Suppose we have a matrix $A = (a_{ij}) \in \operatorname{GL}_n(\mathbb{R})$ with $a_{ij} \in \mathbb{Z}$. I need to show that $A^{-1}$ has entries in $\mathbb{Z}$ if and only if $\det(A) = \pm 1$.
Inverse of integer matrix with determinant $\pm 1$
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matrices
determinant
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0Consider the matrices over the ring $\mathbb Z$. Since $\pm 1$ are the only invertible elements of $\mathbb Z$, $A$ has an inverse if and only if $\det A = \pm1$. – 2017-02-08
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0The inverse operation is not a linear operator, i.e. in general $(M+N)^{-1} \neq M^{-1}+N^{-1}$, and so you might have some problems. – 2017-02-08
1 Answers
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Using Cramer's Rule to obtain the inverse of $A$, we can see that $A^{-1}$ has entries in $\Bbb Z$, if $A$ has entries in $\Bbb Z$ and $\det(A)=\pm 1$. For the other direction of the proof, we use $$\det(A)\det(A^{-1})=\det(AA^{-1})=\det(I)=1$$ If $A$ and $A^{-1}$ have entries in $\Bbb Z$, then $\det(A)\in\Bbb Z$ and $\det(A^{-1})\in \Bbb Z$, and the equation above can hold only if $\det(A)=\det(A^{-1})=\pm 1$.