Math Problem Statement
the folowing matrix has three distinct eigenvalues if and only what values of x and y are true x<k<y for the matrix -1,1,0.-2,5,1.k,0,0.
Solution
To determine the values of and such that the given matrix has three distinct eigenvalues, we need to calculate the eigenvalues and analyze their properties. The matrix is:
-1 & 1 & 0 \\ -2 & 5 & 1 \\ k & 0 & 0 \end{bmatrix}$$ ### Steps to Solve: 1. **Eigenvalues** are the roots of the characteristic polynomial $$\det(A - \lambda I) = 0$$, where $$\lambda$$ is the eigenvalue, and $$I$$ is the identity matrix. 2. Compute $$A - \lambda I$$: $$A - \lambda I = \begin{bmatrix} -1 - \lambda & 1 & 0 \\ -2 & 5 - \lambda & 1 \\ k & 0 & -\lambda \end{bmatrix}$$ 3. Calculate the determinant: $$\det(A - \lambda I) = \begin{vmatrix} -1 - \lambda & 1 & 0 \\ -2 & 5 - \lambda & 1 \\ k & 0 & -\lambda \end{vmatrix}.$$ Using cofactor expansion along the third column: $$\det(A - \lambda I) = -\lambda \begin{vmatrix} -1 - \lambda & 1 \\ -2 & 5 - \lambda \end{vmatrix}.$$ 4. Compute the determinant of the $$2 \times 2$$ submatrix: $$\begin{vmatrix} -1 - \lambda & 1 \\ -2 & 5 - \lambda \end{vmatrix} = (-1 - \lambda)(5 - \lambda) - (1)(-2).$$ Simplify: $$= (-1 - \lambda)(5 - \lambda) + 2 = -5 - \lambda + \lambda^2 - 5\lambda + 2 = \lambda^2 - 6\lambda - 3.$$ 5. Substitute back: $$\det(A - \lambda I) = -\lambda (\lambda^2 - 6\lambda - 3).$$ 6. The eigenvalues are the roots of $$-\lambda (\lambda^2 - 6\lambda - 3) = 0$$. Thus: $$\lambda_1 = 0, \quad \lambda_2, \lambda_3 \text{ are the roots of } \lambda^2 - 6\lambda - 3 = 0.$$ ### Conditions for Three Distinct Eigenvalues To have three distinct eigenvalues: 1. $$\lambda = 0$$ must be distinct. 2. The quadratic equation $$\lambda^2 - 6\lambda - 3 = 0$$ must have two distinct roots. The discriminant of the quadratic is: $$\Delta = b^2 - 4ac = (-6)^2 - 4(1)(-3) = 36 + 12 = 48 > 0.$$ Since the discriminant is positive, $$\lambda_2$$ and $$\lambda_3$$ are distinct. Therefore, the matrix will have three distinct eigenvalues for all $$k$$, as the eigenvalues do not depend on $$k$$ directly. ### Conclusion The matrix has three distinct eigenvalues if and only if $$\lambda^2 - 6\lambda - 3$$ has distinct roots. This condition is satisfied for all $$k$$. Therefore, $$x$$ and $$y$$ can represent the range of the eigenvalues of the quadratic part, depending on the roots. For clarification or further numerical bounds, please confirm if more details are needed. --- #### 5 Related Questions: 1. What are the explicit roots of $$\lambda^2 - 6\lambda - 3 = 0$$? 2. How do changes in $$k$$ affect the eigenvalues of this matrix? 3. Can the eigenvectors be calculated for this matrix? 4. What is the geometric interpretation of having three distinct eigenvalues? 5. How does this result generalize to higher dimensions? #### Tip: To check eigenvalues numerically, always compute the discriminant for quadratic terms and analyze its sign for distinct solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Eigenvalues
Matrix Determinants
Quadratic Equations
Formulas
Characteristic Polynomial: det(A - λI) = 0
Quadratic Formula: λ = (-b ± √(b² - 4ac)) / 2a
Determinant of a 2x2 matrix: det = ad - bc
Theorems
Eigenvalue Theorem
Properties of Determinants
Discriminant Analysis for Quadratic Equations
Suitable Grade Level
Undergraduate (College-level Linear Algebra)