Describe all solutions of Ax = 0 
Let $A = \begin{bmatrix}1&-5&3&-3&-4&-2\\0&0&1&1&0&-5\\0&0&0&0&1&-3\\0&0&0&0&0&0\end{bmatrix}$
Describe all solutions of $Ax = 0$
$x = x_2 \begin{bmatrix}\\\\\\\\\\\end{bmatrix} + x_4 \begin{bmatrix}\\\\\\\\\\\end{bmatrix} + x_6 \begin{bmatrix}\\\\\\\\\\\end{bmatrix}$

I'm not really sure if I'm doing this right but I put the matrix $A$ in RREF form to get:
Performing the matrix operations:
$1) R_1 = R_1 + 4R_3$
$2) R_1 = R_1 + -3R_2$
$\begin{bmatrix}1&-5&0&-6&0&1\\0&0&1&1&0&-5\\0&0&0&0&1&-3\\0&0&0&0&0&0\end{bmatrix}$
But now I'm not sure what to do to get the vectors\matrix they are wanting here?
 A: Notice that there are no pivot columns for columns $2,4,6$, i.e. the columns do not have a leading $1$.
Thus, we let the following corresponding components of the vector $\vec{x}$ be free: $x_2=r,x_4=s,x_6=t$ where $r,s,t\in \mathbb{R}$.
We know solve for $x_1,x_3,x_5$ in terms of these free variables.
\begin{align}
x_1&=5r+6s-t\\
x_3&=-s+5t\\
x_5&=3t
\end{align}
Thus, all solutions to $A\vec{x}=0$ have the form 
$$\vec{x}=r\begin{bmatrix}
5\\
1\\
0\\
0\\
0\\
0
\end{bmatrix}+s\begin{bmatrix}
6\\
0\\
-1\\
1\\
0\\
0
\end{bmatrix}+t\begin{bmatrix}
-1\\
0\\
5\\
0\\
3\\
1
\end{bmatrix}:r,s,t\in\mathbb{R}$$
Note: $x_2=r, x_4=s,x_6=t$
A: From the already row-reduced matrix you can see that $x_2,x_4,x_6$ are free variables because the columns are missing leading $1$'s.
From row $3$, you can get $x_5-3x_6=0$, so $x_5=3x_6$
From row $2$, $x_3+x_4-5x_6=0$, $x_3=-x_4+5x_6$
From row $1$, $x_1-5x_2+3x_3-3x_4-4x_5-2x_6=0$, $x_1=5x_2-3x_3+3x_4+4x_5+2x_6$. Plug in the values of $x_5,x_3$, $$x_1=5x_2+6x_4-x_6$$
Finally turn the results into vector form:$$x = x_2 \begin{bmatrix}5\\1\\0\\0\\0\\0\end{bmatrix} + x_4 \begin{bmatrix}6\\0\\-1\\1\\0\\0\end{bmatrix} + x_6 \begin{bmatrix}-1\\0\\5\\0\\3\\1\end{bmatrix}$$
