# Integer matrices with integer inverses

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?

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.

• 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 $GL(n,\mathbb{Z})$

• 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. – Widawensen Jul 14 '17 at 9:33

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.