Let $M$ be a free $\mathbb{Z}$-module of rank $n$, and $f:M\to M$ be a homomorphism of $\mathbb{Z}$-modules.
$A\in \mathrm{Mat}_n(\mathbb{Z})$ is the matrix representing $f$ with respect to a fixed $\mathbb{Z}$-basis of $M$.
Show that if $\mathrm{Coker}(f):=M/\mathrm{Im}(f)$ is finite, then $|\det A|=|\mathrm{Coker}(f)|$, and for a prime $p$ with $p\nmid |\mathrm{Coker}(f)|$ the induced map $f':M/pM\to M/pM$ is an isomorphism of $\mathbb{Z}/(p)$-modules.
The point is that by the structure theorem for finitely generated modules over a PID, you can make two separate base changes of $M$, so that with respect to the two new bases, the matrix of $f$ looks like $$ D = \begin{bmatrix} d_{1} \\ & d_{2}\\ && d_{3}\\ &&& \ddots\\ &&&&d_{n} \end{bmatrix} $$ for integers $d_{i} > 0$ (as the cokernel is finite, see just below).
Thus $$ \operatorname{Coker}(f) \cong \mathbb{Z} / d_{1} \mathbb{Z} \oplus \dots \oplus \mathbb{Z} / d_{n} \mathbb{Z} $$ has order $$ d_{1} \dots d_{n} = \det(D). $$ Now if $S, T$ are the matrices of the two base changes, we will have $$ S A T = D, $$ so that $$ \det(D) = \pm \det(A), $$ as $S, T$ are invertible integer matrices.
