If $A$ is a singular square matrix, then $Ax = b \neq \vec{0}$ has $0$ or many solutions

Question

I was reading this pdf:

https://www.math.ohiou.edu/courses/math3600/lecture10.pdf

and it tells you that

if $A$ is a singular square matrix, then $Ax = b \neq \vec{0}$ has $0$ or many solutions.

My question is: when does it have $0$ and when it has many solutions? In other words, how does $b$ influence the number of solutions for this linear system of equations in this case?

Note, I'm not asking why such a system of equations doesn't have a unique solution.

Answer 1

The system of equation $Ax=b$ has solutions if and only if $\operatorname{rank}(A)=\operatorname{rank}(Ab)$ (the augmented matrix).

If so, the set of solutions is an affine subspace with codimension $r=\operatorname{rank}(A)$.