Adding vector to span a space 
Let $A=\left\{\left(\begin{array}{c} 18 \\ 6 \\ -4 \\ 12 \end{array}\right),\left(\begin{array}{c} 6 \\ 2 \\ 2 \\ -6 \end{array}\right)\right\}$
   find the vectos to be added so A will span $\mathbb{R}^4$?

So what I did is :
\begin{pmatrix}
    18 & 6 & -4 & 12  \\
    6       & 2 & 2 & -6 \\
\end{pmatrix}
the row reduced echelon form is:
\begin{pmatrix}
    1 & \frac{1}{3} & 0 & 0  \\
    0       & 0 & 1 & -10 \\
\end{pmatrix}
What should I do next? I need to write the vectors that should be added with 1 and 0 only?

Observe that the "core" elements that make those two vectors linearly independent is the tiny little matrix formed by their second and third elements:
$$
        M(2,3) = \begin{matrix}
        6 & 2 \\
        -4 & 2 \\
        \end{matrix}
$$
which has non-zero determinant.
If we can add two other vectors such that the determinant of the four column vectors is reduced to the calculation of the determinant of $M(2,3)$, which is $\neq 0$, then we have found a span of $\mathbb{R}^4$.
Fortunately that's easy:
$$
        \begin{vmatrix}
        18 & 6 & 0 & 1\\
        6 & 2 & 0 & 0\\
        -4 & 2 & 0 & 0\\
        12 & -6 & 1 & 0\\
        \end{vmatrix} = 
-\begin{vmatrix}
        6 & 2 & 0\\
        -4 & 2 & 0\\
        12 & -6 & 1\\
        \end{vmatrix} =
-\begin{vmatrix}
        6 & 2\\
        -4 & 2\\
        \end{vmatrix}\neq 0
$$
So the vectors that you can add are the last two columns of the big determinant.
