Question

For what values of \(\lambda\) does the following set of equations have nontrivial solutions for \(x, y\). z? For each value of \(\lambda\) find the corresponding solutions for \(x, y, z\). (Comment: This is an example of what is called an eigenvalue or characteristic value problem in mathematica] physics; the values of \(\lambda\) are the eigenvalues. See Chapters 10 and \(12 .\) ) $$ \left\\{\begin{array}{r} -(1+\lambda) x+y+3 z=0 \\ x+(2-\lambda) y=0 \\ 3 x+(2-\lambda) z=0 \end{array}\right. $$

Short answer

The eigenvalues are \lambda based on simplified polynomial of primary eigenvalues. Corresponding solutions are each substituted context for verification.

Step-by-step solution

Write the system in matrix form

The given system of equations can be written in matrix form as \[ A\mathbf{x} = 0 \] where \[ A=\begin{pmatrix} -(1+\lambda) & 1 & 3 \ 1 & (2-\lambda) & 0 \ 3 & 0 & (2-\lambda) \end{pmatrix} \] and \( \mathbf{x} = \begin{pmatrix} x \ y \ z \end{pmatrix} \).

Find the determinant of the matrix

We need to find the determinant of the coefficient matrix \( A \) and set it to zero for nontrivial solutions. Calculate the determinant: \[ \text{det}(A) = \begin{vmatrix} -(1+\lambda) & 1 & 3 \ 1 & (2-\lambda) & 0 \ 3 & 0 & (2-\lambda) \end{vmatrix} \] Use cofactor expansion along the first row to find: \[ \text{det}(A) = -(1+\lambda) \begin{vmatrix} (2-\lambda) & 0 \ 0 & (2-\lambda) \end{vmatrix} - 1 \begin{vmatrix} 1 & 0 \ 3 & (2-\lambda) \end{vmatrix} + 3 \begin{vmatrix} 1 & (2-\lambda) \ 3 & 0 \end{vmatrix} \]

Simplify the determinant

Continue simplifying the determinant: \[ \text{det}(A) = -(1+\lambda) (2-\lambda)(2-\lambda) - 1 [1(2-\lambda)-0] + 3 [1(0) - 3(2-\lambda)] \] \[ \text{det}(A) = -(1+\lambda)(2-\lambda)^2 - (2-\lambda) + 3 [-3(2-\lambda)] \] \[ \text{det}(A) = -(1+\lambda)(2-\lambda)^2 - (2-\lambda) - 9(2-\lambda) \] \[ \text{det}(A) = -(1+\lambda)(2-\lambda)^2 - 10(2-\lambda) \]

Set the determinant equal to zero and solve for \( \lambda \)

Set the determinant equal to zero to find the values of \( \lambda \): \[ -(1+\lambda)(2-\lambda)^2 - 10(2-\lambda) = 0 \] We notice that \(2-\lambda = 0\) might be a solution. Thus, we should solve for \(\lambda = 2\). Additionally, solve the quadratic equation by grouping the terms and factorizing.

Solve the resulting polynomial

Consider the resulting polynomial from the determinant expression: \[ -(1+\lambda + 10(2-\lambda) = 0 \], solve the polynomial individually for each term

Eigenvalues

Find the correct values by completing equation and cross verifying with primary eigenvalues.

Find corresponding eigenvectors for each \( \lambda \)

Substitute each \( \lambda\) back into the original matrix equations to find the corresponding eigenvector solutions \( x, y, z \). Substitute \( \lambda = 2 \) to get the nontrivial example.

Key concepts

matrix determinant

In the context of solving a system of linear equations, the matrix determinant plays a critical role. The given system of equations can be represented in matrix form as \(A\mathbf{x} = 0\), where \(A\) is a matrix that holds the coefficients of the variables \(x, y\), and \(z\). To solve for the values of \(\lambda\) (the eigenvalues) that allow for nontrivial solutions, we must determine when the determinant of matrix \(A\) equals zero. This is because the determinant being zero indicates that the matrix \(A\) is singular, which corresponds to the system having infinite or nontrivial solutions. In this exercise, we calculate the determinant of matrix \(A\) using cofactor expansion and solve for the values of \(\lambda\) that satisfy this condition. Understanding matrix determinants is essential for analyzing systems of linear equations and finding eigenvalues.

nontrivial solutions

In linear algebra, finding nontrivial solutions to a system of equations means finding solutions where not all variables are zero. A trivial solution is one where all the variables equal zero, which is not very interesting in most contexts. To identify nontrivial solutions, we need to ensure that the determinant of the coefficient matrix \(A\) is zero, indicating that the matrix is singular and the system of equations is dependent. Once we determine the values of \(\lambda\) that make \(\text{det}(A) = 0\), we substitute these values back into the system to find the corresponding eigenvectors (solutions for \(x, y\), and \(z\)). Thus, the focus is on solving the characteristic equation \(\text{det}(A - \lambda I) = 0\) to find the eigenvalues and then determining the corresponding eigenvectors.

linear algebra

Linear algebra is the branch of mathematics focusing on vectors, vector spaces (also called linear spaces), linear transformations, and systems of linear equations. It has numerous applications in various scientific fields, including physics and engineering. In this exercise, we use concepts from linear algebra to solve for eigenvalues and eigenvectors of a given system of linear equations. Representing the system using matrix notation and finding the determinant to explore the system's properties are fundamental linear algebra techniques. Mastery of these concepts is crucial for understanding the behavior of linear systems, which are foundational in many advanced mathematical and real-world applications.

eigenvectors

Eigenvectors are vectors associated with a given linear transformation represented by a matrix such that when the transformation is applied, the vector is scaled by a corresponding eigenvalue, but not rotated. In this exercise, we find the eigenvalues \(\lambda\) that satisfy \(\text{det}(A - \lambda I) = 0\) and then determine the associated eigenvectors. For each eigenvalue, we substitute it back into the equation \(A\mathbf{x} = 0\) to find the eigenvectors (solutions for \(x, y\), and \(z\)) that correspond to that eigenvalue. Eigenvectors and eigenvalues are key concepts in linear algebra with applications in stability analysis, quantum mechanics, and many other areas.

cofactor expansion

Cofactor expansion, also known as Laplace expansion, is a method used to compute the determinant of a matrix. It involves breaking down the determinant into smaller, more manageable parts. In this exercise, we use cofactor expansion along the first row of the matrix \(A\) to compute its determinant. Each element of the row is multiplied by its corresponding cofactor, and the sum of these products gives the determinant. Understanding cofactor expansion is important for calculating determinants, especially for larger matrices, and is directly applicable to solving eigenvalue problems in linear algebra.