Is vector W in the span of V1,V2,V3 
*

*$v_1=(1,1,0)$

*$v_2=(1,2,1)$

*$v_3=(1,3,2)$

*$w=(2,1,-1)$


Is $w$ in $span(v_1,v_2,v_3)$?
 A: Given $v_1=(1,1,0)^{\top}$, $v_2=(1,2,1)^{\top}$, $v_3=(1,3,2)^{\top}$, $w=(2,1,-1)^{\top}$
If $w \in span\{v_1, v_2, v_3\}$, then there are constants $c_1, c_2, c_3$ such that $w = c_1v_1 + c_2v_2 + c_3v_3$. That is,
$$[v_1\quad v_2 \quad v_3][c_1 \; c_2 \; c_3]^{\top} = w$$
$$\begin{pmatrix}1&1&1\\
1&2&3\\
0&1&2\end{pmatrix}\begin{pmatrix}c_1\\
c_2\\
c_3\end{pmatrix} = \begin{pmatrix}2\\
1\\
-1\end{pmatrix}$$
If there is a non-zero solution to this linear system, then w belongs to the span.
A: Yes. By inspection, observe that $w = 3v_1 - v_2$.
In general, this can be determined by forming an augmented system and using Gaussian Elimination.
A: Form a matrix and reduce: iff the row formed by $\;w\;$ vanishes the answer to your question is yes:
$$\begin{pmatrix}1&1&0\\
1&2&1\\
1&3&2\\
2&1&-1\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&0\\
0&1&1\\
0&2&2\\
0&-1&-1\end{pmatrix}\longrightarrow\begin{pmatrix}1&1&0\\
0&1&1\\
0&0&0\\
0&0&0\end{pmatrix}$$
Thus the answer is yes...and btw, only the first two vectors $\;v_1,v_2\;$ are enough to form $\;Span\,\{v_1,v_2,v_3\}\;$
A: You can easily verify that $v_1, v_2, v_3$  are linearly dependent, since their determinant is $0$.
Thus, you have that $\langle v_1,v_2,v_3\rangle=\langle v_1,v_2\rangle$ and $v_1, v_2$ are linearly independent.
Also, you can easily verify that  $v_1,v_2,w$ are linearly dependent, by checking the $3\times 3$ determinant, which is $0$. 
So, $w$ can be written as a linear combination of $v_1, v_2$.
