In short, row reduced echelon form(RREF) of a matrix $A$ is such that
i) Every leading entry is 1
ii) Any nonzero rows are above zero rows
iii) any leading entry is strictly to the right of any leading entries above that row
iv) any other entry in a column containing a leading entry is 0 except for the leading entry.
So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.
To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?
Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.
Solve:
$x+y=1$
$x+y=2$
Then we have
$\
A =
\left( {\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array} } \right)
$
$\
b =
\left( {\begin{array}{cc}
1 \\
2
\end{array} } \right)
$
and $Ax=b$
If we turn A into RREF, we get
$\
E =
\left( {\begin{array}{cc}
1 & 1 \\
0 & 0
\end{array} } \right)
$
So A has rank 1
and if we put $[A|b]$ into RRE form, we get
$\
E' =
\left( {\begin{array}{cc}
1 & 1 & 0 \\
0 & 0 & 1
\end{array} } \right)
$
So augmented matrix has rank 2. Observe what last row says in terms of equations.
More Generally.
First you are going to want to set this matrix up as an Augmented Matrix
where $Ax=0$.
$1)$ To find the rank, simply put the Matrix in REF or RREF
$\left[\begin{array}{ccc|c}
0 & 0 & 0 &0 \\
0 & 0.5 & -0.5 & 0 \\
0 & -0.5 & 0.5 & 0 \end{array}\right] \longrightarrow RREF \longrightarrow \left[\begin{array}{ccc|c}
0 & 0 & 0 &0\\
0 & 0.5 & -0.5 & 0\\
0 & 0 & 0 & 0 \end{array}\right] $
Seeing that we only have one leading variable we can now say that the rank is 1.
$2)$ To find nullity of the matrix simply subtract the rank of our Matrix from the total number of columns.
So:
Null (A)=3 - 1=2
Hope this is helpful.
Best Answer
The null space is a subspace of the original vector space. Observe that the vector space in question is exactly $N(A)$, the null space of $A$.
As you observed, $rank(A) + null(A) = dim(V)$. So $2 + null(A) = 3$.