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Prove that if $A$ is a square matrix with integer entries and $\det(A)=\pm 1$, then the inverse of $A$ contains all integer entries.

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What do you know? What have you tried? – Julian Kuelshammer Oct 14 '12 at 11:07
up vote 4 down vote accepted

Hint: Consider the inverse written in terms of the adjugate matrix.

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This is one part of the statement that- "The inverse of an integer matrix $M$ is again an integer matrix if and only if the determinant of $M$ is exactly $\pm 1$. Now my question is how to show if $M^{-1}$ has integer entries then $\det(M)=\pm 1$ – Kushal Bhuyan May 9 at 6:03
@KushalBhuyan A general statement is that a matrix over $R$ is invertible (over $R$) if and only if $\det M$ is a unit of $R$. The only units of $R=\mathbb{Z}$ are $\pm1$, so this proves the statement you're looking for. More explicitly, we know that $\det M \cdot \det M^{-1} = 1$. Since $\det M$ and $\det M^{-1}$ are both integers (since $M$ and $M^{-1}$ are integer valued matrices), the only way for this to happen is if $\det M = \det M^{-1} = \pm 1$. – EuYu May 9 at 6:25

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