1. Lets show $x_q$ can't leave the basis
Intuitively
If the row $s$ contains all zeros except in the column of $x_q$, it means the problem contains a constraint of the type
$$x_q = b_s$$
In that case, only solutions with $x_q = b_s$ are feasible, so the variable $x_q$ never leaves the basis.
Mathematically
$x_q$ leaves if there is some non-basic variable $x_r$ that enters because
$$z_r - c_r < 0$$
$$z_r - c_r = \min_j (z_j - c_j)$$
$$\frac{b_q'}{a_{qr}'} = \min_i \{\frac{b_i'}{a_{ir}'} | a_{ir}' > 0 \}$$
It is true, assuming the problem is about maximization.
Now, lets apply this to the case where the row corresponding to $x_q$ contains only zeros except in the column corresponding to $x_q$. You have
$$z_q = 1\times c_q = c_q$$
So $z_q- c_q = 0$, which is normal since $x_q$ is a basic variable.
Now, let $x_r$ be the entering variable. From the asumptions of the problem, we know that $a'_{qr} = 0$. This implies
$$\min_i \{\frac{b_i'}{a_{ir}'} | a_{ir}' > 0 \} \neq \frac{b_q'}{a_{qr}'}$$
This means the variable $x_q$ never leaves the basis.
2. Let's show row $s$ can't change
Intuitively
Lets call $x_r$ the entering variable, and $x_t$ the exiting variable.
Now lets see how the pivot will affect the column of $x_q$. We have
Line $i$ of the new tableau is a linear combinaison of lines $i$ and $t$ of the old tableau. Since these value are 0 for the column of $x_q$, the column of $x_q$ will never change. The reduced cost of variable $x_q$ will always be 0 and $x_q$ will never leave the basis.
Mathematically
Lets call $x_r$ the entering variable, and $x_t$ the exiting variable.
When making the pivot for row $s$, the formulae is (assuming $a'_{ij}$ is the new value in the tableau and $a_{ij}$ is the old value.
$$a'_{si} = a_{si} - \frac{a_{qr}}{a_{tr}} a_{ti}$$
Since $a_{qr} = 0$, the row $s$ stays unchanged.
No, this need not be true. In case of degeneracy, you will change the basis but not improve the objective value. Geometrically this means that you are in a vertex of the polyhedron describing the feasible set and the change of basis does not lead you to a new vertex. Hence, you are stuck in the same vertex for one (or possibly more) iterations and you do not improve the objective value. Degeneracy occurs when there are redundant constraints in the problem.
Best Answer
Suppose you have a maximization problem. When a variable x enters basis and y leaves out of basis, it means that increasing x would increase the objective function. But objective function has a coefficient of
-Mfor artificial variables. So we cannot increase objective function value by increasing the artificial variable. Hence it cannot enter your basis