I've chosen a simple example to help me understand duality and complementary slackness. Suppose we have linear program:
$$Maximize\quad 5x_1+x_2\quad s.t$$ $$2x_1+x_2 \le 6$$ $$x_1+x_2 \le 4$$ $$2x_1+10x_2 \le 20$$ $$x_1,x_2 \ge 0$$
$$Solution \quad is \quad (3,0)$$
Now we have a dual $$Maximize\quad 6y_1+4y_2+20y_3\quad s.t$$ $$2y_1+y_2+2y_3 \ge 5$$ $$y_1+y_2+10y_3 \ge 1$$ $$y_1,y_2 \ge 0$$
In what way can the solution of the first problem help us find the solution to the dual.
$$\texttt{Maximize}\quad x_1+2x_2+3x_3\quad s.t$$ $$x_1+2x_2 +x_3 \le 1$$ $$x_1+x_2 +3x_3 \le 2$$ $$x_1,x_2,x_3 \ge 0$$ $$Optimal solution is (0,\frac{1}{5},\frac{3}{5})$$
$$\texttt{Minimize}\quad y_1+2y_2\quad s.t$$ $$y_1+y_2 \ge 1$$ $$2y_1+y_2 \ge 2$$ $$y_1+3y_2 \ge 3 $$ $$y_1,y_2 \ge 0$$
$$\texttt{Maximize}\quad x_1+2x_2+3x_3\quad s.t$$ $$x_1+2x_2 +x_3 +s_1= 1$$ $$x_1+x_2 +3x_3+s_2= 2$$ $$x_1,x_2,x_3,s_1,s_2 \ge 0$$
$$\texttt{Minimize}\quad y_1+2y_2\quad s.t$$ $$y_1+y_2 -z_1= 1$$ $$2y_1+y_2 -z_2=2$$ $$y_1+3y_2 -z_3= 3 $$ $$y_1,y_2,z_1,z_2,z_3 \ge 0$$
$x_j\cdot z_j=0 \ \forall \ \ j=1,2, \ldots , n \quad (\color{blue} I)$
$y_i\cdot s_i=0 \ \forall \ \ i=1,2, \ldots , m \quad (\color{blue}{ II})$
$x_2=\frac{2}{5}$ so $x_2*z_2=0$ therefore $z_2=0$. Going through and doing this gives me values of $0$ for $y_1$ and $y_2$ since the $x_1,x_2$ have zero slack so what am I doing wrong?