Answer
Since $A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)$ , $\det(A)=1$ and the entries of $A$ are integers, then the entries of $A^{-1}$ must be intergers.
Work Step by Step
Since $A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)$ , $\det(A)=1$ and the entries of $A$ are integers, then the entries of $A^{-1}$ must be intergers.
