Modifying primal constraint in a LP problem

Suppose we have a primal-dual pair in standard form, Add a scalar multiple of one primal constraint to another primal constraint. Does this change the dual solution?

Try

Supose we have primal $$\max cx$$ subjeect to $$Ax = b$$, $$x \geq 0$$ and the $${\bf dual}$$ then is given by $$\min yb$$ subject to $$yA \geq c$$ and $$y \; free$$. Consider rows $$i$$ and $$j$$ of $$A$$ :

$$a^i x_k = b_i \; \; \; \; and \; \; \; \; \; a^j x_k = b_j$$

$$k=1,...,n$$. Let's perform what we are asked: Let $$\alpha$$ be scalar so that

$$(a^i + \alpha a^j) x_k = b_i+\alpha b_j$$

So the ith dual variable coefficient changes. But, is it a multiple of $$\alpha$$ or not?

Let $$E$$ be the corresponding elementary matrix corresponding to adding $$\alpha$$ times the $$j$$-th row to the $$i$$-th row.

Let $$w$$ be the original dual variable, then we have

$$y=E^{-T}w$$

from the working here.

$$E^T$$ be the corresponding elementary matrix corresponding to adding $$\alpha$$ times the $$i$$-th row to the $$j$$-th row.

$$E^{-T}$$ be the corresponding elementary matrix corresponding to adding $$-\alpha$$ times the $$i$$-th row to the $$j$$-th row.

Hence the $$j$$-th entry of the dual would change, $$y_j = w_j-\alpha w_i.$$