Simplex method Tableau Row operations why can't I multiply reduced cost by -1? I am a little confused...
Lets say I have this tableau where I have to make sure my reduced cost (Row 3) is non-negative, why can't I simply multiply Row $3$ by $-1$?
Afterwards, since all my reduced cost is positive, does that mean there is no need to continue the simplex method?
Notice how the reduced cost is all negative:



Pivoting '5':




*

*Why couldn't I just multiply by $-1/10$ for Row 3?

*Then for Row $2$ I'll just divide by $5$

*Then for Row $1$ I'll just $\frac{R1}{3} - Row 2$
 A: In the shown model from the question, if we were to multiply Row Three (the objective function row) by negative one, that would be equivalent to the following:
$$\max z = 8x_1 + 10x_2 + 7x_3 \longrightarrow \min z = -8x_1 - 10x_2 - 7x_3$$
Effectively turning this model from a maximization problem, to a minimization problem. Which we can do, as long as we would remember to multiply the objective function output by $-1$ at the very end of the process. However, since we are doing this way hand, this would be inefficient and cucumber-some to ourselves. ($1$)
In addition, with your proposed approach, we wouldn't really go anywhere more than we just making the starting basis look a little different. It would be much simpler to just go ahead and carryout the Simplex Method on the current tableau as is as we currently have the $s_1$ and $s_2$ variables as our basic variables, which we don't really want as we're not at the optimal point of the model (which is when $x_1=8$, and $x_2=x_3=0$).

($1$) However, in actuality, solvers like CPLEX and LINDO do implicitly do this in the background, as it is much easier to code a program that solves minimization problems, and thus have it turn maximization problems into minimization problems and turn it back into a maximization problem at the very end.
