Is the following matrix $A$ totally unimodular?
$$A = \begin{pmatrix} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 1 & 0 &1 \end{pmatrix}$$
Recall that a matrix is totally unimodular if every square submatrix $B$ has determinant $0$, $1$ or $-1$.
It surely looks like it, since every $B$ of $1 \times 1$ surely is, $3 \times 3$ are only two options and $2 \times 2$ gives really just two cases both which satisfy the required property. Is there a faster way to conlcude this though?