Why the determinant of an invertible matrix $A$ must be equal to $\pm1$? I've been asked the following question: considering that $A$ is an invertible matrix / $A$ and $A^{-1}$ have integer coefficients, why both determinants must be $1$ or $-1$?
We know that, in linear algebra, an $n$-by-$n$ square matrix $A$ is called invertible if there exists an $n$-by-$n$ square matrix $A^{-1}$ such that  $AA^{-1}=I$ where $I$ is the identity matrix. 
So, if we also consider the following properties: $A$ is invertible $\Leftrightarrow$ $\det(A)\not=0$ and that $\det(I)=1$. 
Then, let $A\in\mathbb Z^{n\times n}$ such that $A^{-1}\in\mathbb Z^{n\times n}$ and in consequence, $\det\colon\mathbb Z^{n\times n}\to \mathbb Z$, now we can say:
$\det(A)\cdot\det(A^{-1}) =\det(AA^{-1}) =\det(I) =1$. 
I cannot realize why it could also be $-1$. Any idea or suggestion about how can I prove it? 

If matrices $A$ and $A^{-1}$ have only integer coefficients, that means that both of them must have integer-valued determinant. 
And by Cauchy–Binet formula we get:
$$ det(AA^{-1})=det(A)det(A^{-1})=1=det(I).$$
From here we directly get statement you want to prove.
