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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!

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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 $\pm 2$.

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Ok, thank you. Than seems to be rather easy. However, the 2-2 matrix with determinant 0 would still be TU. And the matrices I have are very large and usually don't contain columns with two consecutuve 1s. – Copi Mar 6 '13 at 9:37
Any square submatrix that has a row or column containing only one $1$ is unimodular iff the sub-submatrix obtained by deleting the row and column with that $1$ is unimodular. So you want to look at square submatrices where all rows and columns have two $1$'s. – Robert Israel Mar 6 '13 at 23:48

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