I have $v_1=(1,0,-1),\; v_2=(2,1,3),\; v_3=(4,2,6)\; \text{ and}\; w=(3,1,2),$
where $v_1,v_2,v_3\;$ and $w$ are all column vectors.
I want to know if $w$ is in the subspace spanned by $(v_1,v_2,v_3).$
I write down:
$$\begin{aligned}x_1+2x_2+4x_3&=3\\x_2+2x_3&=1\\-x_1+3x_2+6x_3&=2\end{aligned}$$
And I augment it to get
$$ \begin{pmatrix} 1 & 2 & 4 &3 \\ 0 & 1 & 2&1 \\ -1 & 3 & 6& 2 \\ \end{pmatrix} $$
Now, from what the professor told us, I should row reduce it to get
$$ \begin{pmatrix} 1 & 2 & 4 &3 \\ 0 & 1 & 2&1 \\ 0 & 0 & 0& 0 \\ \end{pmatrix} $$
And this is my first question : why should I row reduce it only to the last row being all $0$'s?
Moreover, On the book it states :
Since the dimension of the space of the columns of the augmented matrix coincides with the dimension of the space of the columns of the matrix of the coefficients, the system admits a non-trivial solution, and $w\in Span \{v_1, v_2, v_3\}.$
Since I have a form of dyscalculia this statement appears unclear to me. Could somebody please explain in simple words what this means?
What do they mean when they say the dimension of the space of the columns of the augmented matrix coincides with the dimension of the space of the columns of the matrix of the coefficients? Why does this imply that the systems admits a non trivial solution ? What is the non trivial solution here?
If you could help me please please please explain it in the simplest way possible....Thanks for the help guys! You are great.