If $B$ is a square matrix whose entries are integers, then the determinant of $B$ is an integer.

The inverse matrix of $A$ can be computed by the formula
\[A^{-1}=\frac{1}{\det(A)}\Adj(A).\]

Proof.

Let $I$ be the $n\times n$ identity matrix.

$(\implies)$: If $A^{-1}$ is an integer matrix, then $\det(A)=\pm 1$

Suppose that every entry of the inverse matrix $A^{-1}$ is an integer.
It follows that $\det(A)$ and $\det(A^{-1})$ are both integers.
Since we have
\begin{align*}
\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1,
\end{align*}
we must have $\det(A)=\pm 1$.

$(\impliedby)$: If $\det(A)=\pm 1$, then $A^{-1}$ is an integer matrix

Suppose that $\det(A)=\pm 1$. The inverse matrix of $A$ is given by the formula
\[A^{-1}=\frac{1}{\det(A)}\Adj(A),\]
where $\Adj(A)$ is the adjoint matrix of $A$.
Thus, we have
\[A^{-1}=\pm \Adj(A).\]
Note that each entry of $\Adj(A)$ is a cofactor of $A$, which is an integer.

(Recall that a cofactor is of the form $\pm \det(M_{ij})$, where $M_{ij}$ is the $(i, j)$-minor matrix of $A$, hence entries of $M_{ij}$ are integers.)

Therefore, the inverse matrix $A^{-1}$ contains only integer entries.

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