Math Problem Statement
Find all 2 by 2 matrices ([a, b], [c, d]) that are commute with the 2 by 2 matrix ([1, 2], [-1, -1])
Solution
Let the matrix . We are tasked with finding all matrices that commute with , meaning that:
Step 1: Compute
The product of and is:
= \begin{pmatrix} 1 \cdot a + 2 \cdot c & 1 \cdot b + 2 \cdot d \\ -1 \cdot a - 1 \cdot c & -1 \cdot b - 1 \cdot d \end{pmatrix} = \begin{pmatrix} a + 2c & b + 2d \\ -a - c & -b - d \end{pmatrix}$$ ### Step 2: Compute $$BA$$ The product of $$B$$ and $$A$$ is: $$BA = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} a \cdot 1 + b \cdot (-1) & a \cdot 2 + b \cdot (-1) \\ c \cdot 1 + d \cdot (-1) & c \cdot 2 + d \cdot (-1) \end{pmatrix} = \begin{pmatrix} a - b & 2a - b \\ c - d & 2c - d \end{pmatrix}$$ ### Step 3: Set $$AB = BA$$ Now, set the matrices equal to each other: $$\begin{pmatrix} a + 2c & b + 2d \\ -a - c & -b - d \end{pmatrix} = \begin{pmatrix} a - b & 2a - b \\ c - d & 2c - d \end{pmatrix}$$ This gives us the following system of equations by equating corresponding elements: 1. $$a + 2c = a - b$$ (top left entry) 2. $$b + 2d = 2a - b$$ (top right entry) 3. $$-a - c = c - d$$ (bottom left entry) 4. $$-b - d = 2c - d$$ (bottom right entry) ### Step 4: Solve the system of equations #### From equation (1): $$a + 2c = a - b \quad \Rightarrow \quad 2c = -b \quad \Rightarrow \quad b = -2c$$ #### From equation (2): $$b + 2d = 2a - b \quad \Rightarrow \quad (-2c) + 2d = 2a + 2c \quad \Rightarrow \quad 2d = 2a + 4c \quad \Rightarrow \quad d = a + 2c$$ #### From equation (3): $$-a - c = c - d \quad \Rightarrow \quad -a - c = c - (a + 2c) \quad \Rightarrow \quad -a - c = c - a - 2c \quad \Rightarrow \quad -a - c = -a - c$$ This equation is always true, so it gives no new information. #### From equation (4): $$-b - d = 2c - d \quad \Rightarrow \quad (-2c) - (a + 2c) = 2c - (a + 2c) \quad \Rightarrow \quad -3c - a = 0 \quad \Rightarrow \quad a = -3c$$ ### Step 5: Conclusion Now, substitute $$a = -3c$$ into the expressions for $$b$$ and $$d$$: $$b = -2c, \quad d = a + 2c = -3c + 2c = -c$$ Thus, the matrix $$B$$ is given by: $$B = \begin{pmatrix} -3c & -2c \\ c & -c \end{pmatrix} = c \begin{pmatrix} -3 & -2 \\ 1 & -1 \end{pmatrix}$$ where $$c$$ is any real number. Therefore, all matrices that commute with $$A = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}$$ are scalar multiples of $$\begin{pmatrix} -3 & -2 \\ 1 & -1 \end{pmatrix}$$. --- Would you like more details on any part of the solution? Here are 5 related questions you could explore: 1. What are the properties of matrices that commute with a given matrix? 2. How can the concept of commutative matrices be applied in quantum mechanics? 3. Can two different matrices commute with the same matrix? 4. What are the eigenvalues and eigenvectors of the matrix $$A$$? 5. How does the solution change if we consider $$3 \times 3$$ matrices? **Tip:** The condition $$AB = BA$$ is often used to identify diagonalizable or symmetric matrices in linear algebra.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Multiplication
Commutative Property of Matrices
Linear Algebra
Formulas
Matrix product AB = BA for commutativity
System of linear equations derived from matrix entries
Solving for variables in terms of scalar multiples
Theorems
Commutativity of matrices theorem
Properties of matrix multiplication
Suitable Grade Level
Undergraduate (Linear Algebra or Higher Math Courses)
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