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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?

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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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  • $\begingroup$ We find this statement everywhere on the web, but it's only for n > 1. For 1x1 matrices, all matrices are invertible except [0]. $\endgroup$
    – jherek
    Oct 19, 2020 at 9:58
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    $\begingroup$ @jherek: The question is when are matrices with integer coefficients invertible with an inverse that has integer coefficients. The matrix $[2]$ 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? $\endgroup$
    – Arturo Magidin
    Oct 19, 2020 at 11:04
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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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  • $\begingroup$ 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, 2017 at 9:33
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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

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

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

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Just a comment, a picture of a particularly simple one:

enter image description here

Added:
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 $0$.
See also gmres-for-a-non-diagonalizable-matrix on scicomp.stack .

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$$A^{-1} = \frac{1}{|A|}C^T $$ 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 $\infty$.

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