Q:

Reduced cost zero for the two-phase Simplex?

I cannot understand the line -12, -4, -5, 1, 1, -1, 0, 0, 0. Now the formula $\bf c - \bf A ^t \bf y$ when $c=0$ will result into the line. It is just many times a dot product: for example, 0-(1*5+1*1+1*6)=-12. But why is $c=0$ here? 

This problem is a part of two-phase Simplex but probably the same with most Simplexes.
P.s. Page 116 in the Bertsimas -book (1997).

A:

Wrong!

"Now the formula $c−A^ty$ when $c=0$ will result into the line." You would get something like $(1\times 8)-(3\times 8)$ so the matrix dimensions are totally wrong.

Correct

The reduced cost formula is $\bar c_j =c_j - c_B^T B^{-1} A_j$, more about the reduced calculations here. By this you can calculate the top line.
The example
$$\min y_1+y_2+y_3$$ so $c_1=c_2=c_3=c_4=c_5=0$, $c_6=c_7=c_8=1$, $x_B=(x_6,x_7,x_8)$ and $c_B=(1,1,1)$. And $B=I_3$, the 3-identity-matrix. Then for example $$\bar c_1=c_1-c_B^T A_1=-(1,1,1)^T (2,1,2)=-(2+1+2)=-5.$$

Your homework is to read the page 84 in Bertsimas again.
Perhaps related

Optimality conditions and Directions in Simplex method

Reduced cost vector in the phase I of the Two-phase simplex?

Optimum exists but not extreme point in Standard Form LP problem?

