Suppose $A\in\Bbb{GL}_n(\Bbb R)$, (the space of all invertible matrices of order $n$), have integer entries.
If $\det(A)=\pm 1$ then obviously $A^{-1}$ have integer entries. Is the converse true, that is, if $A^{-1}$ have integer entries does it imply that $\det(A)=\pm 1$? Obviously for that $|\det(A)|=\gcd\{|A_{ij}|,1\leq i,j\leq n\}$ where $A_{ij}$ is the$(i,j)^{th}$ cofactor. But is it necessary for the $\gcd$ to be $1$ for a matrix to be invertible?
If $A$ and $A^{-1}$ have integer entries, then $$\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1$$ and as $\det(A)$ and $\det(A^{-1})$ are integers, then $\det(A)=\det(A^{-1})=\pm1$.
Consider this $3 \times 3 $ matrix :
$$A=\begin{pmatrix}1 & 1/2 & 1/3 \\ 1/2 & 1/3 & 1/4 \\ 1/3 & 1/4 &1/5 \end{pmatrix} $$
It is easy to check that
$$A^{-1}=\begin{pmatrix}9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 &180 \end{pmatrix} $$ But $\det(A)=\frac1{2160}$
