My question is very general, but i would be putting a very especific problem just to clarify and avoid doubts.
Suppose i'm using the simplex method to solve the following problem:
Maximize: $20 x_{1} + 10x_{2} + 15x_{3}$
subject to $3x_{1} + 2x_{2} + 5x_{3} \leq 55$
$2x_{1} + x_{2} + x_{3} \leq 26$
$x_{1} + x_{2} + 3x_{3} \leq 30$
$5x_{1} + 2x_{2} + 4x_{3} \leq 57$
$x_{1} , x_{2}, x_{3} \geq 0$
Lets call $z$ to the objective function: When i put the slack variables ($S_{i}$ for i =1,2,3,4) and i write the tableau i get:
$\begin{matrix} z& x_{1}& x_{2}& x_{3}& S_{1}& S_{2}& S_{3}& S_{4}& RHS \\ 1& -20& -10& -15& 0& 0& 0& 0& 0 \\ 0& 3 & 2 & 5 & 1& 0& 0& 0& 55 \\ 0& 2 & 1 & 1 & 0& 1& 0& 0& 26 \\ 0& 1 & 1 & 3 & 0& 0& 1& 0& 30 \\ 0& 5 & 2 & 4 & 0& 0& 0& 1& 57 \\ \end{matrix}$
I know that the first step is to change $x_{1}$ from beying a cero-variable in reeplacement of $S_{4}$, so basicly i want to convert the column:
$\begin{matrix} -20 \\ 3\\ 2\\ 1\\ 5\end{matrix}$
to:
$\begin{matrix} 0 \\ 0\\ 0\\ 0\\ 1\end{matrix}$
The question is:
If i'm allowed to use any elementary row operation, i can use for example the following:
Add $\frac{20}{3}$ times the second row to the first (to convert the $-20$ in $0$), and so on...
The point is that if i do this, the simplex method ends in the next step, because the first row in the tableau would be:
$\begin{matrix} 1 &0 &\frac{10}{3} &\frac{55}{3} &\frac{20}{3} &0 &0 &0 &\frac{1100}{3} \end{matrix}$
Because there isn't a negative value in the first row, the problem is finished. But this is not the solution of the problem. I know that the best decision is to use as a pivot the last row in:
$\begin{matrix} -20 \\ 3\\ 2\\ 1\\ 5\end{matrix}$
But why can i not use other rows? What am i missing?