Let $v_1 = (1,0,-1)^T$, $v_2 = (2, 1, 3)^T$, $v_3 = (4,2,6)^T$ and $w = (7,4,7)^T$
From what I understand, if I set up an augment matrix in the form
\begin{pmatrix}1&2&4&7\\ 0&1&2&4\\ -1&3&6&7\\ \end{pmatrix}
And reduce it using Gaussian elimination, if it is consistent, $w$ is in fact in the span of $\{v_1,v_2,v_3\}$. Else, if it is inconsistent, $w$ does not span $\{v_1,v_2,v_3\}$
This, $w$ does not span $\{v_1,v_2,v_3\}$. Is this true?
On
You are correct. If $w$ were an element of $\operatorname{span}(v_1, v_2, v_3)$, then we could represent $w$ as $w=xv_1 + yv_2 + zv_3$. Thus there'd be a solution to the matrix equation $$\pmatrix{1&2&4\\0&1&2\\-1&3&6}\pmatrix{x \\ y \\ z} = \pmatrix{7 \\ 4 \\ 7}$$
Solving this, we see that $$RREF\left(\begin{array}{ccc|c}1&2&4&7\\ 0&1&2&4\\ -1&3&6&7 \end{array}\right) = \left(\begin{array}{ccc|c}1 & 0 & 0 & 0\\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1\end{array}\right) \\ \iff \begin{cases} x=0 \\ y+2z=0 \\ 0=1\end{cases}$$ but that last equation is never true so this system of equations has no solution.
Thus $w \not\in \operatorname{span}(v_1, v_2, v_3)$.
In other words, @tommy is wrong: $w$ is NOT a linear combination of those $3$ vectors.
yes as it is a linear combination of those 3 vectors