Determining a basis for Col($A$) and a dimension for the null space of $A$ 
Let $A = \begin{bmatrix}1&-1&1&0&-2&1\\1&-1&1&1&0&0\\-1&1&-1&2&5&-1\end{bmatrix}$
a) Determine a basis for Col($A$)
b) What is the dimension of the null space of $A$?

I just want to make sure I am not doing this incorrectly.
part a) I put $A$ in rref form to get: $rref(A) = \begin{bmatrix}1&-1&1&0&-2&1\\0&0&0&1&2&-1\\0&0&0&0&-1&0\end{bmatrix}$ so I got the basis for Col(A) = $\begin{bmatrix}1\\1\\-1\end{bmatrix}, \begin{bmatrix}0\\1\\2\end{bmatrix}$
Then for part b)
I know that nullity($A$) = # of non-pivot columns (or free vars) so I got that dim(null(A)) = 4? Since i have 4 free variables?
 A: Your matrix can be reduced with the following steps


*

*add $-1$ times row 1 to row 2 

*add $1$ times row 1 to row 3 

*add $-2$ times row 2 to row 3 

*scale row 3 by $-1$ 

*add $2$ times row 3 to row 1 

*add $-2$ times row 3 to row 2


This gives
$$
\DeclareMathOperator{rref}{rref}\rref A=
\left[\begin{array}{rrrrrr}
1 & -1 & 1 & 0 & 0 & -3 \\
0 & 0 & 0 & 1 & 0 & 3 \\
0 & 0 & 0 & 0 & 1 & -2
\end{array}\right]
$$
Columns one, four, and five are the pivot columns. Hence
$$
\DeclareMathOperator{Col}{Col}\Col A=\DeclareMathOperator{Span}{Span}\Span\left\{
\left[\begin{array}{r}
1 \\
1 \\
-1
\end{array}\right],
\left[\begin{array}{r}
0 \\
1 \\
2
\end{array}\right],
\left[\begin{array}{r}
-2 \\
0 \\
5
\end{array}\right]
\right\}
$$
To get a feel for what $\Col A$ looks like, you can put these basis vectors into the rows of a matrix
$$
C=
\left[\begin{array}{rrr}
1 & 1 & -1 \\
0 & 1 & 2 \\
-2 & 0 & 5
\end{array}\right]
$$
The rows of $\rref C$ also form a basis for $\Col A$. Here,
$$
\rref C=
\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
Hence
$$
\Col A=\Span\left\{
\left[\begin{array}{r}
1 \\
0 \\
0
\end{array}\right],
\left[\begin{array}{r}
0 \\
1 \\
0
\end{array}\right],\left[\begin{array}{r}
0 \\
0 \\
1
\end{array}\right]
\right\}=\Bbb R^3
$$
So $\dim\Col A=3$ and the rank-nullity theorem implies that 
$$
\dim\DeclareMathOperator{Null}{Null}\Null A=\#\text{ columns of }A-\dim\Col A=6-3=3
$$
