Math Problem Statement
solve the below math problem
Solution
Let's solve Problem 2 step-by-step:
Problem 2: Find the characteristic equation of the matrix and hence compute . Also find the matrix represented by .
Given matrix:
2 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 2 \end{pmatrix}$$ #### Step 1: Find the Characteristic Equation of $$A$$ The characteristic equation of a matrix $$A$$ is found by solving: $$\det(A - \lambda I) = 0$$ where $$\lambda$$ are the eigenvalues of $$A$$, and $$I$$ is the identity matrix. $$A - \lambda I = \begin{pmatrix} 2 - \lambda & 1 & 1 \\ 0 & 1 - \lambda & 0 \\ 1 & 1 & 2 - \lambda \end{pmatrix}$$ Calculate the determinant: $$\det(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 1 & 1 \\ 0 & 1 - \lambda & 0 \\ 1 & 1 & 2 - \lambda \end{vmatrix}$$ Expanding the determinant along the second row: \[ \det(A - \lambda I) = (2 - \lambda) \left[(1 - \lambda)(2 - \lambda) - 1\right] - 1 \left[0 \cdot (2 - \lambda) - 1 \cdot 1\right] + 1 \left[0 \cdot 1 - 1 \cdot (2 - \lambda)\right] \] Simplify the above expression: \[ = (2 - \lambda) \left[(2 - \lambda)(1 - \lambda) - 1\right] - (2 - \lambda) = (2 - \lambda) \left[2 - 3\lambda + \lambda^2 - 1\right] - (2 - \lambda) = (2 - \lambda) \left[\lambda^2 - 3\lambda + 1\right] - (2 - \lambda) = (2 - \lambda)(\lambda^2 - 3\lambda + 1) - (2 - \lambda) = (2 - \lambda)(\lambda^2 - 3\lambda + 1 - 1) = (2 - \lambda)(\lambda^2 - 3\lambda) = (2 - \lambda)\lambda(\lambda - 3) \] Thus, the characteristic equation is: $$\lambda (\lambda - 3)(\lambda - 2) = 0$$ #### Step 2: Compute $$A^{-1}$$ Using the characteristic polynomial, the eigenvalues are $$\lambda = 0, 3, 2$$. To find $$A^{-1}$$, we use the fact that: $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ The determinant $$\det(A) = 6 - 1 = 5$$. Now, compute the adjugate of $$A$$: $$\text{adj}(A) = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 5 & 0 \\ -1 & -1 & 2 \end{pmatrix}$$ Therefore, $$A^{-1} = \frac{1}{5} \begin{pmatrix} 2 & -1 & 1 \\ 0 & 5 & 0 \\ -1 & -1 & 2 \end{pmatrix}$$ #### Step 3: Find the Matrix Represented by $$A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I$$ Using the Cayley-Hamilton theorem, we know that $$A$$ satisfies its own characteristic polynomial: $$A^3 - 5A^2 + 8A - 2I = 0$$ So, $$A^8 = 5A^7 - 7A^6 + 3A^5 - 8A^2 + 2A - I$$ Therefore, $$A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I = 0$$ Thus, the resulting matrix is the zero matrix: $$A^8 - 5A^7 + 7A^6 - 3A^5 + 8A^2 - 2A + I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ --- Would you like the detailed solution for another problem, or do you have any questions on this proof? Here are some options for the next problem to solve: 1. Reduce a given matrix to diagonal form. 2. Reduce a quadratic form to canonical form. 3. Verify a given matrix identity. 4. Solve a system of linear equations by row operations. **Tip:** When working with characteristic polynomials, leveraging the Cayley-Hamilton theorem can significantly simplify matrix computations.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Algebra
Eigenvalues
Characteristic Equation
Cayley-Hamilton Theorem
Formulas
Determinant formula
Adjugate formula
Characteristic polynomial formula
Theorems
Cayley-Hamilton Theorem
Suitable Grade Level
Advanced College
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