\\ -xy & 1+x^{2}\end{array}\right)=1+x^{2}+y^{2},$$ and that $$\det\left(\begin{array}{cc} d & -b\\ -c & a\end{array}\right)=\det\left(\begin{array}{cc} a & b\\ c & d\end{array}\right).$$

This means $G$ is of the form $$G=\frac{1}{ad-bc}\left(\begin{array}{cc} d & -b\\ -c & a\end{array}\right).$$

So the formula for $G$ really looks like the formula for the inverse of a matrix. It has $\frac{1}{\det}$ and the minus signs in the correct spots.

Then we see that

$$G^{-1}=\left(\begin{array}{cc} 1+x^{2} & xy\\ xy & 1+y^{2}\end{array}\right)$$ because applying the inversion formula to $G^{-1}$ will give $G$ directly.

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