21
$\begingroup$

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?

$\endgroup$
17
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ how? Please explain $\endgroup$ – anonymous Jan 30 '11 at 5:27
  • 3
    $\begingroup$ @Chandru1: See the link. $\endgroup$ – Arturo Magidin Jan 30 '11 at 5:29
10
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Widawensen Jul 14 '17 at 9:33
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy