Why the requirement that variables must be non-negative isn't written in the corresponding tableau? Isn't this kind of constraint equivalent to others? For instance consider some LP with 2 variables $x_1 \geq 0, x_2 \geq 0$. I expect, that I can add two extra rows at the end of the tableau: $\left[\begin{array}{ccc|c}... &1& 0& 0\end{array}\right]$ and $\left[\begin{array}{ccc|c}... &0& 1& 0\end{array}\right]$. The reason that caused this question is that I obtain optimal value of LP with negative entries, but it must not by the condition of the exercise and I can't find another solution by the simplex method, only by selection, which isn't convenient.
UPD: I clarify a bit my problem. Firstly I mention, that I learn linear algebra and I encountered simplex method in context with linear algebra's methods, so my method of solving problem may differ a little bit from common. I have an optimization problem: image from textbook
I have put my problem into standard form (changed "=" constraint by two equivalent "<=" constraints and replaced minimize problem with maximize by multiplying objective function by -1):
maximize subject to: -x1-2*x2 +x3
x2 -x3 <= 3 2*x1+2*x2 -x3 <= 1 -2*x1-2*x2 +x3 <= -1 x1+2*x2+2*x3 <= 3 x1, x2, x3 >= 0
Then, as I don't have starting feasible solution (0,0,0), I transform my problem: 1) max -x1-2*x2+x3 ---> max -y 2) I subtract "y" from each constrain that have negative right-side.
maximize subject to: -y
x2 -x3 <= 3 2*x1+2*x2 -x3 <= 1 -2*x1-2*x2 +x3 -y <= -1 x1+2*x2+2*x3 <= 3 x1, x2, x3, y >= 0
This program have obvious feasible solution (0,0,0,1). By pivoting on column of corresponding tableau I have found another solution (1,0,1,0) that gives me optimal value of "-y" equals 0, so I know that initial LP is feasible and at least have solution (1,0,1) (this answer is correct, but isn't supposed to be by the explanation from textbook).
Then I can pivot on columns of tableau corresponding initial LP.
\begin{array}{c|cccc|ccc|c} 1 & 0 & 0 & 0 & 0 & 1 & 2 & -1 & 0 \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1 & 3 \\ 0 & 0 & 1 & 0 & 0 & 2 & 2 & -1 & 1 \\ 0 & 0 & 0 & 1 & 0 & -2 & -2 & 1 & -1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 2 & 2 & 3 \\ \end{array}
\begin{array}{c|cccc|ccc|c} 1 & 0 & 0 & 0 & -1 & 0 & 0 & -3 & -3 \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1 & 3 \\ 0 & 0 & 1 & 0 & -2 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & 1 & 2 & 0 & 2 & 5 & 5 \\ 0 & 0 & 0 & 0 & 1 & 1 & 2 & 2 & 3 \\ \end{array}
\begin{array}{c|cccc|ccc|c} 1 & 0 & 0 & 0 & -1 & 0 & 0 & -3 & -3 \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 1 & -1 & 3 \\ 0 & 0 & -1/5 & 0 & 2/5 & 0 & 2/5 & 1 & 1 \\ 0 & 0 & 0 & 1/5 & 2/5 & 0 & 2/5 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 2 & 2 & 3 \\ \end{array}
\begin{array}{c|cccc|ccc|c} 1 & 0 & -3/5 & 0 & 1/5 & 0 & 0 & 0 & 0 \\ \hline 0 & 1 & -1/5 & 0 & 2/5 & 0 & 7/5 & 0 & 4 \\ 0 & 0 & -1/5 & 0 & 2/5 & 0 & 2/5 & 1 & 1 \\ 0 & 0 & 1/5 & 1/5 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2/5 & 0 & 1/5 & 1 & 6/5 & 0 & 1 \\ \end{array}
From here you can see that (-17/7, 20/7, -1/7) is solution (optimal value -24/7), but contains negative entries. Correct answer written in textbook is (1,0,1) with optimal value 0. I suspect, that I have add x1,x2,x3 >= 0 constraints to tableau. Any help would be appreciated.