Finding null space of matrix. I need to make sure I'm understanding this correctly.
I skipped a few steps to reduce typing, but let me know if I need to clarify something.
Question asks:

Find $N(A)$ for $A$ = \begin{bmatrix}
        -3 & 6 & -1 & 1 & -7 \\
        1 & -2 & 2 & 3 & -1\\
        2 & -4 & 5 & 8 & -4 \\
        \end{bmatrix}

First thing I did was put the augmented matrix into reduced echelon row:

$\begin{bmatrix}
        1 & -2 & 0 & -1 & 3 & 0 \\
        0 & 0 & 1 & 2 & -2 & 0\\
        0 & 0 & 0 & 0 & 0 & 0 \\
        \end{bmatrix}$


$(1)$

So then...

$x=\begin{bmatrix}
x_1\\
x_2 \\
x_3\\
x_4\\
x_5\\
\end{bmatrix} 
= 
\begin{bmatrix}
2x_2 + x_4 - 3x_5\\
x_2 \\
-2x_4 + 2x_5\\
x_4\\
x_5\\
\end{bmatrix}$


$(2) $

Since $x_2, x_4$ and $x_5$ are free variables..

$
x_2
\begin{bmatrix}
2\\
1 \\
0\\
0\\
0\\
\end{bmatrix}
+
x_4
\begin{bmatrix}
1\\
0 \\
-2\\
1\\
0\\
\end{bmatrix}
+
x_5
\begin{bmatrix}
-3\\
0 \\
2\\
0\\
1\\
\end{bmatrix}$


$(3)$

Resulting in..

$N(A)= \left(
\begin{bmatrix}
2\\
1 \\
0\\
0\\
0\\
\end{bmatrix}
,
\begin{bmatrix}
1\\
0 \\
-2\\
1\\
0\\
\end{bmatrix}
,
\begin{bmatrix}
-3\\
0 \\
2\\
0\\
1\\
\end{bmatrix}
\right)$


$(4)$


As mentioned in the comments, provided your arithmetic is accurate, this is the correct response. The idea behind the null space of a matrix is that it is precisely those vectors in the domain being sent to the $\mathbf{0}$ vector in the codomain. So, what you have (correctly) done, is determined the solution set of $A\mathbf{x}=\mathbf{0}$. 
You did this by finding the null space of a reduced row echelon form of $A$, which has the same null space as $A$. That is, if $B$ is the reduced row echelon form for $A$ that you found, $A\mathbf{x}=\mathbf{0}$ if and only if $B\mathbf{x}=\mathbf{0}$. So, $N(B)=N(A)$.
