Are there $0$-$1$-matrices that are not unimodular?

Question

I am just wondering if there are matrices that only consists of $0$s and a few $1$s that are not totally unimodular (TU)? I cannot come up with an example but I am not very experienced with this stuff.

In a specific case, if I have a very sparse $0$-$1$ matrix where every row consist of at most two $1$s and every column consists of at most three $1$s, how can I find out whether this matrix is TU?

I cannot use the four sufficient conditions for $A$ to be totally unimodular (where $B$, $C$ is a disjoint partition of the rows of $A$), because condition $3$ might be violated:

1. Every column of contains at most two non-zero entries;
2. Every entry in is $0, +1$, or $−1$;
3. If two non-zero entries in a column of have the same sign, then the row of one is in $B$, and the other in $C$;
4. If two non-zero entries in a column of have opposite signs, then the rows of both are in $B$, or both in $C$.

Thank you!

Answer 1

There is a (very simple) $2 \times 2$ example where the determinant is $0$. Any $3 \times 3$ example with all rows having two $1$'s that does not have determinant $0$ has determinant $\pm 2$.