TITLE: $A^{-1}$ has integer entries if and only if the ${\rm det}\ (A) =\pm 1$
QUESTION [20 upvotes]: So, $A$ is a $n \times n$ matrix with integer entries.  The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ .
I know that $A^{-1}= {\rm adj}(A)/{\rm det}(A)$ but I have no idea where to go from there for the forward direction.  Any help here would be greatly appreciated.
For the backwards direction, I think I am okay.  I plugged the 1 and the -1 options into $ {\rm adj}(A)/{\rm det}(A)$ So, I know that $A^{-1}= {\rm adj}(A)$ or $-{\rm adj}(A)$.  Then I said that because the ${\rm adj}(A)$ is simply the matrix of co-factors of $A$, and because $A$ has all integer entries then ${\rm adj}(A)$ will have all integer entries, which will mean that $A^{-1}$ will have all integer entries.  Is that okay? Or am I missing something else? 
Thank you!!``

REPLY [2 votes]: You did one direction well. Here the other direction:
If $A$ only contains integers then $\det(A)$ is an integer (computing a determinant only uses $+$ and $\cdot$). Assume $A^{-1}$ is also an integer matrix, then $\det(A^{-1})$ is also integer. Then note that
$$\Bbb Z\ni\det(A^{-1})=\frac 1{\det(A)}$$
implies $\det(A)\in\{-1,1\}$ because no other integers then $\pm 1$ have integer reciprocals.