Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -1 & -12 & 8 \\ 1 & -9 & 4 \\ 1 & -6 & 1 \end{array}\right] \mathbf{y} $$

Short answer

Question: Given a system of linear differential equations represented by the matrix A: $$ A = \begin{pmatrix} -1 & -12 & 8 \\ 1 & -9 & 4 \\ 1 & -6 & 1 \end{pmatrix} $$ Find the general solution of the system of linear differential equations. Answer: To find the general solution, follow the steps described in the analysis and solution sections, including calculating the eigenvalues and eigenvectors for the matrix A and plugging them into the general solution formula. Once done, the general solution will be expressed as: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{x}_1 + c_2 e^{\lambda_2 t} \mathbf{x}_2 + c_3 e^{\lambda_3 t} \mathbf{x}_3 $$ where \(\lambda_i\) are the eigenvalues, \(\mathbf{x}_i\) are the corresponding eigenvectors, and \(c_i\) are arbitrary constants.

Step-by-step solution

Find Eigenvalues and Eigenvectors of Matrix A

First, we need to find the eigenvalues and eigenvectors of the given matrix. To do this, we will compute the determinant of \((A-\lambda I)\), where \(A\) is the given matrix, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. $$ \det(A-\lambda I) = \begin{vmatrix} -1-\lambda & -12 & 8 \\ 1 & -9-\lambda & 4 \\ 1 & -6 & 1-\lambda \end{vmatrix} $$ Next, we will compute the determinant of the above matrix and set it equal to zero and solve for the eigenvalues \(\lambda\). Then, for each eigenvalue, we'll solve the system \((A-\lambda I)\mathbf{x} = 0\) to find the corresponding eigenvectors \(\mathbf{x}\).

Write Down the General Solution

The general solution to a system of linear differential equations of the form \(\mathbf{y}^{\prime} = A\mathbf{y}\) is given by the following expression: $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{x}_1 + c_2 e^{\lambda_2 t} \mathbf{x}_2 + ... + c_n e^{\lambda_n t} \mathbf{x}_n $$ Here, \(\lambda_i\) are the eigenvalues, \(\mathbf{x}_i\) are the corresponding eigenvectors, and \(c_i\) are arbitrary constants, which we can obtain once we're given initial conditions. Once you find the eigenvalues and eigenvectors, plug them into the general solution expression above. The final result will be the general solution to the system of linear differential equations.

Key concepts

Eigenvalues and Eigenvectors

Understanding the concepts of eigenvalues and eigenvectors is crucial when considering the stability and behavior of systems described by matrix equations. Essentially, eigenvalues are special numbers associated with a square matrix, which when multiplied by a non-zero vector (the eigenvector), result in a scalar multiple of that vector.

For a given square matrix, A, eigenvalues are denoted by \( \lambda \) and are found by calculating the determinant of the matrix formed by subtracting \( \lambda \) times the identity matrix from A, represented as \( A - \lambda I \). Setting this determinant to zero produces a characteristic polynomial whose roots are the eigenvalues.

Once we have the eigenvalues, we can find the eigenvectors by substituting each eigenvalue back into the equation \( (A - \lambda I)\mathbf{x} = 0 \) and solving for the vector \( \mathbf{x} \). Eigenvectors provide a sense of direction related to each eigenvalue, and represent the axes along which a transformation given by matrix A operates.

Determinant of a Matrix

The determinant is a scalar attribute of a square matrix that provides important algebraic and geometric insights into the system it represents. It is often denoted as \( \det(A) \) or \( |A| \) for a matrix A.

In practical terms, for a two-dimensional matrix, the determinant gives an indication of the area scaling factor for the geometric transformation, and for a three-dimensional matrix, it refers to the volume scaling factor.

When solving for eigenvalues, as in the given exercise, the determinant is pivotal. It must be set to zero to find the values of \( \lambda \) for which the system \( A - \lambda I \mathbf{x} = 0 \) has non-trivial solutions. A non-zero determinant also implies that the matrix is invertible, while a zero determinant indicates a matrix that is not invertible, which can have implications for the existence and uniqueness of solutions to systems of equations.

System of Linear Differential Equations

A system of linear differential equations consists of multiple equations that relate several functions and their derivatives. In matrix form, these systems can be compactly written as \( \mathbf{y}' = A\mathbf{y} \) where A represents a square matrix and \( \mathbf{y} \) is a vector-valued function.

The solution to such a system is greatly facilitated by finding the eigenvalues and eigenvectors of matrix A, as they form the cornerstone for the general solution. As per the step-by-step solution provided in the textbook, once the eigenvalues \( \lambda_i \) and their corresponding eigenvectors \( \mathbf{x}_i \) have been found, the general solution is a linear combination of terms \( c_i e^{\lambda_i t} \mathbf{x}_i \) where \( c_i \) are arbitrary constants, and \( e^{\lambda_i t} \) signifies that the solutions involve exponential functions.

The arbitrary constants are determined by the initial conditions of the system, which are often given in the problem statement. When all the terms are added together, they form a comprehensive solution that describes the behavior of the system over time.