# Show that the solution space of $Ax=b$ varies continously with $b$, $A$ is $n \times (n+1)$

Suppose $$A$$ is a $$n \times (n+1)$$ matrix of rank $$n$$.

(a) Show that the one dimensional solution space of $$A x =b$$ varies continuously with $$b \in \mathbb{R}^n$$.

(b) Generalize

My attempt:

(a) The nullspace of $$A$$ has dimension 1. If I think of the RREF of $$A$$ obtained by premultiplication with an $$n \times n$$ matrix $$E$$, then the solution to $$Ax= b$$ is

$$x = \begin{pmatrix} Eb \\ 0 \\ \end{pmatrix} + \alpha \begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ where the second term on the right is a general vector in the nullspace of $$A$$.
Thus for RHS $$b_1, b_2$$, (now denoting the nullspace basis vector by $$e_n$$

$$||x_1 - x_2|| \leq ||E|| \cdot||b_1-b_2|| + |\gamma| \cdot ||e_n||, \forall \; \gamma \;\in \mathbb{R}$$.

This would prove the claim if $$\gamma=0$$, but as it stands, I do not see if this is correct. How do I show the claim ?

(b) I can think of

1. Rank of A = n, but dimension of nullspace > 1.

2. Rank of A < n. In this case it is not clear to me how to even begin, since there may be no solution for a given $$b \in \mathbb{R}^n$$

So I am puzzled by the point of this question.