Prove that the matrix is totally unimodular Is there any (theoretic) way I can prove the matrix is totally unimodular? I have tested it by Matlab and know it is TU, however I cannot prove it.
-1 -1 -1 -1  0  0  0  0  0  0  0  0
 0  0  0  0 -1 -1 -1 -1  0  0  0  0
 0  0  0  0  0  0  0  0 -1 -1 -1 -1
 1  1  1  1  0  0  0  0  0  0  0  0
 0  0  0  0  1  1  1  1  0  0  0  0
 0  0  0  0  0  0  0  0  1  1  1  1
 1  0  0  0  1  0  0  0  1  0  0  0
 0  1  0  0  0  1  0  0  0  1  0  0
 0  0  1  0  0  0  1  0  0  0  1  0
 0  0  0  1  0  0  0  1  0  0  0  1

 A: First, observe that since $R_1= - R_4$, $R_2= -  R_5$, $R_3 = - R_6 $, then if any sub matrix uses any of these pairs of rows, then the determinant must be 0. Hence, (WLOG) we may assume that rows $1, 2, 3$ are removed (and replaced with rows 4, 5, 6 respectively in the sub matrix determinant calculation). Now call this matrix $A$.
If you look up the Wikipedia article, you will see a sufficient condition for Totally Unimodular matrices:


*

*Every column of  contains at most two non-zero entries.   

*Every entry in  is 0, +1, or −1   

*If two non-zero entries in a column of $A$ have the same sign, then the row of one is in $B$, and the other in $C$.   

*If two non-zero entries in a column of $A$ have opposite signs, then the rows of both are in $B$, or both in $C$.


Observe that if $B$ is the set of rows 4, 5, 6 and $C$ is the set of rows 7, 8, 9, 10, then this will satisfy the conditions. Hence we are done.
A: Here is another, perhaps easier way, of seing this. Your matrix is:
\begin{equation}
\begin{pmatrix}\begin{pmatrix}-1 & -1 & -1 & -1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \otimes Id \\
Id \otimes \begin{pmatrix}1 & 1 & 1 \end{pmatrix} 
 \end{pmatrix}
\end{equation}
