# $[Solved]$ Find all units in the ring $M_2(\mathbb{Z})$ [duplicate]

my problem here is:

Find all units in the ring $M_2(\mathbb{Z})$

I've done it before but now it's a matrix and I'm lost. First I think on the determinant because I saw some other wrote about it, but I'm not sure. Any help?

Thank you

• Determinants was a good way to go. How can you tell from a determinant whether the matric is invertible? Commented Jan 16, 2018 at 13:23
• If a matrix is invertible, then $det(M)\neq 0$ if we let $M$ be the matrix. Am I right? And we also know that we can get identity matrix when $M$ and $M^{-1}$ is "multiplied" together. I hope it makes sense. Commented Jan 16, 2018 at 13:26
• Yes, but $\det(M)\det(M^{-1})=\det{I}=1$ and $\det(M), \det(M^{-1})$ are integers. What can they be?
– user491874
Commented Jan 16, 2018 at 13:29
• I believe that it can be $\pm 1$ Commented Jan 16, 2018 at 13:29
• Exactly! This is why this is true: $M$ is invertible in $M_2(\mathbb Z)$ if and only if $\det(M)=\pm 1$. Try to prove it.
– user491874
Commented Jan 16, 2018 at 13:30

More generally, for any commutative ring $R$, an $n\times n$ matrix $M$ lies in $\mathbf{GL}(n,R)$ if and only if $\det M\in R^{\times}$ (the units of $R$).
The units of $\mathbf Z$ are indeed only $1$ and $-1)$, simply because if $|u|>1$, you cannot find another integer $v$ such that $uv=1$ since $|uv|=|u||v|>|v|$.