Math Problem Statement
Diagonalize the matrix 2 1 -1 1 1 -2 1-2 1 Find A?
Solution
To diagonalize the matrix , we need to find its eigenvalues and eigenvectors. Then, the matrix can be diagonalized as , where is the diagonal matrix of eigenvalues, and is the matrix whose columns are the corresponding eigenvectors.
Given matrix :
2 & 1 & -1 \\ 1 & 1 & -2 \\ 1 & -2 & 1 \end{pmatrix}$$ ### Step 1: Find the Eigenvalues To find the eigenvalues, we solve the characteristic equation: $$\text{det}(A - \lambda I) = 0$$ where $$\lambda$$ is an eigenvalue and $$I$$ is the identity matrix. $$\text{det}\left(\begin{pmatrix} 2 & 1 & -1 \\ 1 & 1 & -2 \\ 1 & -2 & 1 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\right) = 0$$ This simplifies to: $$\text{det}\begin{pmatrix} 2-\lambda & 1 & -1 \\ 1 & 1-\lambda & -2 \\ 1 & -2 & 1-\lambda \end{pmatrix} = 0$$ ### Step 2: Compute the Determinant Let's calculate the determinant of the matrix: $$\text{det}\begin{pmatrix} 2-\lambda & 1 & -1 \\ 1 & 1-\lambda & -2 \\ 1 & -2 & 1-\lambda \end{pmatrix}$$ I'll calculate this determinant to find the eigenvalues.The characteristic equation is: $$-\lambda^3 + 4\lambda^2 - \lambda - 6 = 0$$ The eigenvalues are: $$\lambda_1 = -1, \quad \lambda_2 = 2, \quad \lambda_3 = 3$$ ### Step 3: Find the Eigenvectors Now, we need to find the eigenvectors corresponding to each eigenvalue by solving $$(A - \lambda I) \mathbf{v} = 0$$ for each $$\lambda$$. I'll calculate the eigenvectors for these eigenvalues.The eigenvectors corresponding to the eigenvalues are: 1. For $$\lambda_1 = -1$$: $$\mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$$ 2. For $$\lambda_2 = 2$$: $$\mathbf{v}_2 = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$$ 3. For $$\lambda_3 = 3$$: $$\mathbf{v}_3 = \begin{pmatrix} -4 \\ -3 \\ 1 \end{pmatrix}$$ ### Step 4: Form the Matrices $$P$$ and $$D$$ Now, we can form the matrix $$P$$ with the eigenvectors as columns and the diagonal matrix $$D$$ with the eigenvalues. $$P = \begin{pmatrix} 0 & 3 & -4 \\ 1 & 1 & -3 \\ 1 & 1 & 1 \end{pmatrix}$$ $$D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$ Thus, the matrix $$A$$ can be diagonalized as $$A = PDP^{-1}$$. Do you want details on any specific part or have any questions? --- Here are some related questions you might find useful: 1. How do you compute the inverse of matrix $$P$$? 2. What does it mean for a matrix to be diagonalizable? 3. Can all matrices be diagonalized? Why or why not? 4. How do you verify that $$A = PDP^{-1}$$? 5. What are the applications of diagonalization in linear algebra? 6. How do eigenvalues and eigenvectors relate to matrix transformations? 7. What is the significance of the diagonal elements in the matrix $$D$$? 8. How does the process differ if the matrix has complex eigenvalues? **Tip:** When finding eigenvectors, it's essential to check for linear independence to ensure that the matrix can be diagonalized.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Eigenvectors
Diagonalization
Formulas
Characteristic equation
Eigenvalue-eigenvector equation
Diagonalization formula
Theorems
Diagonalization theorem
Suitable Grade Level
Advanced Undergraduate