As a rote student, I recently have started to try to learn the 'un-rote' way, but the ways in which the institutions do so is very limited and I'm stuck at the very first classes of linear algebra.

Technically, they say: A free variable is a variable with no pivot column, as if it would make any sense for students to interpret!.

Consider the following matrix:

$$ \begin{pmatrix} 1 & 3 & -5 & 0 \\ 0 & 1 & -1 & -1\\ \end{pmatrix} $$

which is equivalent to:
$$ \begin{pmatrix} 1 & 0 & -8 & -1\\ 0 & 1 & 4 & -1\\ \end{pmatrix} $$

Now the free variable isenter image description here z. But what does it actually represent? As I plot the equations in geogebra, I basically can make no sense out of it:

enter image description here

And this is what all the vectors look like as they are in 2D space:
enter image description here

How can I make sense of what the free variable is geometrically?

EDIT: If not geometrically(as mentioned below, what is it(with a little more sense rather than the non-sensemaking technical above in bold above)?)

  • $\begingroup$ To some extent the concept of a free variable is not intrinsically geometric. The result of row reduction is sensitive to the order in which you present the variables; this in turn affects the location of the pivot columns, and hence also which variables are free. Since the graphs of the equations are not sensitive to the naming of the variables, it should be impossible to read off the free variables by looking at only the graphs. Essentially, a variable doesn't "know" that it's free, because the label is bestowed by the process (though the number of free variables is invariant). $\endgroup$ – Erick Wong Oct 13 '17 at 6:32
  • $\begingroup$ Actually, could you edit your question to explain where the graphs come from? I don't understand why you are graphing $z=2$, which does not come out of any of the equations. $\endgroup$ – Erick Wong Oct 13 '17 at 6:38
  • $\begingroup$ @ErickWong, since z was the free variable, it could have any value(right?), so I just put z=2 $\endgroup$ – mathmaniage Oct 13 '17 at 6:51
  • $\begingroup$ Ah, I think it would be more productive to plot the locus of points $(x,y,z)$ that arise from choosing every possible choice of free variables (i.e. $z$). That’s your actual solution space (which will be a single line in 3D). $\endgroup$ – Erick Wong Oct 13 '17 at 6:53
  • $\begingroup$ That would mean entire R^3? $\endgroup$ – mathmaniage Oct 13 '17 at 6:53

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