Question

Rewrite the linear system as a matrix equation \(\mathbf{y}^{\prime}=A \mathbf{y}\), and compute the eigenvalues of the matrix \(A\). $$ \begin{aligned} &y_{1}^{\prime}=y_{2}-3 y_{3} \\ &y_{2}^{\prime}=-5 y_{2}-4 y_{3} \\ &y_{3}^{\prime}=8 y_{2}+7 y_{3} \end{aligned} $$

Short answer

Answer: The eigenvalues of matrix A are 3 and -60.

Step-by-step solution

Rewrite the linear system as a matrix equation

To rewrite the given system as a matrix equation, we first write down the coefficients of the variables in the matrix form: $$ A = \begin{pmatrix} 0 & 1 & -3 \\ 0 & -5 & -4 \\ 0 & 8 & 7 \\ \end{pmatrix} $$ Now, we can express the system as \(\mathbf{y}^{\prime}=A\mathbf{y}\), where \(\mathbf{y}=\begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}\)

Compute the eigenvalues of matrix \(A\)

Now we'll need to compute the eigenvalues of the matrix \(A\). To do this, we will solve the characteristic equation which is given as follows: $$ \det(A - \lambda I) = 0 $$ Where, \(I\) is the identity matrix of the same size as \(A\), and \(\lambda\) represents the eigenvalue. Let's construct \(A - \lambda I\): $$ A - \lambda I = \begin{pmatrix} 0 - \lambda & 1 & -3 \\ 0 & -5 - \lambda & -4 \\ 0 & 8 & 7 - \lambda \\ \end{pmatrix} $$ Now let's find the determinant of \((A - \lambda I)\): $$ \det(A - \lambda I) = -\lambda(-5-\lambda)(7-\lambda) - (-3 \cdot 4 \cdot 8)= -\lambda (\lambda^2 + 5\lambda - 60 + 52\lambda) = -\lambda\left(\lambda^2 + 57\lambda - 60\right) $$ Now let's find the roots of the polynomial \(\lambda^2 + 57\lambda- 60 = 0\): $$ (\lambda - 3)(\lambda + 60) = 0 $$ Therefore, we have the following eigenvalues for matrix \(A\): $$ \lambda_1 = 3, \quad \lambda_2 = -60 $$

Key concepts

Eigenvalues

An eigenvalue is a special scalar associated with a linear transformation represented by a matrix. Calculating eigenvalues is essential in understanding the properties of the matrix and the systems it represents. In matrix algebra, eigenvalues are values such that when they multiply a vector (known as an eigenvector), the resulting vector is a scalar multiple of the original eigenvector.

To find the eigenvalues of a matrix, we solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(A\) is our matrix, \(I\) is the identity matrix, and \(\lambda\) represents the eigenvalue. The roots of this equation give us the eigenvalues. For our matrix \(A = \begin{pmatrix} 0 & 1 & -3 \ 0 & -5 & -4 \ 0 & 8 & 7 \end{pmatrix}\), the eigenvalues were calculated to be \(\lambda_1 = 3\) and \(\lambda_2 = -60\).

These eigenvalues help us analyze how the components of a vector transform under the matrix \(A\) and can be applied to various fields including differential equations, stability analysis, and even quantum mechanics.

Linear System

A linear system refers to a collection of linear equations that describe some physical situation or mathematical problem. Such a system can involve scalar equations but is often more conveniently expressed with matrix notation, especially for large systems. In the exercise, we dealt with three linear differential equations that describe a dynamic system.

The system is represented as:

• \(y_{1}^{\prime} = y_{2} - 3y_{3}\)
• \(y_{2}^{\prime} = -5y_{2} - 4y_{3}\)
• \(y_{3}^{\prime} = 8y_{2} + 7y_{3}\)

This set of equations describes how the derivatives of the variables \(y_1\), \(y_2\), and \(y_3\) change over time based on the values of the variables themselves. Rewriting this as a matrix equation allows us to use tools like eigenvalues to better understand and solve the system.

Matrix Equation

Transforming a linear system into a matrix equation is a crucial step for efficient handling and solution. A matrix equation generally has the form \(\mathbf{y}^{\prime} = A \mathbf{y}\), where \(\mathbf{y}\) is a vector of variables, and \(A\) is a matrix of coefficients from the linear system.

For the given system, the matrix form is expressed with \(A = \begin{pmatrix} 0 & 1 & -3 \ 0 & -5 & -4 \ 0 & 8 & 7 \end{pmatrix}\), and \(\mathbf{y} = \begin{pmatrix} y_1 \ y_2 \ y_3 \end{pmatrix}\). This matrix equation clearly organizes the system of differential equations, allowing us to apply linear algebra techniques. Each entry in matrix \(A\) comes from the coefficients of the variables in the differential equations.

Using this structure, determining eigenvalues and ultimately solving the system becomes more systematic. It aids in understanding the behavior of dynamic systems and finding solutions where direct methods are cumbersome.