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We know that if all the entries of an invertible matrix $A$ are rational, then all the entries of $A^{-1}$ are also rational.

Now suppose that all entries of an invertible matrix $A$ are integers. Then it's not necessary that entries of $A^{-1}$ are all integers. Question is:

  • What are all the invertible matrices with integer entries such that its inverse is also having integer entries?
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up vote 9 down vote accepted

Exactly those whose determinant is $1$ or $-1$.

See the previous question about the $2\times 2$ case. The determinant map gives necessity, the adjugate formula for the inverse gives sufficiency.

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how? Please explain – anonymous Jan 30 '11 at 5:27
@Chandru1: See the link. – Arturo Magidin Jan 30 '11 at 5:29

The inverse of an integer matrix is again an integer matrix iff if the determinant of the matrix is $\pm 1$. Integer matrices of determinant $\pm 1$ form the General Linear Group $GL(n,\mathbb{Z})$

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Arturo and Sivaram have already given the general condition for integer matrices with integer inverses; here I only note this particular example due to Ericksen that the matrix $\mathbf A$ with entries


where $n$ is an arbitrary nonnegative integer has an integer inverse.

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