Proof involving determinant and inverse of matrices

Question

Let $P$ $\in$ $\Bbb Z^{n,n}$, det$(P)$ $\neq$ $0$.

Prove: det$(P)$ $\in$ $\{$$-1, 1$$\}$ if and only if $P^{-1}$ $\in$ $\Bbb Z^{n,n}$.

Please can anyone help me out here?

Answer 1

If $\det(P) \notin \{\pm 1\}$, then $\det(P^{-1}) = 1/\det(P) \notin \Bbb Z$. So, $P^{-1}$ can't possibly be a matrix with integer entries.

On the other hand: if $\det(P) \in \{\pm 1\}$, then $P^{-1}$ is given by the formula $$ P^{-1} = \frac{1}{\det(P)} \operatorname{adj}(P) $$ Where adj denotes the adjugate matrix. Since $\operatorname{adj}(P)$ is a matrix with integer entries, so is $P^{-1}$.