Let $A\in \mathbb{R}^{n\times n}$ be invertible such that $A$ and $A^{-1}$ have integer entries. Prove that $\det(A)\in\{-1,1\}$.

Question

Let $A\in \mathbb{R}^{n\times n}$ be invertible such that $A$ and $A^{-1}$ have integer entries. Prove that $\det(A)\in\{-1,1\}$.

My work:

$\det(I)=\det(AA^{-1})=(-1)^n\det(AA^{-1})=(-1)^n\det(A)\det(A^{-1}).$

I'm stuck here. Can someone help me?

Answer 1

$det(A)$ and $det(A^{-1})$ are integers, $det(AA^{-1})=1=det(A)det(A^{-1})$ and an integer is invertible for the multiplication if and only if it is equal to $1$ or $-1$.