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? 
 A: Suppose $A$ is an invertible matrix with integer coefficients such that $A^{-1}$ has integer coefficients. Then, 
$$AA^{-1}=I\quad\Rightarrow\quad \mathrm{det}(AA^{-1})=\mathrm{det}(I)=1 $$
But since that for all matrices, $ \mathrm{det}(AB)= \mathrm{det}(A) \mathrm{det}(B)$, we have
$$  \mathrm{det}(A) \mathrm{det}(A^{-1})=1.$$
You may notice that the formula for the determinant of a matrix only contains addition/substraction and multiplication. This means that if all the entries of a matrix are integers, then the determinant of the matrix is an integer. By hypotesis, we thus have
$$\mathrm{det}(A)=m\in\mathbb{Z}^*,\qquad \mathrm{det}(A^{-1})=n\in\mathbb{Z}^*,$$
and $mn=1$. The only solution to this is $m=n=1$ or $m=n=-1$, which is the desired result.
A: 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.
A: $det(A)det(A^{-1})=1$ implies $det(A)$ or $det(A^{-1})$ are divisors of $1$. Hence they can only be 1 or -1.
