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 $Ax=0$.
 A: 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 $a$, $b$, $c$).
A: You did nicely in your row-echelon form!
You have what is called a Underdetermined System because you have fewer equations than unknowns.
For this type of system, you either have a.) no or b.) infinitely many solutions.
Taking your reduced form, we have infinitely many (is it clear why).
Regards
A: The solution would be a vector $(x_1,\ldots,x_6)$ that satisfies the equations the system states:
$$ x_1 -2x_2 + 3x_3 + 29x_5 = 0 $$
$$ x_4 + 4x_5 = 0 $$
$$ x_6 = 0 $$
To begin with you get $(x_1,\ldots,x_6)$. You know you can eliminate three of these since you have three equations. The first one is easy: $x_6=0$, so you have $(x_1,\ldots,x_5,0)$. You can use your second equation to eliminate $x_4$: if you set $x_5 = c$, then $x_4=-4c$, so you have: $(x_1,x_2,x_3,-4c,c,0)$.
You can use the first equation to eliminate another variable, say $x_1$. Set $x_2=a,x_3=b$, and you get that $x_1=2a-3b-29c$, so your general solution, with parameters $a,b,c$ that can be chosen freely, is $(2a-3b+29c,a,b,-4c,c,0)$.
