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?

 A: 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.
A: 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.
A: 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 $GL(n,\mathbb{Z})$
A: Just a comment, a picture of a particularly simple one:

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