What is the intuition behind the way of reading off a dual optimal solution from simplex primal tabular in CLRS?
Answer My Own Question:

I found that the interpretation in Section 8.3 of the textbook "Introduction to Linear Algebra; 4th Edition" by Gilbert Strang is quite intuitive and much easier to understand than that in CLRS.

Note: The textbook of Gilbert Strang takes the "min" program as the primal and the max program as the dual. I follow this convention. Besides, it adopts the matrix form of linear programs, which, in my opinion, helps a lot in understanding the simplex method and the duality theory.

Quoted from the textbook:


  The $m$ inequalities $Ax \ge b$ were changed to equations by introducing the slack variables $w = Ax - b$:
  $$\begin{bmatrix} A &amp; -I \end{bmatrix} \begin{bmatrix} x \\ w \end{bmatrix} = b$$ and
  $$\begin{bmatrix} x \\ w \end{bmatrix} \ge 0.$$
  Every simplex step picked $m$ columns of the long matrix $\begin{bmatrix} A &amp; -I \end{bmatrix}$ to be basic, and shifted them (theoretically) to the front. This produce $\begin{bmatrix} B &amp; N \end{bmatrix}$. The same shift reordered the long cost vector $\begin{bmatrix} c &amp; 0 \end{bmatrix}$ into $\begin{bmatrix} c_{B} &amp; c_{N} \end{bmatrix}$. The stopping condition, which brought the simplex method to an end, was $r = c_N - c_B B^{-1} N \ge 0$. At the stopping moment, the cost was as low as possible:
  $$cx^{\ast} = \begin{bmatrix} c_{B} &amp; c_{N} \end{bmatrix} \begin{bmatrix} B^{-1}b \\ 0 \end{bmatrix} = c_B B^{-1} b.$$
  If we can choose $y^{\ast} = c_B B^{-1}$ in the dual, we certainly have $y^{\ast}b = cx^{\ast}$. The minimum and maximum will be equal.


It turns out that $y^{\ast} = c_B B^{-1}$ is exactly the one read off in the final slack form of the primal program in the way CLRS describes. Its feasibility in the dual is due to the stopping condition $r = c_N - c_B B^{-1} N \ge 0$.