Is the inverse of an invertible totally unimodular matrix also totally unimodular?

My question is learned from here. Let me restate it as follows:

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.

Now suppose an $n\times n$ 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? Any help is much appreciated.

Edit: What I have already known is that the statement is true when $n=2$ and $3$. It follows from the definition of unimodular matrix and the fact that if $A$ is unimodular, then $$A^{-1}=\det A\cdot \mathrm{adj}(A)=\pm \mathrm{adj}(A).$$

Solutions Collecting From Web of "Is the inverse of an invertible totally unimodular matrix also totally unimodular?"

As commented by Jyrki Lahtonen, the statement is true and it is immediately implied by the following relation between the minors of $A^{-1}$ and the minors of $A$.

Proposition: If $A$ is an invertible $n\times n$ matrix, and if $i_1,\dots,i_n$ and $j_1,\dots,j_n$ be two permutations of $1,\dots,n$, then the minor of $A$ corresponding to rows $i_1,\dots,i_k$ and columns $j_1,\dots,j_k$, denoted by $d$, and the minor of $A^{-1}$ corresponding to rows $j_{k+1},\dots,j_n$ and columns $i_{k+1},\dots,i_n$, denoted by $d’$,satisfy that
$$d=\pm d’\det A.$$

Proof: Let $e_1,\dots,e_n$ be a basis of $\mathbb{R}^n$, and let $f_i= A e_i$, $i=1,\dots,n$. Then on the one hand,
$$\omega:=Ae_{j_1}\wedge\cdots\wedge Ae_{j_k}\wedge e_{i_{k+1}}\wedge \cdots\wedge e_{i_n}=\pm d\cdot e_1\wedge\cdots\wedge e_n,$$
on the other hand,
$$\omega=f_{j_1}\wedge\cdots\wedge f_{j_k}\wedge A^{-1}f_{i_{k+1}}\wedge \cdots\wedge A^{-1}f_{i_n}=\pm d’\cdot f_1\wedge\cdots\wedge f_n.$$
Since
$$f_1\wedge\cdots\wedge f_n=\det A\cdot e_1\wedge\cdots\wedge e_n,$$
the conclusion follows.