I'm in linear algebra and am struggling with this concept.
If I had a matrix like so:
|1 0 0 4|
|0 1 0 5|
|0 0 0 0|
Would my solution be x=4, y=5, and z can be any number?
Furthermore, what if I had a matrix like so:
|1 0 3 8|
|0 1 1 4|
|0 0 0 0|
Does this also mean z HAS to be my free variable, or could X OR Y be a free variable as well? I can't seem to wrap my head around what would be allowed to be my free variables in this example.
I'd appreciate some clarification/suggestions - especially on what can be allowed as a free variable in that second matrix.
EDIT:
What about a matrix like this: $\begin{bmatrix}1&1&3&2&8\\0&1&1&1&4\\0&0&0&0&0 \end{bmatrix}$
Could the variable in column 1 be free?
In the first case $z$ is the only variable that can be chosen as free.
For the second system, the solutions have the form \begin{cases} x+3z=8 \\ y+z=4 \end{cases} and $z$ can be chosen as the free variable, but also $x$ or $y$ can.
Indeed, if we solve for $z$, we have $z=(8-x)/3$ and therefore $$ y=4-z=4-\frac{8-x}{3}=\frac{4+x}{3} $$ Try for $y$ as free.
What is the difference? In both systems, the third column is a linear combination of the first two columns. But in the second system any of the first three columns is a linear combination of the other two, which is not the case for the first system.