# Matrix row/column swapping problem

Given an $$n \times n$$ square matrix $$M$$ in which every row and column contains exactly one $$-1$$ and one $$1$$, and is $$0$$ otherwise, I'm trying to calculate whether it's always possible to reach $$-M$$ through successive swapping of rows/columns.

I've worked out that it's at least possible sometimes, simply by trial-and-erroring arbitrary matrices by hand, but of course this is a long way from a proof, and it only proves that it's possible for those specific matrices. I'm not too sure where to start with this.

One thing I noticed is that if $$\det(M) \neq 0$$ then the number of swaps must match the evenness of $$n$$, because every swap will flip the sign of the determinant, and then we also have

$$\det(-M)=(-1)^n \det(M)$$

However I'm not sure if this is relevant at all, it's just the only thing I've managed to deduce so far.

Any pointers?

It is always possible to get $$-M$$ by the procedure below. But first, the determinant is always $$0$$, since a vector with all its coordinates equal will be in the kernel of $$M$$. So that argument will not take you far.
Now use row/column swaps to bring $$M$$ into a standard form. Note that the first row can always be transformed into $$(1, -1, 0, \ldots)$$. From here you can continue and get one of two options for the first two rows. Namely either $$\begin{pmatrix}1&-1&0&\ldots \\ -1&1&0&\ldots \end{pmatrix}$$ or $$\begin{pmatrix}1&-1&0&0&\ldots \\ 0&1&-1&0&\ldots \end{pmatrix}.$$ In the first case the new matrix $$M’$$ is in block diagonal form with a $$2\times 2$$ block in the upper left.
If the dimension of $$M$$ is greater than $$2$$ then use induction to argue that $$M$$ can indeed be transformed into $$-M$$.
So, using induction on the size $$n$$, the only case left is to consider an $$m \times m$$ matrix $$M’$$ for any $$m\geq 2$$ of the form $$M’ = \begin{pmatrix} 1 & -1 & 0 & \cdots & & 0 \\ 0 & 1 & -1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \ddots & & \vdots \\ 0 & \cdots & & & 1 & -1 \\ -1 & 0 & \cdots & & 0 & 1 \end{pmatrix}.$$
For this case, reversing the first $$m-1$$ rows (swap rows $$1$$ and $$m-1$$, rows $$2$$ and $$m-2$$, etc.) and then reversing all columns results in $$-M’$$.
Finally retrace the steps that transformed $$M$$ into $$M’$$ to get $$-M$$.