do the four vectors span R^3? why or why not? (I already have the answer to this question but i just need someone to explain the concept behind the logic being used here.)
v1 = (1,0,2)
v2 = (3,-1,1)
v3 = (2, -1,-1)
v4 = (4,-1, 3)
So my professor told us to write the vectors above in the equation below.
$$\pmatrix{1&3&2&4\\0&-1&-1&-1\\2&1&-1&3}\cdot\pmatrix{x\\y\\z\\w}=\pmatrix{b1\\b2\\b3}$$
(b1, b2, and b3 are arbitrary and can equal ANY vector in R^3)
he then used row reduction to get the solutions for x, y, z and w and we got the matrix below
$$
A= \begin{pmatrix} 1 & 3 & 2 & 4 & |b1\\ 0 & -1 & -1 & -1 & |b2\\0 & 0&0&0 &| b3 - 2b1 - 5b2
\end{pmatrix}
$$
so it is obvious that there is an inconsistency on the last row of the matrix above,
0 = b3 - 2b1 -5b3
but my question is, why does this inconsistency tell me that the vectors do NOT span R^3?
can someone explain to me why the vectors do not span R^3?
 A: The inconsistency in that system shows that there is no solution to $xv_1+yv_2+zv_3+wv_4=b$ provided $b_3-2b_1-5b_2\neq 0$. Since $b\in\Bbb R^3$, can we have that $v_1,v_2,v_3,v_4$ span $\Bbb R^3$?
Addendum: Recall that $$\operatorname{span}(v_1,v_2,v_3,v_4)=\{v\in\Bbb R^3:v=c_1v_1+c_2v_2+c_3v_3+c_4v_4,~c_i\in\Bbb R\}$$ That is, the span of a collection of vectors is the set of linear combinations of those vectors. So the inconsistency in the system you have shows us that there is no solution to $xv_1+yv_2+zv_3+wv_4=b$ for an arbitrary vector $b\in\Bbb R$. Hence, $b$ is not a linear combination of $v_1,v_2,v_3,v_4$. So can we say that $v_1,v_2,v_3,v_4$ span $\Bbb R^3$?
In general, to show some vectors do not span a vector space, we can just show that there is a vector in the space which is not a linear combination of those vectors. Linear dependence does not imply that they do not span $\Bbb R^3$. For example, $e_1,e_2,e_3,e_1+e_2$ span $\Bbb R^3$ however they are clearly linearly dependent. In fact, any collection of more than $3$ vectors will be linearly dependent in $\Bbb R^3$, however they may or may not span $\Bbb R^3$.
A: Notice that
$v1+v2=v4$
$v2-v1=v3$
So with these linear dependencies the largest possible span would be a plane in $\mathbb{R}^3$. If you think if these vectors geometrically, what is happening is that you have all 4 vectors pointing in different directions indeed but they all lie inside the same plane embedded in the three dimensional space. If this is true then there is no way to take a linear combination of them and come out of that plane.  
