Checking if a vector is in a subspace? I have 
v1= $$
    \begin{pmatrix}
    1 \\
    0 \\
    -1\\
    \end{pmatrix}
$$
v2= 
    \begin{pmatrix}
2 \\
1 \\
3\\
\end{pmatrix}
v3=\begin{pmatrix}
4 \\
2 \\
6\\
\end{pmatrix}
and 
w=\begin{pmatrix}
3 \\
1 \\
2\\
\end{pmatrix}
I need to check whether w is in the subspace spanned by (v1,v2,v3)
I know that w is in the subspace spanned by (v1,v2,v3) if x1v1+x2v2+x3v3=w has a solution .
I write:
x1+2x2+4x3=3
x2+2x3=1
-x1+3x2+6x3=2
I write down the augmented matrix, which is 
A= $$
\begin{pmatrix}
1 & 2 & 4&3 \\
0 & 1 & 2&1 \\
-1 & 3 & 6&2 \\
\end{pmatrix}
$$
And row reduce it to get 
$$
\begin{pmatrix}
1 & 2 & 4&3 \\
0 & 1 & 2&1 \\
0 & 0 & 0&0 \\
\end{pmatrix}
$$
On the answer sheet is states:
since the dimension of the space of the columns of the augmented matrix coincides with the dimension of the space of the matrix coefficients, the system admits a non trivial solution and w exists in (v1,v2,v3)
I am new studying matrices and my dyscalculia certainly does not help. My question is, what is the dimension of the space of the columns of the augmented matrix? What is the dimension of the space of the matrix coefficient? How can i show that they are the same?
Moreover, can you show me an example where the space of the columns of the augmented matrix DOES NOT coincide with the dimension of the space of the matrix coefficients?
I would greatly appreciate an answer that is as clear and simple as possible ... thank you guys ! 

The dimension of the space of columns of a matrix is the maximal number of column vectors that are linearly independent. 
In your example, both dimensions are $2$, as the last two columns can be written as a linear combination of the first two columns.
An example where the dimensions are not equal can be given by 
$$\begin{cases}
x_1=0\\ x_1=1
\end{cases}.$$
The augmented matrix is 
$$\begin{pmatrix}1 & 0 \\ 1 & 1 \end{pmatrix}.$$
The dimension of the space of columns of the coefficient matrix is $1$, while that of the augmented matrix is $2$.
