Let
\[
A=\left[\begin{array}{ccc}
6 & -6 & 9 \\
-2 & 2 & -3 \\
8 & -8 & 12
\end{array}\right]
\]
a. A basis for the null space of \( A \) is \( \{\square\} \). You should be able to explain and justify your answer. Enter a coordinate vector, such as \( <1,2,3> \), or a comma separated list of coordinate vectors, such as \( <1,2,3>,<4,5,6> \).
b. The dimension of the null space of \( A \) is \( \square \) because (select all correct answers -- there may be more than one correct answer):
A. \( \operatorname{rref}(A) \) has a pivot in every row.
B. \( \operatorname{rref}(A) \) has one free variable column.
C. Two of the three columns in \( \operatorname{rref}(A) \) have pivots.
D. \( \operatorname{rref}(A) \) is the identity matrix.
E. \( \operatorname{rref}(A) \) has two free variable columns.
F. The basis we found for the null space of \( A \) has two vectors.
G. Two of the three columns in \( \operatorname{rref}(A) \) do not have a pivot.
c. The null space of \( A \) is a subspace of \( \square \) because choose \( \square \) .
d. The geometry of the null space of \( A \) is choose \( \square \)
