Given that I have an augmented matrix in Row Echelon Form or Reduced Row Echelon form, and the bottom row(s) contain only zeroes.

Generally speaking (disregarding fringe cases if any?):

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    $\begingroup$ They don't, just means you have a lineraly dependant vector. Yes, if you have more variables than equations. Yes, because its free, it means it can be anything, hence infinite number of soltuons $\endgroup$ – XRBtoTheMOON Feb 11 '18 at 23:45
  • $\begingroup$ So it's incorrect to say that : if there is a row of zeroes, then there are infinitely many solutions $\endgroup$ – mathguy Feb 11 '18 at 23:46

Why do rows of only zero imply the presence of a free variable?

This is not true, consider matrices with more rows than columns like $$\left(\begin{matrix}1 & 0 \\ 0 & 1 \\ 0 & 0\end{matrix}\right).$$ Free variables are implied by columns without pivot element.

Can you have a free variable without the bottom row(s) containing only zeroes?

Yes, consider $$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0\end{matrix}\right).$$

Does the presence of a free variable always mean you have an infinite number of solutions?

This depends on the size of the base field. If it is infinite (like $\mathbb{Q}$ or $\mathbb{R}$) then yes. Otherwise no.

  • $\begingroup$ What about this thread math.stackexchange.com/questions/309724/… is the accepted answer saying that a row of zeroes implies the presence of free variables? $\endgroup$ – mathguy Feb 12 '18 at 0:08
  • $\begingroup$ I don't see an accepted answer there. But as I wrote: Free variables are implied by columns without pivot element. This is the correct characterization. $\endgroup$ – azimut Feb 12 '18 at 0:11
  • $\begingroup$ Sorry, i meant the top answer, but point taken ty $\endgroup$ – mathguy Feb 12 '18 at 0:12

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