In 2-phases simplex method what kind of operations must be done to get the canonical form tableau?
In this step(phase 2 of 2-phases method) after the remotion of artificial variables columns of auxiliary problem and copy of coefficents of objective functions, I don't know what are the operations performed to get canonical form tableau (as you see in the picture)
the objective function was
To get the matrix back in canonical form, you simply need to make sure that any basic variable has a 0 coefficient in the objective function row. So looking at the matrix
$$\begin{bmatrix} 1&0&-1&-1&0&2 \\ 0&0&-4&-4&1&10 \\ 0&1&3&1&0&4 \\ -2&-1&2&0&0&0 \\ \end{bmatrix}$$
You simply need to add 2 times the first row to the last (objective row) and 1 times the third row to the last (objective row). This will make sure there is a coefficient of 0 in the objective row for the first two columns, since they are the columns that contain the basic variables.
Doing this, you should get that after adding $R4 + 2R1$ you get $$\begin{bmatrix} 1&0&-1&-1&0&2 \\ 0&0&-4&-4&1&10 \\ 0&1&3&1&0&4 \\ -2&-1&0&-2&0&4 \\ \end{bmatrix}$$ This makes sure that the basic variable $x_1$ has a 0 coefficient in the objective function row. Doing the same for the basic variable $x_2$, adding $R4+R3$, you should get:
$$\begin{bmatrix} 1&0&-1&-1&0&2 \\ 0&0&-4&-4&1&10 \\ 0&1&3&1&0&4 \\ -2&0&-3&-1&0&8 \\ \end{bmatrix}$$
Which is the matrix in canonical form.