The variable that is going to enter the basis with revised Simplex already is in it I am looking to solve the following program using the revised Simplex method :
$$
\begin{cases}
\begin{aligned}
\max \ & x_1&+ x_2\\
&x_1&-x_2&\le 1\\
&-2x_1&-x_2&\le -2\\
&-2x_1&+x_2&\le 2\\
\forall i, x_i
\end{aligned}
\end{cases}
$$
Yet, I have a problem, the variable that is going to enter the matrix already is in it!
It all started with the following basis for the first iteration : $J_0=\{x_2;x_3;x_5\}$. I already did an iteration that didn't offered any difficulty and led me to the following bases $J_1=\{x_2;x_4;x_5\}$. It also provided me $\hat{A_4}$ and $\hat b$:
\begin{align*}
\hat{A_{4}}&=(A^{J_0})^{-1}A^4\\
&=\begin{pmatrix}
0 & -1 & 0\\
1 & -1 &0\\
0 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
0\\ 1\\ 0
\end{pmatrix}\\
\hat{A_{4}}&=\begin{pmatrix}
-1\\-1\\1
\end{pmatrix}
\end{align*}
And naturally $\hat b_4=\begin{pmatrix}1\\-2\\2\end{pmatrix}$
Then I follow up with Iteration 1: as far as I took the minimum $>0$ between $\hat A_2/b=-1,\hat A_3/b=1/2,\hat A_5/b=1/2$, taking therfore $x_3$.
I first have 
$$(A^{J_1})^{-1}=D_1(A^{J_0})^{-1}$$
Then I calculate the change of basis matrix :
$$D_1=\begin{pmatrix}1 & 1 & 0\\
0 & -1 & 0\\
0 & 2 & 1\end{pmatrix}$$
Therfore 
\begin{align}(A^{J_1})^{-1}&=
\begin{pmatrix}1 & 1 & 0\\
0 & -1 & 0\\
0 & 2 & 1\end{pmatrix}
\begin{pmatrix}0 & -1 & 0\\
1 & -1 & 0\\
0 & 1 & 1\end{pmatrix}\\
(A^{J_1})^{-1}&=\begin{pmatrix}1 &-2 &0\\
-1 & 1 & 0\\
2 & -1 & 1\end{pmatrix}
\end{align}
Then I calculated 
\begin{align}\Pi&=c(A^{J_1})^{-1}\\
&=\begin{pmatrix} 1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}1 &-2 &0\\
-1 & 1 & 0\\
2 & -1 & 1\end{pmatrix}\\
\Pi&=\begin{pmatrix}1 & -2 & 0\end{pmatrix}
\end{align}
It allows me to calculate the new $\hat c = c - \Pi A$
\begin{align*}
\hat c &=\begin{pmatrix}
1 & 1 & 0 & 0 & 0
\end{pmatrix}-
\begin{pmatrix}
1 & -2 & 0
\end{pmatrix}
\begin{pmatrix}
1 & -1 & 1 & 0 & 0\\
-2 & -1 & 0 & 1 & 0\\
-2 & 1 & 0 & 0 & 1
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 1 & 0 & 0 & 0
\end{pmatrix}-
\begin{pmatrix}
5 & 1 & 1 & -2 & 0
\end{pmatrix}\\
\hat c&=\begin{pmatrix}
-4 & 0 & -1 & 2 & 0
\end{pmatrix}
\end{align*}
Theferore $x_4$ entered the matrix. Yet, this is impossible because it already is!

for $J_0=\{x_2;x_3;x_5\}$, the determinant is
$$\begin{vmatrix}
-1&1&0\\
-1&0&0\\
1&0&1
\end{vmatrix}=-1$$ this base is feasible.
Yet, for $J_1=\{x_2;x_4;x_5\}$ $$\begin{vmatrix}
-1&1&0\\
-1&0&1\\
1&0&0
\end{vmatrix}=0$$ It is therfore unfeasible.
