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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.

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    $\begingroup$ If $b\in \mathcal R(A)$, then the equation has infinitely many solution, else there's no solution. $\endgroup$ – BigbearZzz May 1 '16 at 13:08
  • $\begingroup$ I don't get your question. What do you mean with $b$ influencing the number of solutions? Can you give an example of an answer you might expect (even if it is wrong)? $\endgroup$ – Git Gud May 1 '16 at 13:12
  • $\begingroup$ @GitGud I suppose that he was referring to the situation where $b$ is (or isn't) in the range of $A$, a singular matrix. $\endgroup$ – BigbearZzz May 1 '16 at 13:19
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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 $r=\operatorname{rank}(A)$.

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