Math Problem Statement

Find all the matrices B = [a b; c d] that commute with the matrix A = [-320 -192; 216 208]. Give your answer in vector form.

Solution

We are asked to find all matrices B=[abcd]B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} that commute with the matrix A=[320192216208]A = \begin{bmatrix} -320 & -192 \\ 216 & 208 \end{bmatrix}. This means we need to solve for the values of a,b,c,da, b, c, d such that AB=BAAB = BA.

Step-by-Step Explanation:

  1. Compute ABAB: = \begin{bmatrix} -320a - 192c & -320b - 192d \\ 216a + 208c & 216b + 208d \end{bmatrix}$$
  2. Compute BABA: = \begin{bmatrix} a(-320) + b(216) & a(-192) + b(208) \\ c(-320) + d(216) & c(-192) + d(208) \end{bmatrix} = \begin{bmatrix} -320a + 216b & -192a + 208b \\ -320c + 216d & -192c + 208d \end{bmatrix}$$
  3. Set the two results equal to each other (since AB=BAAB = BA): = \begin{bmatrix} -320a + 216b & -192a + 208b \\ -320c + 216d & -192c + 208d \end{bmatrix}$$
  4. Form a system of equations by comparing corresponding elements:
    • For the top-left entry: 320a192c=320a+216b-320a - 192c = -320a + 216b 192c=216borc=98b\Rightarrow -192c = 216b \quad \text{or} \quad c = -\frac{9}{8}b

    • For the top-right entry: 320b192d=192a+208b-320b - 192d = -192a + 208b

      \quad \text{or} \quad -192d = -192a + 528b \quad \Rightarrow d = a - \frac{11}{4}b$$
    • For the bottom-left entry: 216a+208c=320c+216d216a + 208c = -320c + 216d Using c=98bc = -\frac{9}{8}b and d=a114bd = a - \frac{11}{4}b, this equation simplifies, but confirms consistency in terms of aa and bb.

    • For the bottom-right entry: 216b+208d=192c+208d216b + 208d = -192c + 208d simplifies to a tautology, meaning no new information is added.

General Solution:

The matrix BB has two free parameters, aa and bb, and the other elements can be expressed in terms of them:

  • c=98bc = -\frac{9}{8}b
  • d=a114bd = a - \frac{11}{4}b

Vector Form:

The answer in vector form is: [ab98ba114b]=s[1001]+t[0198114]\begin{bmatrix} a \\ b \\ -\frac{9}{8}b \\ a - \frac{11}{4}b \end{bmatrix} = s \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + t \begin{bmatrix} 0 \\ 1 \\ -\frac{9}{8} \\ -\frac{11}{4} \end{bmatrix} where ss and tt are free parameters.

Would you like further clarification or more details on any step?


Here are five related questions you might find useful:

  1. How can I find the inverse of a matrix?
  2. What are the conditions for two matrices to commute?
  3. How does one solve a system of linear equations derived from matrices?
  4. What is the significance of matrix commutativity in linear algebra?
  5. How can eigenvalues help in finding commuting matrices?

Tip: When working with matrix equations, always check for consistency across all rows and columns to ensure the system is fully solved.

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Math Problem Analysis

Mathematical Concepts

Matrix Algebra
Commutativity of Matrices
Linear Systems

Formulas

Matrix multiplication: AB = BA
Solving linear systems from matrix equations

Theorems

Commutativity condition for matrices: AB = BA

Suitable Grade Level

College or Advanced High School (Linear Algebra)