# 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 matrices?

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.

• We find this statement everywhere on the web, but it's only for n > 1. For 1x1 matrices, all matrices are invertible except . Oct 19, 2020 at 9:58
• @jherek: The question is when are matrices with integer coefficients invertible with an inverse that has integer coefficients. The matrix $$ may be invertible, but the inverse does not have integer coefficients. What integers have an integer multiplicative inverse? Exactly $1$ and $-1$. So your statement is either incorrect (if you think you are correcting what I wrote) or irrelevant (answering the question of which matrices have inverses, instead of which integer matrices have integer matrix inverse). Which is it? Oct 19, 2020 at 11:04

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. Jul 14, 2017 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.

Just a comment, a picture of a particularly simple one: Iterative solvers have a hard time solving $$Ax = b$$ with $$b$$ a point source $$[0 0 0 ... 1 ... 0 0 0]$$:
$$A^{-1} b$$ has to pick out a column of $$A^{-1}, \pm \, [1 \, 0 \, -1 \, 0 ...]$$. Also, this $$A$$ is far from positive-definite -- its eigenvalues are symmetric about $$0$$.
$$A^{-1} = \frac{1}{|A|}C^T$$ Where $$|A|$$ denotes the determinant of matrix $$A$$ and $$C$$ is the matrix of minors. Each entry in $$C^T$$, $$c_{ji}$$, represents the minor removing just the $$i$$th row and the $$j$$th column. Each of these determinants is the sum, difference and multiple of integers, so each $$c_{ji}$$ is integer. This means, of course, that $$|A|$$ must be $$\pm 1$$. Since matrices may be of any integer dimension and there are infinite integers, just listing the identity matrices gets us to $$\infty$$.