Determine whether the vectors span $\mathbb{R}^3$ They want us to determine whether it span $\mathbb{R}^3$ and they gave us these vectors 
$V_1=(1,2,6), V_2=(3,4,1), V_3=(4,3,1), V_4=(3,3,1)$
and the answer is that the vectors span $\mathbb{R}^3$ . But I don't know the reason 
 A: You just need to show
$$v_4 = \alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \alpha_3 \cdot v_3$$
A: A way to do this is that there exist a linear combinaison of these vectors that gives you the canonical basis.
Here 
$$V_3 - V_ 4 = (1,0,0) = e_1$$
$$V_2 - V_ 4 = (0,1,0) = e_2$$
$$\frac{1}{6}V_1 - e_1 - 2 e_2 = (0,0,1) = e_3$$
Hence, 
$$(\alpha, \beta , \gamma )= \alpha(V_3 - V_ 4) + \beta (V_2 - V_ 4) + \gamma ( \frac{1}{6}V_1 - (V_3 - V_ 4) - 2 (V_2-V_4) ) $$
$$= \frac{\gamma}{6} V_1 + (\beta - 2 \gamma)V_2 + (\alpha - \gamma) V_3 + (-\alpha - \beta +3 \gamma) V_4$$
A: There are a few things you can do. For example, for constants $c_1,c_2,c_3,c_4$ solve the system of equations
\begin{equation*}
(x,y,z)=c_1(1,2,6)+c_2(3,4,1)+c_3(4,3,1)+c_4(3,3,1)
\end{equation*}
where $x,y,z$ are arbitrary. Then check that there is a choice of $x,y,z$ for which the equation is consistent which will imply that the vectors span $\mathbb{R}^3.$ In particular, there must always be a solution. 
You could use Gaussian elimination. Write the vectors in the form
\begin{pmatrix}
1 & 2 & 6 \\ 
3 & 4 & 1 \\ 
4 & 3 & 1 \\ 
3 & 3 & 1
\end{pmatrix}
You can conclude the vectors span $\mathbb{R}^3$ if the rank (number of linearly independent rows or columns) is $3.$
Another method (based on the first method) is checking that one of your vectors is a linear combination of the others. 
