Math Problem Statement

Please help me solve this problem

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

Solution

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

A=(3221),B=(133k)A = \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 3 \\ -3 & k \end{pmatrix}

Step 1: Calculate ABAB

To find ABAB, we multiply matrix AA by matrix BB:

AB=(3221)×(133k)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: 3×1+2×(3)=36=33 \times 1 + 2 \times (-3) = 3 - 6 = -3
  • First row, second column: 3×3+2×k=9+2k3 \times 3 + 2 \times k = 9 + 2k
  • Second row, first column: 2×1+1×(3)=23=5-2 \times 1 + 1 \times (-3) = -2 - 3 = -5
  • Second row, second column: 2×3+1×k=6+k-2 \times 3 + 1 \times k = -6 + k

Thus, we get:

AB=(39+2k56+k)AB = \begin{pmatrix} -3 & 9 + 2k \\ -5 & -6 + k \end{pmatrix}

Step 2: Calculate BABA

Now, we calculate BABA:

BA=(133k)×(3221)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: 1×3+3×(2)=36=31 \times 3 + 3 \times (-2) = 3 - 6 = -3
  • First row, second column: 1×2+3×1=2+3=51 \times 2 + 3 \times 1 = 2 + 3 = 5
  • Second row, first column: 3×3+k×(2)=92k-3 \times 3 + k \times (-2) = -9 - 2k
  • Second row, second column: 3×2+k×1=6+k-3 \times 2 + k \times 1 = -6 + k

Thus, we get:

BA=(3592k6+k)BA = \begin{pmatrix} -3 & 5 \\ -9 - 2k & -6 + k \end{pmatrix}

Step 3: Set AB=BAAB = BA

For AB=BAAB = BA, the corresponding elements of the two matrices must be equal.

  1. First row, first column: 3=3-3 = -3 (True)
  2. First row, second column: 9+2k=59 + 2k = 5
  3. Second row, first column: 5=92k-5 = -9 - 2k
  4. Second row, second column: 6+k=6+k-6 + k = -6 + k (True)

Step 4: Solve the system of equations

From the second equation: 9+2k=52k=592k=4k=29 + 2k = 5 \quad \Rightarrow \quad 2k = 5 - 9 \quad \Rightarrow \quad 2k = -4 \quad \Rightarrow \quad k = -2

From the third equation: 5=92k5+9=2k4=2kk=2-5 = -9 - 2k \quad \Rightarrow \quad -5 + 9 = -2k \quad \Rightarrow \quad 4 = -2k \quad \Rightarrow \quad k = -2

Both equations give k=2k = -2.

Conclusion:

The value of kk that makes AB=BAAB = BA is k=2k = -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=BAAB = BA have more than one solution for kk?
  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 AA and BB?

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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Matrix Multiplication
Matrix Commutativity

Formulas

Matrix Multiplication: C[i][j] = sum(A[i][k] * B[k][j])
Solving System of Equations

Theorems

Commutativity of Matrices

Suitable Grade Level

Grades 10-12