Finding linear combinations and verifying using the augmented matrix. I was attempting this question.

Is $w=(-12,20)$ a linear combination of $v_1 =(-1,2)$ and $v_2=(4,-6)$?

I understand that you have do the following:
$$w = c_1v_1 +c_2v_2 \\
(-12,20)= c_1 (-1,2) + c_2 (4,-6)$$ and put it in a system like:
\begin{cases}-c_1 + 4c_2 = -12 \\
2c_1- 6c_2 = 20
\end{cases}
and to make sure the system has one solution and is consistent and finally to verify the system has a unique solution, the augmented matrix must be used to find $c_1 \text{and} c_2$, however, this is where I'm stuck on, I have used the augmented matrix and have done the row operations but can someone help me  extracting the values from the augmented matrix
The row operations i did was :
-1 2 |1 0
4 -6 |0 1
R1=3R1+R2
1 0 | 3 1
4 -6| 0 1
R2=4R1-R2
1 0 | 3 1
0 -6| 12 3
R2=R2/-6
1 0 | 3 1
0 1 | -2 -1/6  
Thank You in advance! 
 A: When writing the system
$$\tag{1}
\eqalign{
\color{maroon}{-1}\cdot c_1+\color{darkgreen}4c_2&=\color{darkblue}{-12}\cr
\color{maroon}2c_1\color{darkgreen}{-6}c_2&=\color{darkblue}{20}
}
$$
as an augmented matrix:
$$\tag{2}
\left[ \matrix{\color{maroon}{ -1}&\color{darkgreen}4 \cr\color{maroon} 2&\color{darkgreen}{-6}} \biggl| \matrix{\color{darkblue}{-12}\cr \color{darkblue}{20}}\right]
$$
You're essentially just writing the original system down, but you are not writing the variables or the equality sign. In $(2)$, the red terms are the coefficients of the variable $c_1$ in the system of equations $(1)$. So column one of $(2)$ is the ''$c_1$-column". Similarly, column two of $(2)$ is the ''$c_2$-column" and the third column corresponds to the constants on the right hand side of the equalities of  the system of equations $(1)$.
Performing a row operation on the matrix $(2)$ corresponds to performing the same operation on the system. 
A reduced form of the augmented matrix $(2)$ is
$$\tag{3}
\left[ \matrix{\color{maroon}{ -1}&\color{darkgreen}4 \cr\color{maroon} 0&\color{darkgreen}{2}} \biggl| \matrix{\color{darkblue}{-12}\cr \color{darkblue}{-4}}\right]
$$
This was obtained by adding twice row one to row two. (Note that this corresponds to adding twice equation one to equation two in the system of equations $(1)$.)
Translating the matrix $(3)$ back to the corresponding system of equations:
Row one of $(3)$  gives $\color{maroon}{-1}c_1+\color{darkgreen}4c_2=\color{darkblue}{-12}$
Row two of $(3)$ gives $\color{maroon} 0c_1+\color{darkgreen}2c_2=\color{darkblue}{-4}.$

You could reduce $(3)$ further to
$$\tag{4}
\left[ \matrix{\color{maroon}{  1}&\color{darkgreen}0 \cr\color{maroon} 0&\color{darkgreen}{1}} \biggl| \matrix{\color{darkblue}{ 4}\cr \color{darkblue}{-2}}\right]
$$
And the corresponding system would be
$\eqalign{\color{maroon}1c_1 +\color{darkgreen}0 c_2&=\color{darkblue}4\cr 
\color{maroon}0c_1+\color{darkgreen}1c_2&=\color{darkblue}{-2} }$, or simply $c_1=4$, $c_2=-2$.

Just keep in the back of your mind that writing
$$ 
\left[ \matrix{\color{maroon}{ a}&\color{darkgreen}b \cr\color{maroon} c&\color{darkgreen}{d}} \biggl| \matrix{\color{darkblue}{ e}\cr \color{darkblue}{f}}\right]
$$
is shorthand notation for writing
$$
\eqalign{\color{maroon}a\,c_1 +\color{darkgreen}b\, c_2&=\color{darkblue}e\cr 
\color{maroon}c\,c_1+\color{darkgreen}d\,c_2&=\color{darkblue}{f} }.
$$
A: Notice that your system of linear equations in matrix equation $AX=B$ would look like: 
$$\begin{bmatrix}
-1 & 4 \\2 & -6 
\end{bmatrix} 
\begin{bmatrix}
c_1 \\c_2
\end{bmatrix}=\begin{bmatrix}-12\\20
\end{bmatrix}$$ 
The augmented matrix $[A,B]$ should look like: 
\begin{bmatrix}
-1 & 4 &-12\\2& -6 &20
\end{bmatrix}
Notice that the elementary row transformation $R_2 \leftrightarrow R_2+2R_1$ takes the matrix to Row-Echelon form: 
$$\sim \begin{bmatrix}
-1 &4 &-12\\0 &2 &-4
\end{bmatrix}$$
Now, notice that, your matrix equation is equivalent to $$\begin{bmatrix} -1 & 4  \\0 & 2  
\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 
\end{bmatrix} = \begin{bmatrix}
-12\\-4
\end{bmatrix}$$
Can you take it from here?

So, $c_2=-2$ and hence $c_1=4$. This means that $(-12,20)$ is a linear combination of $(-1,2)$ and $(4,-6)$. $\blacksquare$

