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To find a basis for the column space of the matrix \( A \), we need to determine which columns are linearly independent. The matrix \( A \) is given as:

\[
A = \begin{bmatrix} 2 & 4 & -6 & 2 & -4 \\ -1 & -2 & 3 & 1 & 8 \\ 0 & 0 & 0 & 2 & 6 \end{bmatrix}
\]

### Steps to find the basis for the column space:
1. **Set up the matrix for row reduction:** We will use row operations to bring \( A \) to its row echelon form (or reduced row echelon form if needed).
2. **Identify the pivot columns:** The columns that contain leading ones in the row echelon form are the linearly independent columns and will form the basis for the column space.
3. **Extract basis vectors:** The original columns of \( A \) corresponding to the pivot columns in the row echelon form are the basis vectors.

### Solution

Let's go through the row reduction of \( A \) (detailed steps are generally worked out on paper or with software for precision, but here’s the process):

#### Row Echelon Form Calculation
Upon reducing \( A \), we get:

\[
\text{Row Echelon Form} = \begin{bmatrix} 1 & 2 & -3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}
\]

From this form:
- The pivot columns are the first and fourth columns.

### Basis for the Column Space
The basis vectors for the column space of \( A \) are the columns of \( A \) corresponding to the pivot columns (1st and 4th columns):

\[
\text{Basis} = \begin{bmatrix} 2 & 2 \\ -1 & 1 \\ 0 & 2 \end{bmatrix}
\]

So, the answer is:

\[
\begin{bmatrix} 2 & 2 \\ -1 & 1 \\ 0 & 2 \end{bmatrix}
\]

This matrix represents the basis for the column space of \( A \).

Would you like further details on the row reduction process, or do you have any other questions?

### Related Questions
1. How do you find the basis for the null space of a matrix?
2. What is the significance of the row echelon form in determining the rank of a matrix?
3. How can you determine if a set of vectors forms a basis for a given space?
4. What are the applications of finding the column space of a matrix in real-world problems?
5. Can a matrix have more than one possible basis for its column space?

### Tip
Remember that a basis for the column space of a matrix must include linearly independent vectors only; any dependent columns are omitted from the basis.

Give a basis for the column space of the matrix A, where A is given by:

\[
A = \begin{bmatrix} 2 & 4 & -6 & 2 & -4 \\ -1 & -2 & 3 & 1 & 8 \\ 0 & 0 & 0 & 2 & 6 \end{bmatrix}
\]

Input the answer as a matrix with a basis vector in each column and with spaces between the numbers on each row.

This problem guides students through finding a basis for the column space of a matrix A. It covers key linear algebra concepts, including row reduction to echelon form, identifying pivot columns, and determining linearly independent vectors. Suitable for introductory linear algebra courses.

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