Question

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

Short answer

Answer: The eigenvalues of the matrix $A$ are 1, 4, and 5.

Step-by-step solution

Rewrite the linear system as a matrix equation

First, we rewrite the given linear system as a matrix equation by identifying the coefficients in each equation and arranging them in a matrix. This results in: $$ \begin{pmatrix} y_{1}^{\prime} \\ y_{2}^{\prime} \\ y_{3}^{\prime} \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 2 & 2 \end{pmatrix} \begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix} $$ So, our matrix equation becomes \(\mathbf{y}^{\prime}=A \mathbf{y}\), where $A=\begin{pmatrix} 5 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 2 & 2 \end{pmatrix}$.

Compute the eigenvalues of matrix \(A\)

To find the eigenvalues of the matrix \(A\), we need to solve the characteristic equation: $$\det(A-\lambda I)=0$$, where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix. First, subtract \(\lambda\) from the diagonal of matrix \(A\) then compute the determinant: $$ \det(A-\lambda I) = \det \begin{pmatrix} 5-\lambda & 0 & 0 \\ 0 & 1-\lambda & 3 \\ 0 & 2 & 2-\lambda \end{pmatrix} $$ Computing the determinant, we get: $$ (5-\lambda)((1-\lambda)(2-\lambda) - (3)(2)) = 0 $$ Which simplifies to: $$ (5-\lambda)(\lambda^2 - 3 \lambda - 4) = 0 $$ The roots of this equation (the eigenvalues) are \(\lambda = 1\), \(\lambda = 4\) and \(\lambda = 5\).

Key concepts

Eigenvalue Calculation

Calculating the eigenvalues of a matrix is like finding the hidden powers of a matrix. These special numbers tell us about the matrix's essential characteristics. Imagine you have a matrix \( A \). To find its eigenvalues, we need to solve the characteristic equation:

• Subtract \( \lambda \) from each diagonal element of \( A \)
• Form a new matrix \( A-\lambda I \)
• Take the determinant of this new matrix
• Set it equal to zero and solve for \( \lambda \)

In our example, after setting up \( A-\lambda I \) and computing its determinant, we get a quadratic equation. The solutions to this equation are the eigenvalues \( \lambda = 1 \), \( \lambda = 4 \), and \( \lambda = 5 \). Each eigenvalue shows how the matrix scales and transforms space, much like unique fingerprints for the matrix.

Matrix Representation

Matrix representation is a powerful way to organize systems of linear equations. Instead of dealing with multiple equations, a matrix allows us to compactly write and solve them all at once. In the problem we've tackled, the system of equations for the variables \( y_1 \), \( y_2 \), and \( y_3 \) is rewritten using a matrix \( A \). Here's how:

• Identify coefficients of each variable in the equations
• Organize these coefficients into a matrix form

This setup lets us use matrix operations to solve the system quickly and efficiently. The given system was transformed into the matrix equation \( \mathbf{y}^{\prime} = A \mathbf{y} \). This compact representation makes it easier to solve the system using techniques like finding eigenvalues and eigenvectors.

Differential Equations

Differential equations involve unknown functions and their derivatives. They are crucial in modeling how things change over time or space. For systems of differential equations, matrices play a vital role in representing and solving them. In our exercise, the matrix equation \( \mathbf{y}^{\prime} = A \mathbf{y} \) is actually a set of linear differential equations. By using matrices, we streamline solving these equations, making it simpler to understand the changes in all variables \( y_1 \), \( y_2 \), and \( y_3 \). Differential equations often arise in physics, engineering, and many other fields to describe dynamic systems. When we rewrite these systems in matrix form, we facilitate more efficient and effective computations.

Linear Systems

Linear systems are collections of linear equations that describe relationships between different variables. They can appear daunting at first, with multiple equations and variables, but they become manageable using matrix techniques. For the linear system in our exercise, we have three equations representing how \( y_1 \), \( y_2 \), and \( y_3 \) change. Each equation involves a linear combination of variables, which is a hallmark of linear systems.

• By assembling these equations into a matrix, we transform the problem into a single matrix equation
• This matrix equation allows for neat algebraic manipulations and the application of linear algebra tools

Understanding and representing these systems as matrix equations form the foundation for studying more complex mathematical problems and utilizes the power of matrices to solve real-world issues efficiently.