Is there a way to switch specific elements of a matrix thru matrix multiplication, from left or right, and by one matrix or multiple.
Just as example (the size and the elements may differ) : How to switch the elements of the matrix $$ \begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9 \end{bmatrix} $$ to get : $$ \begin{bmatrix} 1&2&-3\\ 4&-5&6\\ 7&8&9 \end{bmatrix} $$ There is the Hadamard product, but it doesn't allow to calculate the determinant of the product from the determinants of the two (or more) other matrices.
Let's look at the $ 2 \times 2$ case. You'd like to have $$ \pmatrix{p & q \\ r & s}\pmatrix{a & b \\ c & d} = \pmatrix{a & -b \\ c & d} $$ for a fixed $p, q, r, s$ and all $a, b, c, d$. So $$ pa + qc = a $$ for all $a, c$, which says that (by picking $c = 0$ and $a = 1$) that $p = 1$ and $q = 0$. Similarly, $$ rb + sd = d $$ tells us that $r = 0$ and $s = 1$. But in this case, the left matrix is the identity, so the condition on $b$ is not met.
Short summary: flipping the sign of $b$ cannot be done merely by left multiplication. Perhaps, though, it can be done by a combination of left and right multiplication. Let's suppose, for the sake of contradiction, that it can be done by matrices $P$ and $T$. Then we'd have $$ P A T = A', $$ where $A'$ denotes the matrix $A$ with its upper right entry negated, for every matrix $A$. Plugging in $A = I$, we see that $T$ must be $P^{-1}$, hence $P$ and $T$ are invertible. Hence we must have $$ P A P^{-1}= A' $$ for every matrix $A$, i.e., $$ PA = A' P $$ for every matrix $A$. Writing that out in individual entries, we have $$ \pmatrix{p & q \\ r & s}\pmatrix{a & b \\ c & d} = \pmatrix{a & -b \\ c & d}\pmatrix{s & -q \\ -r & p} $$ for every $a, b, c, d$. Picking $a = 1, b = c= d = 0$, we find from the upper left entry that $pa = as$, and since $a = 1$, this says that $p = s$. So now we have $$ \pmatrix{p & q \\ r & p}\pmatrix{a & b \\ c & d} = \pmatrix{a & -b \\ c & d}\pmatrix{p & -q \\ -r & p} $$ Looking at the $1,1$ entry of the two sides, we have $$ pb + qd = -aq - bp $$ for every $a, b, c, d$. Again, setting $a = 1, b = c = d = 0$, we get $$ 0 = -q $$ hence $q = 0$. Setting $b = 1, a = c = d = 0$ gives $p = 0$. But now we'd have to have $$ P = \pmatrix{0 & 0 \\ r & 0} $$ which is not invertible. That's a contradiction.
In short: even in the $2 \times 2$ case, there's no left-and-right matrix multiplication that flips the sign of the $(1,1)$ entry of an arbitrary matrix.