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The given problem asks to find the value(s) of \( k \), if any, that will make \( AB = BA \), where:

\[
A = \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 3 \\ -3 & k \end{pmatrix}
\]

### Step 1: Calculate \( AB \)

To find \( AB \), we multiply matrix \( A \) by matrix \( B \):

\[
AB = \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 3 \\ -3 & k \end{pmatrix}
\]

Multiplying row by column:

- First row, first column: \( 3 \times 1 + 2 \times (-3) = 3 - 6 = -3 \)
- First row, second column: \( 3 \times 3 + 2 \times k = 9 + 2k \)
- Second row, first column: \( -2 \times 1 + 1 \times (-3) = -2 - 3 = -5 \)
- Second row, second column: \( -2 \times 3 + 1 \times k = -6 + k \)

Thus, we get:

\[
AB = \begin{pmatrix} -3 & 9 + 2k \\ -5 & -6 + k \end{pmatrix}
\]

### Step 2: Calculate \( BA \)

Now, we calculate \( BA \):

\[
BA = \begin{pmatrix} 1 & 3 \\ -3 & k \end{pmatrix} \times \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}
\]

Multiplying row by column:

- First row, first column: \( 1 \times 3 + 3 \times (-2) = 3 - 6 = -3 \)
- First row, second column: \( 1 \times 2 + 3 \times 1 = 2 + 3 = 5 \)
- Second row, first column: \( -3 \times 3 + k \times (-2) = -9 - 2k \)
- Second row, second column: \( -3 \times 2 + k \times 1 = -6 + k \)

Thus, we get:

\[
BA = \begin{pmatrix} -3 & 5 \\ -9 - 2k & -6 + k \end{pmatrix}
\]

### Step 3: Set \( AB = BA \)

For \( AB = BA \), the corresponding elements of the two matrices must be equal.

1. First row, first column: \( -3 = -3 \) (True)
2. First row, second column: \( 9 + 2k = 5 \)
3. Second row, first column: \( -5 = -9 - 2k \)
4. Second row, second column: \( -6 + k = -6 + k \) (True)

### Step 4: Solve the system of equations

From the second equation:
\[
9 + 2k = 5 \quad \Rightarrow \quad 2k = 5 - 9 \quad \Rightarrow \quad 2k = -4 \quad \Rightarrow \quad k = -2
\]

From the third equation:
\[
-5 = -9 - 2k \quad \Rightarrow \quad -5 + 9 = -2k \quad \Rightarrow \quad 4 = -2k \quad \Rightarrow \quad k = -2
\]

Both equations give \( k = -2 \).

### Conclusion:

The value of \( k \) that makes \( AB = BA \) is \( k = -2 \).

Would you like further details on this solution, or any clarifications?

### 5 Additional Questions:
1. How does matrix multiplication work, and why is it non-commutative?
2. Can a matrix equation like \( AB = BA \) have more than one solution for \( k \)?
3. What are some real-world applications of matrix commutativity?
4. How can you generalize this solution for larger matrices?
5. What happens if we change elements of matrices \( A \) and \( B \)?

### Tip:
Always check the dimensions of matrices before multiplying them, as matrix multiplication is only defined when the number of columns in the first matrix equals the number of rows in the second matrix.

What value(s) of k, if any, will make AB = BA for matrices A = [[3, 2], [-2, 1]] and B = [[1, 3], [-3, k]]?

Learn how to solve for the value of k that makes AB = BA with two 2x2 matrices. This problem involves matrix multiplication and solving a system of equations. Ideal for students in Grades 10-12 studying advanced algebra and matrix theory.

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