Column Space of A Suppose we have the matrix 
$A = \begin{bmatrix}
4 &6 \\
-8 &-12\\
\end{bmatrix}
$
and the vector $b = \begin{bmatrix}
3\\
-6 
\end{bmatrix}
$.
So I am tying to determine whether b lies in the column space of A. My answer for this problem is yes since the row 2 of the matrix and the vector is a multiple of -2 giving us infinitly many answers since it is parallel. First, is my answer correct? If so, what I am confused with is with 3x3 matrix. For the following matrix, 
$A = \begin{bmatrix}
2 &-6 &-6\\
4 &2 &-8\\
6 &-2 &-2\\
\end{bmatrix}
$
and the following vector
$b = \begin{bmatrix}
-4 \\
-14\\
4 \\
\end{bmatrix}
$
, can someone show me how you decide whether b lies in the column space of A? For the above matrix, I made the problem up myself. So feel free to demonstrate you point in any matrix you feel like. What I am having a bit of trouble at the moment is grasping the mechanics of how in general this is determined. Any inutitive explanation would be much appreciated. 
 A: You're correct that $b$ in the first case is a multiple of the second column of your first matrix: $1/2$ Column 2 = b. So essentially $b = 0\cdot \begin{pmatrix} 4 \\ -8 \end{pmatrix} + \frac 12 \begin{pmatrix} 6 \\ -12\end{pmatrix}$. Indeed, we see that any vector of the form $\langle t, -2t\rangle $ is in the column space of the matrix.
Hint: Solve the augmented Matrix $[A \mid b]$ to see if a solution exists for the constants $c_1, c_2, c_3$. If so, you will know that $b$ is in the column space of $A$.
$$b = c_1\begin{pmatrix} 2 \\ 4 \\ 6 \\ \end{pmatrix}
 + c_2\begin{pmatrix} -6 \\ 2 \\ -2 \\ \end{pmatrix}
+ c_3\begin{pmatrix}-6 \\ -8 \\ -2 \\ \end{pmatrix}$$
$$[A\mid b] =
  \left[\begin{array}{rrr|r}
    2 &-6 &-6 &-4 \\
    4 &2 &-8 &-14\\
    6 &-2 &-2 &4
  \end{array}\right]
$$
A: If $b$ is in Col$A = \text{Col}\begin{bmatrix}
2 &-6 &-6\\
4 &2 &-8\\
6 &-2 &-2\\
\end{bmatrix}
$
then $b= c_1\begin{bmatrix}
2 \\
4 \\
6 \\
\end{bmatrix}
 + c_2\begin{bmatrix}
-6 \\
2 \\
-2 \\
\end{bmatrix}
+ c_3\begin{bmatrix}
-6 \\
-8 \\
-2 \\
\end{bmatrix}
= \begin{bmatrix}
c_1 \\
c_2 \\
c_3\\
\end{bmatrix}\begin{bmatrix}
2 &-6 &-6\\
4 &2 &-8\\
6 &-2 &-2\\
\end{bmatrix}
$
For some $c_1, c_2,$ and $c_3$
so being in the column space is the same as having a solution to the system $Ax=b$
