22
$\begingroup$

If all 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 all the entries of $A^{-1}$ are integers. My question is:

What are all the invertible integer matrices such that their inverses are also integer?

  • 2
    http://en.wikipedia.org/wiki/Unimodular_matrix2011-11-23

3 Answers 3

18

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.

  • 3
    @Chandru1: See the $l$ink.2011-01-30
10

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 $GL(n,\mathbb{Z})$

  • 0
    Very interesting statement. Is it not possible to generate integer inverse for matrix with determinant for example $2$ ? Quite surprising but probably the statement is true.2017-07-14
4

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

$a_{ij}=\binom{n+j-1}{i-1}$

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