Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.