= Solution
The <adjugate matrix> is the transpose of the cofactor <matrix>: $\operatorname{adj}(A)_{ij}=(-1)^{i+j}\det A_{\widehat j,\widehat i}$. Cofactor expansion gives the <adjugate identity>
$$
A\operatorname{adj}(A)=\operatorname{adj}(A)A=(\det A)I.
$$
For an invertible <matrix> this yields $A^{-1}=\operatorname{adj}(A)/\det A$. Integer entries of $A$ imply integer cofactors, so $\det A=\pm1$ makes its inverse integral. Conversely, if $A$ and $A^{-1}$ both have integer entries, their <determinants> are integers and $(\det A)(\det A^{-1})=1$. The only integer units are $\pm1$. Hence \b[an invertible integer <matrix> has an integer inverse exactly when it is a <unimodular matrix>], with <determinant> $\pm1$. Invertibility is implicit whenever $A^{-1}$ is used.
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