What does it mean for a variable to disappear in a linear system? Let's say we have this matrix:
\begin{matrix}
1 & 3 & 0 & -1 &|& 2 \\
2 & 6 & -1 & -6 &|& 1 \\
1 & 3 & 1 & 3 &|& 5 
\end{matrix}
Through row operations, we obtain the matrix
\begin{matrix}
1 & 3 & 0 & -1 &|& 2 \\
0 & 0 & 1 & 4 &|& 3 \\
0 & 0 & 0 & 0 &|& 0 
\end{matrix}
so $x_1 = 2-3x_2+x_4$, $x_2$ is free, $x_3 = 3-4x_4$, $x_4$ is free.
We know this is an equivalent system because I performed valid algebraic operations on this system, and we have values for all variables, and there's no intermingling: $x_1$ is a function of strictly $x_2,x_4$, and $x_3$ is strictly a function of $x_4$. This makes complete sense and I typically am just like: "Ok, good, let's move on." But after staring at this perhaps longer than I should have, I was wondering why we know for a fact that this system must have 2 free variables: i.e. is it possible to express all variables in terms of one variable? For example, can $x_2$ be a function of $x_4$. When I stare at my solution, I think "obviously not: the $x_2$ magically disappears and thus cannot have any constraints" - I guess I'm wondering why this works and what this means.
Edit: After reading my post, I realize it may come across as confused ... I think I'm basically trying to shed light on a process that's been drilled into my head - produce leading one, clean columns, express basic variables in terms of free variables, and bam that's the answer - without me actually thinking of what's going on. I think the point is that what we have is an algebraically equivalent system, and so it just must work, and there must be two free variables because that's just how the algebra worked out - I guess I want some more light on what "it's just how the algebra worked out" means, if that's possible?
 A: $\DeclareMathOperator{\rref}{rref}$The linear relationships you see between the columns of $A$ are exactly the same relationships between the columns of $\rref(A)$.
In $A$, the second column is 3 times the first. In $\rref(A)$ the second column is 3 times the first.
In $A$, the fourth column is 4 times the third minus the first. Same for $\rref(A)$. The only difference is that it is easier to see this for $\rref(A)$.
Given a list of vectors $v_1, v_2, v_3, \dots, v_m$, we can form the matrix $\rref(v_1,\dots,v_m)$ by asking a sequence of questions:

*

*is $v_1 = 0$? If so, then the first column of $\rref(v_1,\dots,v_m)$ is $0$, otherwise it's $(1,0,0,\dots,0)^\top$.


*is $v_2 = a v_1$ for some $a$? If so then the second column of $\rref(v_1,\dots,v_m)$ is $(a,0,0,\dots,0)^\top$, otherwise it's your second pivot, $(0,1,0,0,\dots)^\top$.
If the $i$-th column is a pivot, let us call $v_i$ a pivot vector.


*is $v_3$ a linear combination of the pivot vectors among $v_1, v_2$? E.g. if $v_3 = av_1 + bv_2$ then the third column is $(a,b,0,0,\dots,0)^\top$. Otherwise $v_3$ is the next pivot vector.

and just continue like this


*is $v_{k + 1}$ a linear combination of the pivot vectors among $v_1,\dots,v_k$? E.g. if the pivot vectors are $v_3, v_4, v_6, v_7$ and $v_8 = 2v_3 + 0v_4 + v_6 - 4v_7$ then the $8$-th column becomes $(2,0,1,-4,0,0,0,\dots)^\top$. Otherwise $v_{k+1}$ is the next pivot vector.


Let $v_1,v_2,v_3,v_4,v_5$ be the columns of your first matrix—including the last column. Then

*

*$v_1 \neq 0$ so it's a pivot:

\begin{pmatrix} 1 &&&& \\ 0 &&&& \\ 0 &&&& \end{pmatrix}

*

*$v_2 = 3v_1$:

\begin{pmatrix} 1 & 3 &&& \\ 0 & 0 &&& \\ 0 & 0 &&& \end{pmatrix}

*

*$v_3$ is not a scalar multiple of the pivot vector $v_1$, so $v_3$ is the next pivot vector:

\begin{pmatrix} 1 & 3 & 0 && \\ 0 & 0 & 1 && \\ 0 & 0 & 0 && \end{pmatrix}

*

*the pivot vectors are $v_1, v_3$ and $v_4 = -v_1 + 4v_3$:

\begin{pmatrix} 1 & 3 & 0 & -1 & \\ 0 & 0 & 1 & 4 & \\ 0 & 0 & 0 & 0& \end{pmatrix}

*

*the pivot vectors are still $v_1, v_3$ and $v_5 = 2v_1 + 3v_3$:

\begin{pmatrix} 1 & 3 & 0 & -1 & 2 \\ 0 & 0 & 1 & 4 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}
The pivot vectors are linearly independent, and every column is either a pivot vector or a linear combination of pivot vectors. I.e. the pivot vectors form a basis (spanning + linearly independent) for the column space.
Which vectors are free and which are pivots depends on the order of your columns. If we swap $v_1$ and $v_2$ then $v_2$—which now comes first—is a pivot vector and $v_1 = \frac13 v_2$ is no longer a pivot because it comes after $v_2$ in the list. So if you swap the vectors around you can get a different basis. But, remember that different basis for the same space have the same number of vectors. You cannot have 2 pivot vectors in one basis and 3 pivot vectors in another.
Another way to look at it is as a parameterized set. You have pivot variables $x_1$ and $x_3$ which are linear functions of your two free variables $x_2$ and $x_4$—and also $x_2$ and $x_4$ are functions of $x_2$ and $x_4$. So your solution set is parameterized by these functions $x_1,x_2,x_3,x_4$. So the solution set is
$$\{f(x_2, x_4) = (2-3x_2+x_4, x_2, 3-4x_4, x_4) : (x_2, x_4) \in \mathbb{R}^2\}$$
Thus the solution set is the image of some function in two variables. It must be a plane. If there is only one variable, the image cannot be a plane because you cannot have a linear function in one variable hit every point in a plane. Functions in one variable give you lines.
A: You have 4 columns, i.e. 4 unknown variables $x_1,x_2,x_3,x_4$. After performing row operations you see that only $2$ non-zero rows remained in the matrix, the matrix has rank $2$, i.e. you have $2$ independent conditions only. I.e. $4-2= 2$ free parameters have to be introduced, for example $x_2=a,x_4=b$. Unknown values of $x_1,~x_3$ can be determined as a function of these parameters, because the remaining matrix is regular.
\begin{align}
x_1&=2-3x_2+x_4=2-3a+b,\\
x_3&=3-4x_4=3-4b.
\end{align}
All together:
\begin{align}
x_1&=2-3a+b,\\
x_2&=a,\\
x_3&=3-4b,\\
x_4&=b.\\
\end{align}
where $a,b\in\mathbb{R}$.
