Finding the linear combination of a vector that is in a span So say we have
Span S =
$
        \begin{Bmatrix}
        \begin{bmatrix}
        1 \\
        0 \\
        1 \\
        \end{bmatrix}
        \begin{bmatrix}
        -1 \\
        1 \\
        1 \\
        \end{bmatrix}
        \begin{bmatrix}
        1 \\
        1 \\
        3 \\
        \end{bmatrix}
        \end{Bmatrix}
$
I know that 
$
        \begin{bmatrix}
        -1 \\
        4 \\
        7 \\
        \end{bmatrix}
$
is in the span because the reduced row echelon form of [A v] is consistent. Now 
how do I find what exactly the linear combination that makes the vector?
 A: $$\begin{pmatrix}-1\\4\\7\end{pmatrix}=3\begin{pmatrix}1\\0\\1\end{pmatrix}+4\begin{pmatrix}-1\\1\\1\end{pmatrix}.$$
$$\begin{pmatrix}-1\\4\\7\end{pmatrix}=-5\begin{pmatrix}1\\0\\1\end{pmatrix}+4\begin{pmatrix}1\\1\\3\end{pmatrix}.$$
A: Your three vectors $u_1,u_2,u_3$  are not independent since
$$u_3-u_2=2u_1$$
thus we will write $v=(-1,4,7)$ as a combination of $u_1$ and $u_2$.
from here, it is easy to see that
$$v=4u_2+3u_1=u_1+3u_2+u_3.$$
A: You can read the solutions in the reduced row echelon form of the augmented matrix:
\begin{align}
\begin{bmatrix}1&-1&1&-1\\0&1&1&4\\1&1&3&7\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&-1&1&-1\\0&1&1&4\\0&2&2&8\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&-1&1&-1\\0&1&1&4\\0&0&0&0\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&2&3\\0&1&1&4\\0&0&0&0\end{bmatrix}
\end{align}
The last form says the coefficients $x,y,z$ of a linear combination satisfy the relations:
$$x=-2z+3,\quad y= -z+4,\quad\text{or, in matrix form}\quad
\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\4\\0\end{bmatrix}-z\begin{bmatrix}2\\1\\-1\end{bmatrix}.$$
There remains to choose the value of $z$ at your convenience.
