Linear Algebra - Invertible matrices and determinants

Question

Let $A$ be any $n \times n$ invertible matrix, defined over the integer numbers. Let assume that $A^{-1}$ (Inverse of A) is also defined over the integer numbers.

Prove that $\det A\in\{-1,+1\}$.

Any help will be appreciated, Thanks.

Answer 1

$det(A)det(A^{-1}) = det(I) = 1$. You also know that $det(A) \in \mathbb{Z}$ and $det(A^{-1}) \in \mathbb{Z}$. See what you can do with that.

Answer 2

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