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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.

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

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    $\begingroup$ Isn't $A = B = -I$ a counterexample to this? $\endgroup$ – Nathaniel Mayer May 5 '18 at 6:51
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    $\begingroup$ @gimusi Its inverse is no integer matrix. $\endgroup$ – Michael Hoppe May 5 '18 at 6:51
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    $\begingroup$ It is not true. The group $GL(n,\mathbf{Z})$ contains all matrices with determinant $\pm 1$. $\endgroup$ – Michal Adamaszek May 5 '18 at 6:55
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    $\begingroup$ 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. $\endgroup$ – mol3574710n0fN074710n May 5 '18 at 7:06
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    $\begingroup$ It's false because [7 2; 3 1] is invertible. $\endgroup$ – Cross Ratio May 5 '18 at 8:28
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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.

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    $\begingroup$ This example is nice because it shows it is insufficient for one of the matrices to have non-negative entries (also illustrated by @CrossRatio's Comment on the OP). $\endgroup$ – hardmath May 5 '18 at 14:01

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