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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?

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  • $\begingroup$ Just check hilbert matrix and its inverse. Inverse of hilbert matrix have integer entries but matrix is not itself integer matrix.It is rational entry matrix. $\endgroup$ Nov 15, 2017 at 5:55
  • $\begingroup$ To refer to what maths student had already said, related question and explanation why the inverse of the Hilbert matrix has integer coefficients. $\endgroup$
    – PinkyWay
    Jul 6, 2020 at 7:45
  • $\begingroup$ This is only true for n > 1 ! $\endgroup$
    – jherek
    Oct 19, 2020 at 10:03
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    $\begingroup$ @jherek: No, it's not. If you think otherwise, please provide a counterexample. $\endgroup$ Oct 19, 2020 at 19:11

2 Answers 2

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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 $\det(A)=\det(A^{-1})=\pm1$.

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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 $\det(A)=\frac1{2160}$

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