A unimodular matrix $M$ is a square integer matrix having determinant $+1$ or $−1$.
A totally unimodular matrix (TU matrix) is a matrix for which every square non-singular submatrix is unimodular. A totally unimodular matrix need not be square itself. Obviously, any totally unimodular matrix has only $0$, $+1$ or $−1$ entries.


Now suppose a $n\times n$ non-singular matrix $A$ is totally unimodular. Can we prove that
$A^{-1}$ is also totally unimodular? Or if it is not correct, can we have a counterexample?



Inverse of a totally unimodular matrix
The answer is yes, because if $B=A^{-1}$, then we have an equality between minors:
$$B(I,J)=\pm\frac{A(J^c,I^c)}{\det A},$$
for every subsets $I,J\subset[[1,n]]$ of same cardinals. This formula generalizes that giving the entries of $A^{-1}$ in terms of minors of $A$. The $\pm$ sign is not essential to prove the stability of the TU class under inversion.


