Why does putting linearly dependent vectors in a matrix and row reducing yield their coordinates? Say we have the vectors $a = (1,2,3,4),$ $b = (2,-1,1,1),$ $c = (-1,8,7,10)$ and $d = (-5, 5, 0, 1).$ They are linearly dependent and span a plane. We can prove this by putting them as columns in a matrix $$(a \:\:\: b \:\:\: c \:\:\: d) = \begin{pmatrix}1&2&-1&-5\\2&-1&8&5\\3&1&7&0\\4&1&10&1
\end{pmatrix},$$ and row reducing to find its rank. Row reducing yields $$ \begin{pmatrix} 
1 & 0 & 3 & 1\\
0 & 1 & -2& -3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{pmatrix},$$ we see the rank is two and that $a$ and $b$ span the plane.
My question is the following: if we let $a$ and $b$ be a basis for the plane that these four vectors span, we see that the first two rows tell us the coordinates with respect to this (ordered) basis, for example $c$ with respect to the basis $a$ and $b$, has coordinates $(3,-2)$ since $c = 3a -2b$. Why is it the case that this shows up in this row reduced matrix? I've been thinking about it for a while and I can't seem to prove it or even explain it with intuition.

Starting with the matrix $A$ as follows
$A = [ v_1 , v_2, ..., v_m ] $
Reduce it to reduced-row echelon form $\widetilde{A} $, this is equivalent to premultiplying $A$ with the matrix $E$
$ \widetilde{A} = E A = [E v_1, Ev_2, ..., E v_m] $
where $E v_k$ is either an elementary unit vector $e_i$ or a linear combination of the other $e_i$'s.  Thus if
$ v_k = \displaystyle \sum_{i=1, i \ne k}^m \alpha_i v_i $
Then
$ E v_k = \displaystyle \sum_{i=1, i \ne k}^m \alpha_i E v_i $
Thus we can identify the vectors that are linearly independent, by identifying those columns that reduce to different $e_i$'s, while the others are linearly dependent on them.  And the coefficients of this dependency are the entries in the reduced row echelon form for that column.
