Find whether vector w belongs in the span $$v_1=[1,0,1,2]$$
$$v_2 = [0,1,1,3]$$
$$v_3 = [2,1,3,7]$$
$$w = [1,2,3,4]$$
We are supposed to determine if $w$ is in $\operatorname{span}(v_1,v_2,v_3)$.
 A: Create a matrix $A=[v_1 | v_2 | v_3 | w]:$
$A =\left[\begin{array}{cccc}
1 & 0 & 2 & 1 \\
0 & 1 & 1 & 2 \\
1 & 1 & 3 & 3 \\
2 & 3 & 7 & 4
\end{array}\right]$
Now just calculate the rank of $A$.
If the rank comes out to be $4$, then $w$ cannot be written as a linear combination of $v_1,v_2,v_3$.
Otherwise, we can say that w is in the span of $v_1,v_2,v_3$.
Correct me if wrong.
A: The vector $w$ will be in the span of the given set of vectors if you can write $w$ as a linear combination of the vectors. That is, provided that $w$ is in the span, you will have
$$w=c_1v_1+c_2v_2+c_3v_3$$
$w$ will be in the span if you can find at least one set of solutions for the coefficients.
A: Let us take the vectors $v_1$, $v_2$ and $v_3$ and perform elementary row operations until we get reduced row echelon form.
$$
\begin{pmatrix}
  1 & 0 & 1 & 2 \\
  0 & 1 & 1 & 3 \\
  2 & 1 & 3 & 7
\end{pmatrix}\sim
\begin{pmatrix}
  1 & 0 & 1 & 2 \\
  0 & 1 & 1 & 3 \\
  0 & 1 & 1 & 3
\end{pmatrix}\sim
\begin{pmatrix}
  \boxed{1} & 0 & 1 & 2 \\
  0 & \boxed{1} & 1 & 3 \\
  0 & 0 & 0 & 0
\end{pmatrix}$$
Since elementary row operations do not change the row space, the non-zero rows of the last matrix span the same subspace. So now we have a rather simple basis consisting of the vectors $u_1=[1,0,1,2]$ and $u_2=[0,1,1,3]$ and we are asking whether $w\in\operatorname{span}(u_1,u_2)$.
Now it suffices to look at the numbers in the first and second position in the vector $w$, since this is where the pivots are in the rref.
So $w$ can only be a linear combination of $u_1$ and $u_2$ if it is equal to
$$u_1+2u_2=[1,2,3,8].$$
Therefore we see that $w\notin \operatorname{span} (u_1,u_2)= \operatorname{span}(v_1,v_2,v_3)$.
See also this similar question: Finding whether a vector is in the span of a set of vectors
A: The row reduced form:
$$
\begin{align}
  \mathbf{A} &\to \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{ccc}
 1 & 0 & 2 \\
 0 & 1 & 1 \\
 1 & 1 & 3 \\
 2 & 3 & 7 \\
\end{array}
\right]
%
 &\to
%
\left[
\begin{array}{ccc}
 1 & 0 & 2 \\
 0 & 1 & 1 \\
 0 & 0 & 0 \\
 0 & 0 & 0 \\
\end{array}
\right]
%
\end{align}
$$
Therefore the matrix rank is $\rho = 2$.
We have two elementary columns, 1 and 2. If $w$ is in the span of the column space then we must have
$$
  \left[
\begin{array}{c}
 1 \\
 2 \\
 3 \\
 4
\end{array}
\right]
= 
\alpha
  \left[
\begin{array}{c}
 1 \\
 0 \\
 1 \\
 2
\end{array}
\right]
+
\beta 
  \left[
\begin{array}{c}
 0 \\
 1 \\
 1 \\
 3
\end{array}
\right]
$$
The first two elements reveal the answer: $w$ is not in the span.
The vector $w$ can be resolved into $\color{blue}{range}$ and $\color{red}{null}$ space components
$$
\begin{align}
  w &= 
   \color{blue}{w_{\mathcal{R}(\mathbf{A})}} + 
   \color{red}{w_{\mathcal{N}(\mathbf{A})^{*}}} \\
%
\left[
\begin{array}{c}
 1 \\
 2 \\
 3 \\
 4
\end{array}
\right]
&=
\frac{1}{17}
%
\color{blue}{\left[
\begin{array}{c}
 21 \\
 50 \\
 71 \\
 56 \\
\end{array}
\right]}
+
\frac{1}{17}
\color{red}{\left[
\begin{array}{r}
 -4 \\
 -16 \\
 -20 \\
 12 \\
\end{array}
\right]}
% 
\end{align}
$$
