When solving the following matrix equation:
$\begin{bmatrix} 1 & 3 & 3 & 2 \\ 2 & 6 & 9 & 7 \\ -1 & -3 & 3 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
I first turn $A$ into reduced row echelon form:
$R= \begin{bmatrix} 1 & 3 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
And then I write the pivot variables ($x_1$ and $x_3$) in terms of the free variables ($x_2$ and $x_4$):
$x_1 = -3x_2 + x_4 \\ x_3 = -x_4$
Which then becomes the solution to the equation:
$x= \begin{bmatrix} -3x_2 + x_4 \\ x_2 \\ -x_4 \\ x_4 \end{bmatrix} \\ x= x_2 \begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 1 \\ 0 \\ -1 \\ 1 \end{bmatrix}$
I understand that pivot variables can be written in terms of the free variables only. However, my question is why does writing the free variables in terms of the pivot variables invalid?
I understand that free variables cannot be expressed in terms of the pivot variables as other free variables get involved as well. However, why does it not work? I'm struggling to think of a reason for why it shouldn't work.
For instance, if I were to write the free variables in terms of the pivot variables:
$3x_2=-x_1+x_4\Rightarrow x_2 = -\frac{1}{3}x_1 + \frac{1}{3}x_4\\x_4=-x_3$
$x= \begin{bmatrix} x_1 \\ -\frac{1}{3}x_1 + \frac{1}{3}x_4 \\ x_3 \\ -x_3 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ -\frac{1}{3} \\ 0 \\ 0 \end{bmatrix} +x_3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ -1 \end{bmatrix} +x_4 \begin{bmatrix} 0 \\ \frac{1}{3} \\ 0 \\ 0 \end{bmatrix}$
I hope this question makes sense.'
Edit: I'm really sorry. I made a typo in my initial matrix $A$. The entry in row 1 and column 4 should be a $2$ not a $3$.