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We are tasked with finding the following for the given matrix \( A \):

\[
A = \begin{bmatrix} 0 & 0 & 1 & -2 \\ 4 & 4 & 8 & -12 \end{bmatrix}
\]
1. A basis for the column space (the space spanned by the columns of \( A \)).
2. A basis for the null space (the space of solutions to \( A \mathbf{x} = 0 \)).
3. The rank of the matrix (the number of linearly independent columns of \( A \)).

### Step 1: Row Reduction to Echelon Form
To determine the rank, we first perform row reduction to bring the matrix to row echelon form (REF).

Starting with:
\[
A = \begin{bmatrix} 0 & 0 & 1 & -2 \\ 4 & 4 & 8 & -12 \end{bmatrix}
\]

We can make the first pivot in row 2 by dividing the second row by 4:
\[
\text{Row 2} \to \frac{1}{4} \cdot \text{Row 2} = \begin{bmatrix} 0 & 0 & 1 & -2 \\ 1 & 1 & 2 & -3 \end{bmatrix}
\]

Now, the matrix is in row echelon form. From here, we can easily identify the rank and proceed with further analysis.

### Step 2: Determining the Rank
The matrix now looks like this:
\[
\begin{bmatrix} 0 & 0 & 1 & -2 \\ 1 & 1 & 2 & -3 \end{bmatrix}
\]

Both rows are non-zero, so the rank of the matrix \( A \) is 2. This means there are 2 linearly independent columns.

### Step 3: Basis for the Column Space
The column space is spanned by the columns that correspond to the pivot columns in the row-reduced form. These pivot columns are the third and second columns of the original matrix. Thus, the basis for the column space is given by:
\[
\text{Basis for column space} = \left\{ \begin{bmatrix} 1 \\ 8 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \end{bmatrix} \right\}
\]

### Step 4: Basis for the Null Space
To find the null space, we solve \( A \mathbf{x} = 0 \) (i.e., find the solution to the homogeneous system).

We return to the original matrix:
\[
\begin{bmatrix} 0 & 0 & 1 & -2 \\ 4 & 4 & 8 & -12 \end{bmatrix}
\]
and row-reduce to the following system:
\[
\begin{bmatrix} 0 & 0 & 1 & -2 \\ 1 & 1 & 2 & -3 \end{bmatrix}
\]
This translates to the system of equations:
\[
x_3 - 2x_4 = 0 \quad (1)
\]
\[
x_1 + x_2 + 2x_3 - 3x_4 = 0 \quad (2)
\]

From equation (1), we get \( x_3 = 2x_4 \). Substituting this into equation (2):
\[
x_1 + x_2 + 2(2x_4) - 3x_4 = 0
\]
\[
x_1 + x_2 + 4x_4 - 3x_4 = 0
\]
\[
x_1 + x_2 + x_4 = 0
\]
Thus, we have:
\[
x_1 = -x_2 - x_4, \quad x_3 = 2x_4
\]
So the general solution to the system is:
\[
\mathbf{x} = \begin{bmatrix} -x_2 - x_4 \\ x_2 \\ 2x_4 \\ x_4 \end{bmatrix} = x_2 \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -1 \\ 0 \\ 2 \\ 1 \end{bmatrix}
\]
Thus, the basis for the null space is:
\[
\text{Basis for null space} = \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 2 \\ 1 \end{bmatrix} \right\}
\]

### Final Answer:
- **Rank of the matrix**: 2
- **Basis for the column space**: \( \left\{ \begin{bmatrix} 1 \\ 8 \end{bmatrix}, \begin{bmatrix} 0 \\ 4 \end{bmatrix} \right\} \)
- **Basis for the null space**: \( \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 2 \\ 1 \end{bmatrix} \right\} \)

Would you like any further clarification or details on any of the steps? 

Here are 5 related questions to expand the concept:
1. How do we determine linear dependence between columns in a matrix?
2. Can the rank of a matrix ever exceed the number of rows?
3. What is the geometric interpretation of the null space?
4. How do elementary row operations affect the determinant of a matrix?
5. What is the relationship between the rank of a matrix and its invertibility?

**Tip**: When solving for the null space, remember to express the free variables clearly to avoid confusion when finding the basis.

Find a basis for the column space, a basis for the null space, and the rank of the matrix A = [[0, 0, 1, -2], [4, 4, 8, -12]]

This linear algebra problem involves finding the basis for the column space and null space of a matrix and determining its rank. The solution uses Gaussian elimination and concepts like linear independence and the rank-nullity theorem.

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