# Recovering a dependent column vector from a matrix after row reducing that matrix

This is an old exam question I remember how to do, but not why it makes sense. Take a matrix $A$:

$A=\left(\begin{array}{rrrr} 1 & 6 & 2 & -4 \\ -3 & 2 & -2 & -8 \\ 4 & -1 & 3 & 9 \end{array}\right)$

Then row reduce it:

$\text{rref}(A)=\left(\begin{array}{rrrr} 1 & 0 & \frac{4}{5} & 2 \\ 0 & 1 & \frac{1}{5} & -1 \\ 0 & 0 & 0 & 0 \end{array}\right)$

But then if you want the rightmost (in this case) linearly dependent vector from $A$ expressed as linear combination of column vectors from $A$, you can use the rightmost column of rref$(A)$ as weights;

$\left(\begin{array}{rrr} 1 & 6 & 2 \\ -3 & 2 & -2 \\ 4 & -1 & 3 \end{array}\right)$$\left(\begin{array}{r} 2 \\ -1 \\ 0 \end{array}\right)= \left(\begin{array}{r} -4 \\ -8 \\ 9 \end{array}\right)$

And you can use the second rightmost column of rref$(A)$ as weights if you want to express the second dependent column vector as a linear combination of column vectors from $A$.

Edit: With 'weights' I mean viewing the matrix multiplication above as a linear combination of the column vectors, where the vector you multiply with contains the weights, a la $2\bf{v_1}-1\bf{v_2}+0\bf{v_3}$, where $\bf{v_1,v_2,v_3}$ are the first three column vectors of $A$.

I've been trying to remember why this is obvious (is it obvious?) but I'm not getting anywhere. Thanks for any replies.

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$A=\left(\begin{array}{rrrr} 1 & 6 & 2 & -4 \\ -3 & 2 & -2 & -8 \\ 4 & -1 & 3 & 9 \end{array}\right)$
As an augmented matrix $A\vec{x}=\vec{y}$, with $\vec{y}={(-4,-8,9)}^T$. The the row reduction then solves for exactly the weights needed to create $\vec{y}$ out of $\bf{v_1,v_2,v_3}$.