Finding a basis for the nullspace $A$=\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
I need to find the null space for this matrix. After performing row operations $R_3 + 3R_1$, then $R_{new row 3}$ + $R_2$, I got the reduced row echelon form. (the variable I chose were $x$, $y$, $z$, $w$). I identified the pivot variables as $x$ and $y$, after expressing both of these in terms of the free variables $z$ and $w$, I got $$\begin{bmatrix}
x\\
y\\
z\\
w\\
\end{bmatrix} = z \begin{bmatrix}
23/2\\
4\\
1\\
0\\
\end{bmatrix} + w\begin{bmatrix}
-11/2\\
-2\\
0\\
1\\
\end{bmatrix}$$, and then chose both the column vectors alongside $z$ and $w$ as the basis for the nullspace. However the answer given is $$\begin{bmatrix}
23\\
8\\
2\\
0\\
\end{bmatrix}\begin{bmatrix}
9\\
4\\
0\\
2\\
\end{bmatrix}$$. What am I doing wrong?

If matrix
$$
\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
Row reduced form of matrix is
$$
\begin{bmatrix}1 & 0 & -\frac{23}{2} & 0\\ 0 & 1 & -4 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
So basis is
$$
\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix}
$$
If matrix is 
$$
\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & -2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
Row reduced form of matrix is
$$
\begin{bmatrix}1 & 0 & -\frac{23}{2} & -\frac{9}{2}\\ 0 & 1 & -4 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 &0 &0\end{bmatrix}
$$
So basis is
$$
\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix}
\begin{bmatrix}9 \\ 4 \\ 0 \\ 2\end{bmatrix}
$$
