Reduced Echelon Form Question Suppose we transform an augmented matrix ( originated form a system Ax=b ) into reduced row echelon form like in this example (original image here)
$$\begin{array}{lr}
\left[\begin{array}{cccc|c}
\color{red}{1}&2&0&3&2\\
0&0&\color{red}{1}&-2&5\\
0&0&0&0&0
\end{array}\right]=\operatorname{rref}(A)&&&
\begin{align*}
\color{red}{x_1}+2x_2\quad\,+3x_4&=2\\
\color{red}{x_3}-2x_4&=5
\end{align*}\\ \hline
\text{Red shows pivot variables; the others are free.}
\end{array}$$
Why we need to solve for $x_1$ and $x_3$ ? what so special aobut this variables ? Why cant we solve for $x_2$ and $x_3$ for example ? or for $x_2$ and $x_4$ ?      
Why do some columns show up with pivot and we call the corresponding variables the free variables ? Why every variable doesn't have the same importance as the others ?     
Thanks
 A: There is nothing special about $x_1$ and $x_3$. By elementary row operations you can change $x_2$ and $x_4$ to be pivot variables with $x_1$ and $x_3$ as free variables by the following:
$$\begin{array}{lr}
\left[\begin{array}{cccc|c}
\frac{1}{2}&\color{red}{1}&\frac{3}{4}&0&\frac{19}{4}\\
0&0&-\frac{1}{2}&\color{red}{1}&-\frac{5}{2}\\
0&0&0&0&0
\end{array}\right]&&&
\begin{align*}
\frac{1}{2}x_1+\color{red}{x_2}+\frac{3}{4}x_3\quad\quad&=\frac{19}{4}\\
-\frac{1}{2}x_3+\color{red}{x_4}&=-\frac{5}{2}
\end{align*}\\ \hline
\end{array}$$
However it is not possible to choose arbitrary variables as pivot variables. We can write $A$ in the form $[a_1,...,a_4]$, where $a_i$'s are the column vectors of $A$. Solving they system $Ax=b$ is essentially to find a linear combination of  $(a_1,...,a_4)$ such that 
$$
x_1a_1+...+x_4a_4=b.
$$
The textbook method is to find a particular solution $x^p$ such that
$$
x^p_1a_1+...+x^p_4a_4=b
$$
and a homogeneous solution $x^h$ such that
$$
x^h_1a_1+...+x^h_4a_4=0.
$$
Summing them together we get
$$
(x^p_1+x^h_1)a_1+...+(x^p_4+x^h_4)a_4=b
$$
The general solution is given $x=x^p+x^h$. What determines the pivot variables is the homogeneous equation. The Gaussian elimination reduces the problem to the echelon form. Let $\text{rref}(A)=(a_1^*,...,a_4^*)$, solving the homogeneous equation is equivalent to finding 
$$
x^h_1a_1^*+...+x^h_4a_4^*=0.
$$
Now  we need to use linear combination of $a_1^*,...,a_4^*$ to form the $0$ vector. We can choose two vectors, say $a_2^*,a_3^*$, and allow their coefficients $x_2,x_3$ to vary. Now 
$$
x_2a_2^*+x_3a_3^*\neq 0
$$
in general. To get the homogeneous equation, we need to set the coefficients of $a_1^*$ and $a_4^*$ to vary accordingly such that
$$
x_1a_1^*+x_4a_4^*=-(x_2a_2^*+x_3a_3^*)
$$
Thus we have chosen $x_1$ and $x_4$ to be pivot variables and $x_2$ and $x_3$ to be free variables. Now you can see why $x_1$ and $x_2$ cannot be pivot variables. As in the reduced echelon form
$$
x_1a_1^*+x_2a_2^*=-(x_3a_3^*+x_4a_4^*)\Leftrightarrow x_1\begin{bmatrix}1\\0\\0\end{bmatrix}+x_2\begin{bmatrix}2\\0\\0\end{bmatrix}=-x_3\begin{bmatrix}0\\1\\0\end{bmatrix}-x_4\begin{bmatrix}3\\-2\\0\end{bmatrix}
$$ 
cannot be satisfied since the left hand side only has nonzero first coordinate and cannot be manipulated to match the right hand side.
A: There is nothing special about $x_1$ and $x_3$. If you want to make $x_2$ a pivot variable instead of $x_1$ then do the following:
$$ x_1 + 2x_2 + 3x_4 = 2 $$
$$ 2x_2 + x_1 + 3x_4 = 2 $$
Divide by 2
$$ x_2 + \frac{1}{2}x_1 + \frac{3}{2}x_4 = 1 $$
Now $x_2$ and $x_3$ are the pivot variables.
A: As I mentioned in chat, there's nothing particularly special about the pivot variables, other than they happen to be the terms that, by our construction, end up with unity coefficients.
This simplifies things in that you don't need to do any division to get the result you want. But in the end, $x_1, x_2, x_3, x_4$ are all independent variables in your system. Writing $x_3$ in terms of $x_4$ is no better or worse than writing $x_4$ in terms of $x_3$ in this case.
There is some notion that the columns that these variables appear in form a basis (indeed, in reduced row echelon form, the standard basis) for your vector space. But once you know how to change bases, then this is of minimal value, except in terms of economy of solution-finding.
