following an example from here : https://en.wikipedia.org/wiki/Augmented_matrix
Given $A=\pmatrix{1 \;\;\;1 \;\;\;2 \\ 1 \;\;\;1 \;\;\;1 \\ 2\;\;\;2 \;\;\;2}, X=\pmatrix{x\\y\\z}, B=\pmatrix{3\\1\\2}$ forming the linear system: $ AX=B$
Performing Gaussian elimination (by hand), starting with this augmented matrix:
$\pmatrix{1 \;\;\;1 \;\;\;2 \;\;\;3 \\ 1 \;\;\;1 \;\;\;1 \;\;\;1 \\ 2\;\;\;2 \;\;\;2 \;\;\;2}$
After doing the following Row operations:
L3=L3-2L2
L2=L2-L1
L2=-L2
I stopped at this point: $\pmatrix{1 \;\;\;1 \;\;\;2 \;\;\;3 \\ 0 \;\;\;0 \;\;\;1 \;\;\;2 \\ 0\;\;\;0 \;\;\;0 \;\;\;0}$
it is clear that z=2 I decided to set x as a free variable and let x=t then i get: $\{x=t,y=1-t,z=2\}$
E.g, let $t=10$, then $X_{10}= \pmatrix{10\\ 1-10 \\ 2} =\pmatrix{10 \\ -9 \\ 2}$
and $AX_{10}=\pmatrix{5\\ 3 \\ 6}$
The true RREF matrix is one step later with L1=L1-2L2: $\pmatrix{\;\;\;\;1 \;\;\;\;\;\; 1 \;\;\;\;\; 0 \;\;\;\; -1 \\ 0 \;\;\;\;\; 0 \;\;\;\;\; 1 \;\;\;\;\; 2 \\ 0 \;\;\;\;\; 0 \;\;\;\;\; 0 \;\;\;\;\; 0}$
then again if x is the free param. we have this time: $\{x=t,y=-1-t,z=2\}$
(Please, note that i get this time $y=-1-t$ instead of $y=1-t$ before!)
E.g, let $t=10$, then $X_{{10}_{RREF}}= \pmatrix{10\\ -1-10 \\ 2} =\pmatrix{10 \\ -11 \\ 2}$
and this time 'it works' as $AX_{{10}_{RREF}} = \pmatrix{3\\1\\2}$ which is indeed equal to $B$.
My question is: why does it make a difference to go to the FULL (or true?) RREF form? Why can't I stop 1 step before? It seems very strange, like if the system was 'wrong' at the point before L1=L1-2L2 ... I am probably missing some fundamental point, but I hope that an answer to this will allow me to understand more about linear algebra (i know i could just say 'ok one must do the full RREF' but i m interested in the 'Why?')