Show that $u, v, w$ are in the span of $\{u+v, 2u+3v, 4v+6w\}?$ I know this has to do with linear combinations, namely that you would set out to solve the following set of equations to show that $c_{1}, c_{2}$, and  $c_{3}$ exist and are not all 0, but I'm unclear as to how I actually solve for those in this case. 
That is, I know I should have these equations:
$u = c_{1}(u+v) + c_{2}(2u+3v) + c_{3}(4v+6w)$
$v = c_{1}(u+v) + c_{2}(2u+3v) + c_{3}(4v+6w)$
$w = c_{1}(u+v) + c_{2}(2u+3v) + c_{3}(4v+6w)$
Do you not need to solve for c to do this proof?
 A: HINT: Notice that $(2u+3v)-2(u+v)=v$.
A: HINT: What is the rank of the following matrix?
$$
\begin{pmatrix}
  1 & 2 & 0 \\
  1 & 3 & 4 \\
  0 & 0 & 6
\end{pmatrix}
$$
A: If you write the matrix $\begin{bmatrix}u & v & w\end{bmatrix}$ with $u,v,w$ as columns, then the linear combination $au + bv + cw$ can be written as
$$au + bv + cw = \begin{bmatrix}u & v & w\end{bmatrix} \begin{bmatrix}a \\ b \\ c\end{bmatrix}.$$
Therefore,
$$\begin{bmatrix}u+v & 2u+3v & 4v+6w\end{bmatrix}
=
\begin{bmatrix}u & v & w\end{bmatrix}
\begin{bmatrix}1 & 2 & 0\\1 & 3 & 4\\0 & 0 & 6\end{bmatrix}.
$$
Now $u,v,w$ are linear combinations of $u+v, 2u+3v, 4v+6w$ if and only if there is some $3 \times 3$ matrix $A$ such that
$$
\begin{bmatrix}u & v & w\end{bmatrix}
=
\begin{bmatrix}u+v & 2u+3v & 4v+6w\end{bmatrix}
A
=
\begin{bmatrix}u & v & w\end{bmatrix}
\begin{bmatrix}1 & 2 & 0\\1 & 3 & 4\\0 & 0 & 6\end{bmatrix}
A.
$$
Does such a matrix exist?
A: The solution I got is as follows:
$u = 3(u+v) + -1(2u+3v) + 0(4v+6w)$
$v = -2(u+v) + 1(2u+3v) + 0(4v+6w)$
$w = \frac{4}3(u+v) + \frac{-2}3(2u+3v) + \frac{1}6(4v+6w)$
