Chapter 10: Problem 43
Find the general solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & -1 & 2 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right) \mathbf{x} $$
Short Answer
Expert verified
Solve the characteristic equation for eigenvalues and eigenvectors to express the solution as a combination of exponential functions.
Step by step solution
01
Write the matrix differential equation
The given system can be expressed in matrix form as \( \mathbf{X}' = A \mathbf{x} \), where \( A \) is the matrix \( \begin{pmatrix} 1 & -1 & 2 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{pmatrix} \). To solve the system, we need to find the eigenvalues and eigenvectors of matrix \( A \).
02
Find the characteristic equation
The characteristic equation is obtained by solving \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculate \( A - \lambda I \) and find the determinant.
03
Solve for eigenvalues
Calculate \( \det(A - \lambda I) \) which results in the polynomial equation \(-\lambda^3 + 3\lambda^2 - 2 = 0\). Factoring gives \( (\lambda - 1)^2(\lambda - 2) = 0 \). Thus, the eigenvalues are \( \lambda = 1 \) with multiplicity 2, and \( \lambda = 2 \).
04
Find eigenvectors for each eigenvalue
For each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = 0 \) to find the eigenvector \( \mathbf{v} \). Start with \( \lambda = 1 \). The matrix \( A - I \) simplifies to \( \begin{pmatrix} 0 & -1 & 2 \ -1 & 0 & 0 \ -1 & 0 & 0 \end{pmatrix} \). Solve to find eigenvectors. Repeat this for \( \lambda = 2 \) using \( A - 2I \).
05
Find the general solution
Using the eigenvalues and eigenvectors, write the general solution. The solution will be a linear combination of \( \mathbf{x}_1(t) = e^{\lambda_1 t} \mathbf{v}_1 \) and \( \mathbf{x}_2(t) = e^{\lambda_2 t} \mathbf{v}_2 \), where \( \lambda \) is an eigenvalue and \( \mathbf{v} \) is a corresponding eigenvector.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eigenvalues and Eigenvectors
In the realm of matrix differential equations, eigenvalues and eigenvectors play a crucial role in finding solutions. An eigenvalue is a scalar that, when multiplied by an eigenvector, leaves the direction of the vector unchanged after a linear transformation. Eigenvectors are non-zero vectors associated with a particular eigenvalue which, when multiplied by the transformation matrix, result in a scalar multiple of the vector itself.
To better understand this, consider a square matrix \( A \). If there exists a non-zero vector \( \mathbf{v} \) and a scalar \( \lambda \) such that:
Knowing the eigenvalues and eigenvectors of a matrix allows us to express complex transformations in simplified terms, which is crucial for solving systems of differential equations. In our example, eigenvalues \( \lambda = 1 \) with multiplicity 2, and \( \lambda = 2 \) were found, each with their respective eigenvectors. These concepts are essential in writing the general solution of the matrix differential equation.
To better understand this, consider a square matrix \( A \). If there exists a non-zero vector \( \mathbf{v} \) and a scalar \( \lambda \) such that:
- \( A\mathbf{v} = \lambda \mathbf{v} \)
Knowing the eigenvalues and eigenvectors of a matrix allows us to express complex transformations in simplified terms, which is crucial for solving systems of differential equations. In our example, eigenvalues \( \lambda = 1 \) with multiplicity 2, and \( \lambda = 2 \) were found, each with their respective eigenvectors. These concepts are essential in writing the general solution of the matrix differential equation.
Characteristic Equation
The characteristic equation is a polynomial equation in terms of \( \lambda \) that one must solve to find the eigenvalues of a matrix. Finding it is a key step in solving matrix differential equations.
To derive it, you need to find the determinant of the matrix \( A - \lambda I \), where \( I \) is the identity matrix of the same size as \( A \). This leads to the characteristic equation:
Factoring this polynomial resulted in:
To derive it, you need to find the determinant of the matrix \( A - \lambda I \), where \( I \) is the identity matrix of the same size as \( A \). This leads to the characteristic equation:
- \( \det(A - \lambda I) = 0 \)
Factoring this polynomial resulted in:
- \((\lambda - 1)^2(\lambda - 2) = 0\)
Matrix Differential Equations
Matrix differential equations are a class of differential equations where the unknown function is a vector and the system is described using matrices. The general form of a linear matrix differential equation is:
Using these, the solution to the equation can be expressed as a linear combination of exponential functions of the eigenvalues and their associated eigenvectors:
Mastering the methods related to matrix differential equations is essential for solving complex systems found in various scientific and engineering fields.
- \( \mathbf{X}' = A\mathbf{x} \)
Using these, the solution to the equation can be expressed as a linear combination of exponential functions of the eigenvalues and their associated eigenvectors:
- \( \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v_1} + c_2 e^{\lambda_2 t} \mathbf{v_2} + \dots \)
Mastering the methods related to matrix differential equations is essential for solving complex systems found in various scientific and engineering fields.