I want perform a simple check for total unimodularity. 


**Question:**  

what, if anything, can be concluded from the fact, that $$det(A)=1,\ a\_{ij}\in\{-1,0,+1\}\ \wedge\ a\_{ij}^{-1}\in\{-1,0,+1\}$$ 
where $a\_{ij}^{-1}$ denotes the entries of $A^{-1}$? 



Pragmatic Test for Total Unimodularity
There won't be any counterexamples to total unimodularity for $n=3$
since when $\det A=1$ the entries of $A^{-1}$ are up to sign the
$2\times 2$ minors of $A$. On the other hand, for $n=4$ let
 $$ A=\left[ \begin{array}{rrrr} 1 & 0 & 1 & 1\\ 0 & 1 & -1 & 1\\
 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array} \right]. $$
Then $A$ satisfies the conditions of the problem but is not totally
unimodular. In general, the conditions of the problem are just saying
that the $1\times 1$ and $(n-1)\times(n-1)$ minors are $0,\pm 1$,
while the the $n\times n$ minor (i.e., $\det A$) is 1.


