I want to show that this matrix is totally unimodular:
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}
I do laplace expansion by first row and I get:
\begin{bmatrix} 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}
Then laplace expansion by last row and get:
\begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \end{bmatrix}
So it sufficces to show that this matrix is totally unimodular.
Do I need to show that every square submatrix is unimodular?
I.e. this matrix has 4x4=16 submatrices of size (n-1) which can be computed by Sarrus rule?
What I did incorrectly was, I went one more laplace expansion, to get 3x3 and then Sarrused it, but I guess it is not correct, am I right?
How can I show that this 4x4 is totally unimodular? ( I know because R told me)
Could you show it by partitioning it to B,C or somehow with incidence matrices of Graphs?
I will try to answer your questions in the order that you asked them:
You do need to show that every square sub-matrix of your original matrix (before the Laplace expansions) has a determinant = $\pm1,$ or 0. Notice that if the -1 in your original matrix were a -5, the matrix would no longer be TUM, despite not changing the determinant of the entire matrix, so even though the $4 \times 4$ matrix is TUM, it is not sufficient proof that the larger matrix is TUM.
I am not familiar with Sarrus's rule, so I can't comment there.
As for the final 4 $\times$ 4 matrix, you can consider it to be the incidence matrix of a 4 node graph. Partition the nodes into two groups, $B,C$ such that there are no edges between any nodes in $B$, and there are no edges between any nodes in $C$. Doing so shows that your graph is bipartite, and thus its adjacency matrix is TUM.