Solving LP from tableau $$\begin{array}{cccccc}
      &  x1 & x2 & x3 & x4&  x5 \\
    -4&  2  & 0&   -2 &  0&   3\\
     3 & 1 &  0 &  -1&   1 &  3\\
     2  &0&   1  & 0&   0.5 & 2\\

\end{array}$$
When I learned about solving LP represented by the tableau in class, I thought I need to select the column that has negative value at the top. According to that, x3 is the only column that has -2 value. However, it's subsequent values -1 in row 1 and 0 in row 2 are neither positive, which I need it to pivot the row.
Am I missing something here? 
The top row with z =-4, x1=2, x3=-2 is the cost function

$$\begin{array}{cccccc}
      &  x1 & x2 & x3 & x4&  x5 \\
    -4&  2  & 0&   -2 &  0&   3\\
     3 & 1 &  0 &  -1&   1 &  3\\
     2  &0&   1  & 0&   0.5 & 2\\

\end{array}$$
To begin with a bfs
$$\begin{array}{cccccc}
       & x1 & x2 & x3 & x4 & x5 \\
    -10&  0 &  0 &  0 & -2 & -3 \\
     3 &  1 &  0 & -1 &  1 &  3 \\
     2 &  0 &  1 &  0 &0.5 &  2 \\
\end{array}$$
Suppose we pick column4, then row 2 is most profitable
$$\begin{array}{cccccc}
       & x1 & x2 & x3 & x4 & x5 \\
    -2 &  0 &  4 &  0 &  0 &  5  \\
    -1 &  1 & -2 & -1 &  0 & -1\\
     4 &  0 &  2 &  0 &1 &  4 \\
\end{array}$$
Now all cost are positive and we are done. Optimal value of z is -2.
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Did you miss your objective functions? And you need to find a basis(basic feasible solution) first.
