Question

Find the general solution. $$ \mathbf{y}^{\prime}=\frac{1}{3}\left[\begin{array}{rrr} 1 & 1 & -3 \\ -4 & -4 & 3 \\ -2 & 1 & 0 \end{array}\right] \mathbf{y} $$

Short answer

Based on the given problem and the step-by-step solution, the general solution for the first-order linear system of differential equations is: $$\mathbf{y}(t) = c_1 \left[\begin{array}{r} 1\\ 1\\ 1 \end{array}\right] + c_2 e^{-t} \left[\begin{array}{r} -1\\ 1\\ 2 \end{array}\right]$$ where \(\mathbf{y}(t)\) is given by: $$\mathbf{y} = \left[\begin{array}{rrr} y_1\\ y_2\\ y_3 \end{array}\right]$$ and \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Find the eigenvalues of matrix A

We start by finding the characteristic equation for matrix \(A\): $$|A - \lambda I|= 0$$ where, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix of size 3x3. Using the given matrix, we get: $$\begin{vmatrix} \frac{1}{3}-\lambda & \frac{1}{3} & -1 \\ -\frac{4}{3} & -\frac{4}{3}-\lambda & 1 \\ -\frac{2}{3} & \frac{1}{3} & -\lambda \end{vmatrix}= 0$$ Which simplifies to the following characteristic equation for eigenvalues: $$ -\lambda^3 - \lambda^2 = 0 $$

Find the eigenvectors corresponding to the eigenvalues

Now, we'll find the eigenvalues by factoring the previous equation: $$\lambda^2 (\lambda + 1) = 0$$ We get eigenvalues of \(\lambda_1 = 0\), \(\lambda_2 = -1\). For each eigenvalue, we need to find corresponding eigenvectors. Let's start by finding the eigenvector for \(\lambda_1=0\): $$(A-0I)\mathbf{v}=\mathbf{0}$$ Which is equivalent to solving the following system of equations: $$\left[\begin{array}{rrr} \frac{1}{3} & \frac{1}{3} & -1 \\ -\frac{4}{3} & -\frac{4}{3} & 1 \\ -\frac{2}{3} & \frac{1}{3} & 0 \end{array}\right]\left[\begin{array}{rrr} v_{1}\\ v_{2}\\ v_{3} \end{array}\right] = \left[\begin{array}{rrr} 0\\ 0\\ 0 \end{array}\right]$$ After solving this system of equations, we get the eigenvector: $$\mathbf{v_1} = k\left[\begin{array}{r} 1\\ 1\\ 1 \end{array}\right]$$ where \(k\) is a constant. Next, we find the eigenvector for \(\lambda_2 = -1\): $$(A + I)\mathbf{v}=\mathbf{0}$$ Equivalent to solving the following system of equations: $$\left[\begin{array}{rrr} \frac{4}{3} & \frac{1}{3} & -1 \\ -\frac{4}{3} & -\frac{1}{3} & 1 \\ -\frac{2}{3} & \frac{1}{3} & 1 \end{array}\right]\left[\begin{array}{rrr} v_{1}\\ v_{2}\\ v_{3} \end{array}\right] = \left[\begin{array}{rrr} 0\\ 0\\ 0 \end{array}\right]$$ After solving this system of equations, we get the eigenvector: $$\mathbf{v_2} = k\left[\begin{array}{r} -1\\ 1\\ 2 \end{array}\right]$$ where \(k\) is a constant.

Construct the general solution

Now that we have the eigenvalues and the corresponding eigenvectors, we can construct the general solution for the differential equation system. For the given system, the general solution is given by: $$\mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v_1} + c_2 e^{\lambda_2 t} \mathbf{v_2}$$ Therefore, the general solution is: $$\mathbf{y}(t) = c_1 \left[\begin{array}{r} 1\\ 1\\ 1 \end{array}\right] + c_2 e^{-t} \left[\begin{array}{r} -1\\ 1\\ 2 \end{array}\right]$$

Key concepts

General Solution

In mathematics, solving a differential equation often involves finding a *general solution* that satisfies the equation for a broad range of initial conditions. For system of equations in linear algebra of the form \( \mathbf{y}' = A \mathbf{y} \), we utilize the concepts of eigenvalues and eigenvectors to construct this solution.

The *general solution* is crucial because it provides a way to describe all possible solutions using arbitrary constants, in this case denoted by \( c_1 \) and \( c_2 \). This means when initial conditions are specified, we can efficiently determine the particular solution.

For example, if our system is dependent on the found eigenvectors and eigenvalues, the general solution takes the form:

• \( \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v_1} + c_2 e^{\lambda_2 t} \mathbf{v_2} \)
• Where \( \lambda_1 \) and \( \lambda_2 \) are the eigenvalues, and \( \mathbf{v_1} \) and \( \mathbf{v_2} \) are the corresponding eigenvectors.

This comprehensive solution considers all potential behavior of the system, governed by the differential equation.

Characterization Equation

The *characterization equation* is an important step in finding eigenvalues for a given matrix. It is derived from the determinant expression \( |A - \lambda I| = 0 \), where \( A \) is our matrix, and \( I \) is the identity matrix.

By solving the expression, we can find the roots which are precisely the eigenvalues \( \lambda \). These eigenvalues are critical as they determine the system's behavior.

For example, from the original solution:

• We formed the equation using the matrix \( A \), leading to \( -\lambda^3 - \lambda^2 = 0 \)
• This expresses the relationship between the matrix and its inherent properties.

By factoring the polynomial, we found \( \lambda_1 = 0 \) and \( \lambda_2 = -1 \), providing insight into the solutions and dynamic modes of the system under study.

Matrix Theory

*Matrix Theory* is a vital area in linear algebra dealing with the study of matrices and their use in systems like linear equations, transformations, and now in differential equations. Matrices serve as a compact representation of a system of equations where solutions can be efficiently computed using various algebraic techniques.

In the context of eigenvalues and eigenvectors;

• The matrix \( A \) given in the original exercise is a starting point.
• It's properties like determinants and inverses help evaluate system behaviors like stability and solution existence.

More specifically, tasks like solving \((A-\lambda I)\mathbf{v}=\mathbf{0}\) to determine eigenvectors are directly applicable. Understanding matrix traits facilitates solving systems involved in many scientific and engineering scenarios.

Differential Equations

*Differential equations* are mathematical equations that involve functions and their derivatives, representing a large array of phenomena including physics, engineering, and economics.

In linear algebra, when a differential equation involves matrices, it typically looks like \( \mathbf{y}' = A \mathbf{y} \). The solutions to these equations help predict the change of systems over time under various conditions. They can describe dynamics such as population growth, thermal conduction, or electrical circuits.

We found that by solving for eigenvalues and eigenvectors, you can construct solutions to such differential equations. Crucial steps are:

• Finding the characterization equation and solving for \( \lambda \).
• Determining the corresponding eigenvectors.
• Using them to form the exponential solution involving initial conditions.

This comprehensive method allows for an in-depth understanding of the system dynamics.