Math Problem Statement
Diagonalize the matrix 2 1 -1 1 1 -2 -1-2 1 Find A?
Solution
To diagonalize the matrix, we first need to find its eigenvalues and corresponding eigenvectors, and then express the matrix in the form , where is a diagonal matrix consisting of the eigenvalues of , and is the matrix whose columns are the corresponding eigenvectors.
Given the matrix :
2 & 1 & -1 \\ 1 & 1 & -2 \\ -1 & -2 & 1 \end{pmatrix}$$ ### Step 1: Find the Eigenvalues The eigenvalues are found by solving the characteristic equation: $$\text{det}(A - \lambda I) = 0$$ where $$I$$ is the identity matrix and $$\lambda$$ is the eigenvalue. Let's compute this determinant. $$\text{det}(A - \lambda I) = \text{det} \begin{pmatrix} 2 - \lambda & 1 & -1 \\ 1 & 1 - \lambda & -2 \\ -1 & -2 & 1 - \lambda \end{pmatrix}$$ We can now find this determinant and solve for $$\lambda$$. I'll calculate it.The characteristic equation is: $$-\lambda^3 + 4\lambda^2 + \lambda - 4 = 0$$ The eigenvalues of the matrix $$A$$ are $$\lambda_1 = -1$$, $$\lambda_2 = 1$$, and $$\lambda_3 = 4$$. ### Step 2: Find the Eigenvectors Next, we find the eigenvectors corresponding to each eigenvalue by solving the system $$(A - \lambda I) \mathbf{v} = 0$$. I'll calculate the eigenvectors for each eigenvalue.The eigenvectors corresponding to the eigenvalues are as follows: 1. For $$\lambda_1 = -1$$, the eigenvector is $$\mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$$. 2. For $$\lambda_2 = 1$$, the eigenvector is $$\mathbf{v}_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$$. 3. For $$\lambda_3 = 4$$, the eigenvector is $$\mathbf{v}_3 = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix}$$. ### Step 3: Form the Matrix $$P$$ and $$D$$ The matrix $$P$$ is formed by placing the eigenvectors as its columns, and the matrix $$D$$ is a diagonal matrix with the corresponding eigenvalues. $$P = \begin{pmatrix} 0 & 2 & -1 \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{pmatrix}$$ $$D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}$$ ### Step 4: Calculate $$P^{-1}$$ Finally, we find $$P^{-1}$$ to express $$A$$ as $$A = PDP^{-1}$$. I'll calculate $$P^{-1}$$ next.The matrix $$P$$ and its inverse $$P^{-1}$$ are: $$P = \begin{pmatrix} 0 & 2 & -1 \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{pmatrix}$$ $$P^{-1} = \begin{pmatrix} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{3} & -\frac{1}{6} & \frac{1}{6} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{1}{3} \end{pmatrix}$$ ### Final Answer: Diagonalization of $$A$$ Thus, the matrix $$A$$ can be diagonalized as: $$A = PDP^{-1}$$ where $$D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix}$$ Would you like more details or have any further questions? --- **Related Questions:** 1. What is the significance of diagonalizing a matrix? 2. Can all square matrices be diagonalized? 3. How does the diagonalization process change if eigenvalues are repeated? 4. What are the applications of diagonalization in differential equations? 5. How does the concept of eigenvectors apply to other fields, like quantum mechanics? 6. What if a matrix has complex eigenvalues? How does diagonalization work then? 7. How does diagonalization relate to matrix exponentiation? 8. What is the geometric interpretation of eigenvectors and eigenvalues? **Tip:** When solving for eigenvectors, make sure they are linearly independent to correctly form the matrix $$P$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Diagonalization
Eigenvalues
Eigenvectors
Formulas
Characteristic equation
Theorems
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Suitable Grade Level
Advanced College Level
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