<s>[INST] Describe all solutions of $Ax=0$. Describe all solutions of $Ax=0$ in parametric vector form, where $A$ is row equivalent to the given matrix.
$$  \begin{bmatrix}
    1 & -2 & 3 & -6 & 5 & 0\\
    0 & 0 & 0 & 1 & 4&-6\\
0&0&0&0&0&1\\
0&0&0&0&0&0
  \end{bmatrix}$$

I know that I should get this into row reduced echelon form, but I'm having trouble doing so.  I attempted it below.
$$  \begin{bmatrix}
    1 & -2 & 3 & -6 & 5 & 0\\
    0 & 0 & 0 & 1 & 4&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0
  \end{bmatrix}$$
$$  \begin{bmatrix}
    1 & -2 & 3 & 0 & 29 & 0\\
    0 & 0 & 0 & 1 & 4&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0
  \end{bmatrix}$$
I'm not quite sure where to go from here, also I don't know how I would describe all solutions of $Ax=0$.
 [/INST] Think about what $Ax=0$ means for your $A$:
$$\begin{bmatrix}
    1 & -2 & 3 & -6 & 5 & 0\\
    0 & 0 & 0 & 1 & 4&-6\\
0&0&0&0&0&1\\
0&0&0&0&0&0
  \end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\end{bmatrix},$$
that is,
\begin{align}
x_1-2x_2+3x_3-6x_4+5x_5+0x_6&=0,\\
x_4+4x_5-6x_6&=0,\\
x_6&=0.
\end{align}
Clearly the last equation says $x_6=0$ and then the second equation says $x_4=-4x_5$ while the first equation tells you $x_1=2x_2-3x_3+6x_4-5x_5 \implies x_1=2x_2-3x_3-29x_5$.
Putting this in vector form, you can say
$$
x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix}
=\begin{bmatrix}2x_2-3x_3-29x_5\\ x_2\\x_3\\-4x_5\\x_5\\0\end{bmatrix},$$
where $x_2,x_3,x_5$ are viewed as parameters (sometimes renamed $a$, $b$, $c$).
</s>