transforming a $3{\,\times\,}3$ table by a sequence of allowable moves Question: A 3x3 table is filled with +'s and -'s. In a single move, you can either change all the signs in any row or all the signs in any column. By repeating these moves, is it possible to make all the signs equal?
Now this one seemed straight forward at first before I read the question properly and I assumed I could create any configuration I want and prove that I can make them all equal but now under further inspection, it doesn't seem that's the case. I feel the question is a little vague but requires some proof to show it either works for all configurations or have an example where you couldn't possibly do it and then say it's not possible for all configurations?
I've spent a while trying many configurations and personally feel it isn't possible for all configurations but don't know how to prove it or actually as to whether it is possible and I haven't tried all possible ways yet. 
What I did note was that there are $9$ "cells", each move changes the signs of $3$ cells and that there and $12$ different moves you can make. ($3$ rows + $3$ columns) I also tried representing the $+$'s and $-$'s in binary but that didn't seem to help me at all.
Can someone please shed some light on how to answer this question, thanks in advance.
 A: Hint, but not a complete solution
You might want to turn your matrices into vectors, so that 
$$
\pmatrix{a & b & c \\ 
d & e & f \\
g & h & k}
$$
becomes 
$$
\pmatrix{a & b & c  & d & e & f & g & h & k},
$$
and further treat each "$+$" as a $0$ and each "$-$" as a $1$, where these $0$s and $1$s lie in the field of two elements, $F_2$, where $1 + 1 = 0$. 
Now "flipping" the first row amounts to adding the vector 
$$
r_1 = \pmatrix{1 & 1 & 1  & 0 & 0 & 0 & 0 & 0 & 0},
$$
to the current "state". There are similar vectors $r_2, r_3, c_1, c_2, c_3$. 
Exercise to think about: Does it matter which order you do row- or column-flips in? Is flipping row1, then flipping column2, the same as flipping column2 and then flipping row1? 
Now what you're asking is whether, for any initial state $s$, there's some linear combination of the $r$s and $c$s (with $F_2$ coefficients) that gives you $s$. (Or, equivalently, whether there's some combination of the $r$ and $c$ vectors which, when added to $s$, gives you the vector of all zeroes. Exercise to think about: Why are these equivalent?)
A: It is very easy to describe which grids can be solved. 

It is possible to make all the symbols equal if and only if every row is either equal to or the exact opposite of the first row. 

Proof: Given a matrix where every row is equal or opposite to the first row, here is how you can succeed. First, reverse all rows which are opposite the first row, so now all rows are equal. Then, flip all negative columns. 
This shows the condition is sufficient. Conversely, it is necessary, because you can show that the property "all rows are equal or opposite to the first row" is preserved by both row and column flips. Therefore, if a matrix has does not have that property, it will never have that property, and therefore never be all equal.

For example, you cannot succeed from the below board, because the third row is neither equal to the first row $(+-+)$ nor its opposite $(-+-)$. 
$$
\begin{bmatrix}+-+\\-+-\\++-\end{bmatrix}
$$

There is a perhaps even simpler description of the same property; a matrix can be made all equal if and only if each of the $2\times 2$ subsquares has an even number of $+$'s. There are four such subsquares, one in each corner. In the above example, the lower left subquare is highlighted below in red. Since there are an odd number $(3)$ of $+$'s, the matrix is unsolvable. 
$$
\begin{bmatrix}+-+\\\color{red}{-+}-\\\color{red}{++}-\end{bmatrix}
$$
A: Flipping a row or column twice is the same as not flipping at all.

Also, actions are commutative, so the order doesn't matter.

Note that if position $A$ can be transformed to position $B$ by a sequence of flips, then performing the same sequence of flips on $B$ transforms $B$ to $A$.

Let $P$ denote the all-plus-signs position, and let $M$ denote the all-minus-signs position.

Starting at position $P$, if $P$ is transformed to position $Q$ by a finite sequence of moves, there are only two possible actions that have happened for each row (flipped or not flipped), and two possible actions that have happened for each column (flipped or not flipped).

Since there $3$ rows and $3$ columns, each of which can be independently flipped or not flipped, there are at most $2^6$ possible results for position $Q$.

But since a $3{\,\times\,}3$ matrix has $9$ entries, each with two possible values, there are $2^9$ different positions.

Since $2^9 > 2^6$, it follows that some positions can't be reached from $P$.

Let $X$ be a position that can't be reached from $P$.

Since $X$ can't be reached from $P$, it follows that $P$ can't be reached from $X$.

Then $M$ also can't reached from $X$, else if we continue from $M$ by flipping all $3$ rows, we would reach $P$.

Hence $X$ can't be transformed to an all-the-same-signs position.
$$
{
\bullet\qquad
\bullet\qquad
\bullet\qquad
\bullet\qquad
\bullet\qquad
\bullet\qquad
\bullet\qquad
\bullet\qquad
}
$$
For an explicit example of such an $X$, let $D$ be the position with minus symbols on the main diagonal and plus symbols everywhere else.

Claim:$\;P$ can't be reached from $D$.

Proof:$\;$Suppose instead that $P$ can be reached from $D$.

We can assume the transformation from $D$ to $P$ is achieved by a sequence of flips such that no row or column is flipped more than once.


*

*Since the top left corner cell must flip, it follows that the first row or the first column must flip, but not both.$\\[4pt]$

*Since the center cell must flip, it follows that the second row or the second column must flip, but not both.$\\[4pt]$ 

*Since the bottom right corner cell must flip, it follows that the third row or the third column must flip, but not both.


Thus, in the transition from $D$ to $P$, the total number of row or column flips is exactly $3$.

At the cell level, when $D$ is fully transformed into $P$, exactly $3$ cells will have flipped (the ones on the main diagonal)

Hence the $3$ row or column flips can't be all row flips or all column flips, else all $9$ cells would flip, contradiction.

Thus the $3$ row or column flips are either two row flips and one column flip, or two column flips and one row flip. In either case, it's easy seen that exactly $5$ cells would flip, contradiction.

It follows that $P$ can't be reached from $D$, as claimed.

As previously discussed, since $P$ can't be reached from $D$, it follows that $M$ also can't be reached from $D$.

Therefore $D$ can't be transformed to an all-the-same-signs position.
