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It is well known that for $A \in \mathbb{K}^{n \times m}$ and $b \in \mathbb{K}^n$ the system $$ Ax = b $$ is solvable if and only if $\operatorname{rank}(A) = \operatorname{rank}(A \mid b)$.

I suppose, in practice one would just use Gauß-Algorithm. However, is there any criterion for $\operatorname{rank}(A) = \operatorname{rank}(A \mid b)$?

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$ {\rm rank}( A ) = {\rm rank}( A | b ) $ if and only if $ b $ is in the column space of $ A $. Is that what you are looking for?

Your question is hard to interpret.

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  • $\begingroup$ Hmm, true, I guess it probably doesn‘t go much better than this obvious criterion. And yeah, I like the question to have room for interpretation. Thanks. $\endgroup$ – Qi Zhu Oct 8 '18 at 6:21

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