det(A)  is always an integer and will come in denominator in A-1 . Therefore, A to be unimodular, det(A) should be 1.

Clearly, you need to check whether |det(A)|=1. This is not automatically true, even if all columns are different, as the following (singular) matrix shows:

1 1 0

1 0 1

0 1 1

Your example is a special case, since the matrix is singular (clearly column 2 is obtained by subtracting columns 1 and 3). Can this be generalized to non-singular matrices? Do you know any example of 0-1 non-singular square matrix A for which |det(A)| > 1?

The following matrix A has det(A)=2:

1 1 0 0

1 0 1 0

1 0 0 1

1 1 1 1

Thank you! Now I am convinced.

@Anselmo: But what was wrong with my argument. Only I did not write  | det A|=1.

As I understand it, these were in fact two questions:

i) When is a (0-1-) matrix A invertible over Z?

ii) Is the 0-1-matrix always invertible if all columns are pair-wise distinct?

And the answers are

i) Iff |det(A)|=1.

ii) No.