Consider the following problem.

We have a table of four by four, and in top left corner there is

a sign minus and in all other cells,

there are plus signs.

And in one step we can do the formula.

You can pick one row of this table, or one column,

and we can switch signs, all signs,

in this column, or in this row.

And the question is can we after several steps obtain a table consisting of all pluses?

Okay. We can try to do it and we can see that it somehow doesn't work.

So, now we will try to show that it is impossible.

And for this we will observe the following thing.

If we try to do this,

if we try to switch signs of columns and rows,

we can see that the number of signs always stays odd.

So let's just show that it is always the case,

if a number of signs is odd in the beginning then it will be odd all the time.

And for this we should do the following: we should show that if

a number of minuses is odd before one step,

it will be odd after one step.

And for this, we will have to consider some row,

and we will have to consider several cases: what is in this row,

and we will see that the number of minuses stays odd.

Okay, so consider the first case,

where this row is filled with pluses.

And by the way, the case of column is the same,

so still four cells in the column, four cells and row as we see.

So suppose there are four pluses in this row.

So after the step there will be four minuses,

or the number of minuses increased by four,

if it was odd, it stays odd.

Okay, now consider the next case.

There are three pluses and one minus.

After the step, we will have three pluses,

three minuses and one plus.

So the number of minuses increased by two and it also stays odd.

By the way, note that the case of one plus and three minuses is symmetrical.

After this step we will obtain three pluses and one minus.

So the number of minuses decreased by two now,

and also stays odd.

In the last case is the following: we have two pluses and two minuses,

after the step we still have two pluses and two minuses,

just in different positions.

So the number of minuses doesn't change at all.

So, indeed after the step,

the number of minuses stays odd.

And so this is an invariant.

It is odd in the beginning and it is odd after each step.

And so, this means that we cannot switch to all pluses in the table since,

if we do this,

the number of minuses becomes even, equal to zero.

Zero is an even number. So, we can't obtain such a position.

So, thus we solve the problem.