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?

  • it seems that invertible TU matrices also form a group like unimodular matrices...but I don't have a proof yet. – Suvrit Apr 19 '13 at 19:13
  • Doesn't this follow from looking at the adjoint or cofactor marix? Gerhard "Ask Me About Binary Matrices" Paseman, 2013.04.19 – Gerhard Paseman Apr 19 '13 at 22:01
  • 1
    S. Sra: Good luck with that proof. You might consider 1 1;0 1 meanwhile. Gerhard "Not Grouplike Under Matrix Multiplication" Paseman, 2013.04.19 – Gerhard Paseman Apr 19 '13 at 22:11
  • @Gerhard: The inverse of the matrix that you've mentioned is also TU...so I don't understand your comment??? – Suvrit Apr 19 '13 at 22:25
  • S. Sra, I am suggesting that I don't know what group structure you are placing on the set of matrices. For the two notions of multiplication I considered, the matrix I gave does not help form a group. Gerhard "How Do You Multiply Them?" Paseman, 2013.04.19 – Gerhard Paseman Apr 19 '13 at 22:52
up vote 14 down vote accepted

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.

  • 1
    ah, good old Cramer's rule (essentially!) – Suvrit Apr 19 '13 at 23:03
  • 1
    Thank you very much. But I think the formula should be $B(I,J) = \pm \frac{A(J^c, I^c)}{\det A}$. – qianchi Apr 20 '13 at 2:50
  • @qianchi. Of course you're right. I'll edit. – Denis Serre Apr 20 '13 at 3:39

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.