Question

Find the general solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 4 & 12 \\ -3 & -8 \end{array}\right] \mathbf{y} $$

Short answer

Answer: The general solution of the given matrix differential equation represents the expression for the function $\mathbf{y}(t)$ that satisfies the equation, which can be expressed as a linear combination of the eigenvector solutions associated with the eigenvalues: $\mathbf{y}(t) = c_1 e^{3t} \begin{bmatrix}2 \\ 1\end{bmatrix} + c_2 e^{-2t} \begin{bmatrix}1 \\ 1\end{bmatrix}.$

Step-by-step solution

Identify the given matrix

The given matrix differential equation is: $\mathbf{y}^{\prime}=\left[\begin{array}{rr}2 & -4 \\\ -1 & -1\end{array}\right] \mathbf{y}$ And the matrix we need to analyze is: \(A = \left[\begin{array}{rr}2 & -4 \\\ -1 & -1\end{array}\right]\)

Find the eigenvalues of the matrix

To find the eigenvalues, we will use the following equation: \(|A - \lambda I| = 0\) Using this, we get: \(|\begin{array}{cc}2 - \lambda & -4 \\ -1 & -1 - \lambda\end{array}| = 0\) Calculate the determinant of the matrix: \((2 - \lambda)(-1 - \lambda) - (-1)(-4) = \lambda^2 - \lambda - 2 - 4 = \lambda^2 - \lambda - 6 = 0\) The eigenvalues are the roots of this equation: \(\lambda_1 = 3, \lambda_2 = -2\)

Find the eigenvectors of the matrix

To find the eigenvectors, we need to solve the following system of linear equations for each eigenvalue: \((A - \lambda I) \mathbf{v} = 0\) For \(\lambda_1 = 3\): \((A - 3I) \mathbf{v} = \left[\begin{array}{cc}-1 & -4 \\ -1 & -4\end{array}\right] \mathbf{v} = \mathbf{0}\) Eigenvector for \(\lambda_1=3\) is: \(\mathbf{v_1} = \begin{bmatrix}2 \\ 1\end{bmatrix}\) For \(\lambda_2 = -2\): \((A + 2I) \mathbf{v} = \left[\begin{array}{cc}4 & -4 \\ -1 & 1\end{array}\right] \mathbf{v} = \mathbf{0}\) Eigenvector for \(\lambda_2=-2\) is: \(\mathbf{v_2} = \begin{bmatrix}1 \\ 1\end{bmatrix}\)

Write the general solution of the matrix differential equation

The general solution of the matrix differential equation is a linear combination of solutions associated with each eigenvalue. In this case, the general solution can be written as: \(\mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v_1} + c_2 e^{\lambda_2 t} \mathbf{v_2}\) Substituting the eigenvalues and eigenvectors: \(\mathbf{y}(t) = c_1 e^{3t} \begin{bmatrix}2 \\ 1\end{bmatrix} + c_2 e^{-2t} \begin{bmatrix}1 \\ 1\end{bmatrix}\) Where \(c_1\) and \(c_2\) are arbitrary constants to be determined by initial conditions.

Key concepts

Eigenvalues and Eigenvectors

Understanding the concepts of eigenvalues and eigenvectors is essential for solving many types of problems in linear algebra, including matrix differential equations. An eigenvalue, often denoted as \(\lambda\), is a scalar that indicates a magnitude of stretching or compressing that an eigenvector undergoes during a linear transformation represented by a matrix.

Let's break it down with a simple analogy: Imagine a fabric, and a force is applied along particular lines of the fabric, altering its shape. The lines that stay aligned with the force and only change in length, not in direction, identify the direction of the eigenvectors, and the factor by which they stretch or shrink represents their associated eigenvalues.

To find the eigenvalues of a matrix, we solve the characteristic equation \(\vert A - \lambda I \vert = 0\) where \(A\) is the matrix and \(I\) is the identity matrix of the same dimension. The eigenvectors are then determined by solving the equation \( (A - \lambda I) \mathbf{v} = \mathbf{0} \) for each eigenvalue \(\lambda\), where \(\mathbf{v}\) is the eigenvector. In the context of differential equations, each eigenvector represents a direction along which the system evolves, and each eigenvalue corresponds to the rate of evolution in that direction.

General Solution of Differential Equations

The general solution to a system of differential equations describes all the possible behaviors of the system under various initial conditions. In a matrix differential equation of the form \( \mathbf{y}' = A\mathbf{y} \), where \( A \) is a square matrix, we can find this general solution by incorporating the eigenvalues and eigenvectors of the matrix \( A \).

The general solution is a sum of terms, each involving an exponential function of time scaled by a constant and multiplied by one of the eigenvectors. This looks like \( \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v_1} + c_2 e^{\lambda_2 t} \mathbf{v_2} \), where \(c_1\) and \(c_2\) are constants that depend on initial conditions, and \( \lambda_1, \lambda_2 \) are the eigenvalues with corresponding eigenvectors \( \mathbf{v_1}, \mathbf{v_2} \).

Such representation makes it clear and manageable to understand how the system's state evolves over time. This is especially useful in fields like physics, engineering, and economics, where systems often change in a predictable way over time.

Systems of Linear Equations

In algebra, a system of linear equations is a collection of two or more linear equations involving the same set of variables. We encounter such systems frequently when dealing with eigenvectors, where the system takes the form \( (A - \lambda I) \mathbf{v} = \mathbf{0} \). Solving this system effectively is pivotal as it uncovers the eigenvectors, which are crucial to finding the general solution of a matrix differential equation.

To solve a system of linear equations, we can use methods like substitution, elimination, or matrix-based approaches such as finding the inverse or applying row reduction to echelon form. In the example included in the original exercise, we can notice how we move from a matrix equation to find specific vectors that satisfy it.

Handling Redundancies

When working with systems, it may seem that we have many equations, but in cases like the eigenvector calculation, some of them are redundant, meaning they don't all give us new information. Recognizing and managing such redundancies is a vital skill when working through these systems.