Answer
Basis for column space: $\begin{bmatrix}
1\\2\\-3\\3
\end{bmatrix}$, $\begin{bmatrix}
2\\5\\-9\\10
\end{bmatrix}$, $\begin{bmatrix}
0\\4\\-7\\11
\end{bmatrix}$ and dim Col A = 3
Basis for null space: $\begin{bmatrix}
9\\-2\\1\\0\\0
\end{bmatrix}$, $\begin{bmatrix}
-5\\3\\0\\-2\\1
\end{bmatrix}$ and dim Nul A = 2
Work Step by Step
The basis for the column space are the pivot columns
Basis for column space: $\begin{bmatrix}
1\\2\\-3\\3
\end{bmatrix}$, $\begin{bmatrix}
2\\5\\-9\\10
\end{bmatrix}$, $\begin{bmatrix}
0\\4\\-7\\11
\end{bmatrix}$ and dim Col A = 3
The reduced row echelon form is:
$\begin{bmatrix}
1&0&-9&0&5\\
0&1&2&0&-3\\
0&0&0&1&2\\
0&0&0&0&0\\
\end{bmatrix}$
Basis for null space: $\begin{bmatrix}
9\\-2\\1\\0\\0
\end{bmatrix}$, $\begin{bmatrix}
-5\\3\\0\\-2\\1
\end{bmatrix}$ and dim Nul A = 2
