Is the inverse of a unimodular matrix unimodular? Is the inverse of a unimodular matrix unimodular?
I saw this question: Is the inverse of an invertible totally unimodular matrix also totally unimodular?
And what is the case if the matrix is not square?
 A: From http://en.wikipedia.org/wiki/Unimodular_matrix : one definition of a unimodular matrix is an invertible matrix of integers whose inverse is also a matrix of integers.  So it is required to be square, in order to be invertible.
So, if $M$ is unimodular, then $M^{-1}$ is a matrix of integers$\ldots$ , and its inverse (which is $M$) is also a matrix of integers, so $M^{-1}$ is also unimodular.
(In fact the unimodular matrices form a subgroup of $GL_n(\mathbb{R})$.)
If you're more interested in non-square matrices, then you probably want to consider total unimodularity, as in the link you provided.
A: In fact, a stronger statement is true:
A square matrix A with integer entries is invertible if and only if $A^{-1}$ is unimodular.
Proof:
It is enough to show one direction, because we can apply the theorem to $A^{-1}$.
Assume $A$ is unimodular. Then, since
$$\det(A A^{-1}) = \det(I) = 1 $$ and $$\det(A A^{-1}) = \det(A) \det(A^{-1})$$
But since $A$ is unimodular, we know that $\det(A) = \pm 1$, so we can conclude that $\det(B) = \pm 1$.
We now just need to show that $B^{-1}$ has integer entries, and the proof will be finished.
There may be a cleaner solution out there, but I will use the cofactor matrix to show this. The cofactor matrix of $A$, denoted $cof(A)$ is related to $A$ by the following equality:
$$A^{-1} = \frac{1}{\det(A)} cof(A)^T$$
And from the definition of the cofactor, we can see that $B$ integral $\Rightarrow cof(B)$ integral.
From this, we can immediately see that $\det(A) = \pm 1 \Rightarrow A^{-1} = \pm cof(A)^T$, so $A$ is integral.
