# Show $G$ is a group, when there doesn't seem to be an inverse?

I would like to show that $G=\left\{\begin{bmatrix} a&a\\a&a \end{bmatrix}\mid a\in\mathbb{R}\setminus\{0\}\right\}$ is a group under matrix multiplication.

I've already verified that associativity holds and that the identity element exists, which is $I_2$. However I'm having trouble understanding why this is a group, since I don't see how to get the inverse, because the determinant of the matrix is $0$.

Any help would be appreciated. Thanks.

• How did you show that $I_2\in G$? Or perhaps what are you calling $I_2$, since what it usually is isn’t in $G$? Oct 15, 2014 at 3:54
• Ah, you're right. I was thinking $I_2$ as the $2\times 2$ identity matrix, but the identity has some entries are $0$, so its not part of the group. Thanks for that catch. I have to find the correct identity element now too. Oct 15, 2014 at 3:56
• I think $\begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}\end{bmatrix}$ works as the identity. Oct 15, 2014 at 4:00
• @SujaanKunalan it seems you have it, then. Oct 15, 2014 at 4:03

Hint: note that $$\pmatrix{1/2&1/2\\1/2&1/2} \pmatrix{a&a\\a&a} = \pmatrix{a&a\\a&a} \pmatrix{1/2&1/2\\1/2&1/2} = \pmatrix{a&a\\a&a}$$ Now, $$\pmatrix{a&a\\a&a} \pmatrix{b&b\\b&b} = 2 \pmatrix{ab&ab\\ab&ab}$$ For what $b$ does $2ab = 1/2$?
Alternatively: consider the map $$a \mapsto a \pmatrix{1/2&1/2\\1/2&1/2} = \pmatrix{a/2&a/2\\a/2&a/2}$$