# Invertible matrices with integer entries has to be permutation matrices [duplicate]

Let $A$ and $B$ be $n\times n$ matrices with integer entries. Show that if $B=A^{-1}$ then $A$ and $B$ are permutation matrices (matrix obtained by permuting rows of the identity matrix).

If the entries are non negative integers then it is quite easy to show, but if the entries also involve negative integers then I couldn't show it.

## marked as duplicate by Dietrich Burde, GNUSupporter 8964民主女神 地下教會, polfosol, Jose Arnaldo Bebita-Dris, SaadMay 5 '18 at 14:35

• Isn't $A = B = -I$ a counterexample to this? – Nathaniel Mayer May 5 '18 at 6:51
• It is not true. The group $GL(n,\mathbf{Z})$ contains all matrices with determinant $\pm 1$. – Michal Adamaszek May 5 '18 at 6:55
• I think you actually mean permutations and sign shifts. Negative entries would not result from permutations alone. Here is a start: We know that $\det A \cdot \det B = 1$ and as both are polynomial expressions of integers we can conclude $\det A = \det B = \pm 1$. Then I think we can work with cofactor expansion of the determinant... But I have not elaborated this yet. – mol3574710n0fN074710n May 5 '18 at 7:06
If the matrices may have negative integer entries, then the statement is not true. For example, $$A= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$$ both have determinant $1$, and so are invertible over the integers, and in fact $AB = I$. But they are not permutation matrices.