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Assume $A$ is an $n\times n$ matrix where every entry is an integer.

Suppose $A$ is invertible, and that every entry in $A^{-1}$ is also an integer. Why must $\det(A)$ be only $1$ or $-1$?

It's clear that any matrix $A$ consisting only of integers will produce a determinant that is an integer, but I am unsure of how to show why that integer can only be $1$ or $-1$ in this case. Is there a specific equation I should consider? Any push in the right direction is appreciated.

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    $\begingroup$ Hint:$\;\det(A)\det(A^{-1})=\;??$. $\endgroup$
    – quasi
    Oct 27, 2018 at 22:37
  • $\begingroup$ If $ab = 1$ and $a, b$ are both integers, then both $a, b$ are $\pm 1$. $\endgroup$
    – Joppy
    Oct 27, 2018 at 22:37

1 Answer 1

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We know that

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

And by the assumption, both $\det A$ and $\det A^{-1}$ are integers.

Hence, $\det A$ should be $1$ or $-1$.

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