Problem 15 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(\left[\begin{array}{rrr}-17 & 0 & -42 \\ -7 & 4 & -14 \\ 7 & 0 & 18\end{array}\right]\) [Hint: The eigenvalues of \(A \text { are } \lambda=4,-3 .]\)

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x'} = A\mathbf{x}\) with the given matrix A is: \[\mathbf{x}(t) = c_1e^{4t}\begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} + c_2e^{-3t}\begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix} \] Where \(c_1\) and \(c_2\) are arbitrary constants.

Step 1: Eigenvector for \(\lambda = 4\)

We start with the given matrix A, and subtract \(\lambda I\) to find the matrix \(A - 4I\): \[A - 4I= \begin{bmatrix} -21 & 0 & -42 \\ -7 & 0 & -14\\ 7 & 0 & 14 \end{bmatrix} \] Now we want to find the nullspace of \(A - 4I\), which are the eigenvectors corresponding to \(\lambda = 4\). Row reduce the matrix to its echelon form: \[ \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix} \] The eigenvector \(\textbf{v}_1\) is the nullspace of this matrix: \[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = t \cdot \begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix}, t \in \mathbb{R} \]

Step 2: Eigenvector for \(\lambda = -3\)

We repeat the process to find the eigenvector for \(\lambda = -3\). We compute \(A + 3I\): \[A + 3I = \begin{bmatrix} -14 & 0 & -42 \\ -7 & 7 & -14 \\ 7 & 0 & 21\end{bmatrix} \] Row reduce the matrix to its echelon form: \[ \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \] The eigenvector \(\textbf{v}_2\) is the nullspace of this matrix: \[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = r \cdot \begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix}, r \in \mathbb{R} \] #Step 2: Compose the general solution#

Step 3: General solution

With the eigenvectors found, we can now write the general solution to the linear system \(\mathbf{x'} = A\mathbf{x}\) as: \[\mathbf{x}(t) = c_1e^{4t}\begin{bmatrix} -2 \\ -3 \\ 1 \end{bmatrix} + c_2e^{-3t}\begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix} \] Where \(c_1\) and \(c_2\) are arbitrary constants. This is the general solution to the given linear system.

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is pivotal in solving systems of linear differential equations like the one in this exercise. The eigenvalues of a matrix are special numbers that give us insight into the matrix's structure and behavior, particularly when it is applied to vector spaces. An eigenvalue, denoted usually by \( \lambda \), satisfies the equation \( A \mathbf{v} = \lambda \mathbf{v} \), where \( A \) is a square matrix, \( \mathbf{v} \) is a nonzero vector known as an eigenvector. This vector essentially points in the direction where the matrix operates by simply stretching or compressing, without changing direction. For the matrix given in the exercise, eigenvalues are \( \lambda = 4 \) and \( \lambda = -3 \). These values help us to find corresponding eigenvectors, which form the foundation to constructing solutions in the space spanned by these vectors.

Matrix Row Reduction

One of the fundamental skills in linear algebra is the ability to manipulate matrices through row reduction. This technique transforms a given matrix into a simpler form, making it much easier to identify solutions or properties such as eigenvectors. To row-reduce a matrix, we apply a series of elementary row operations aiming to convert the original matrix into its row-echelon or even reduced row-echelon form. In the context of this exercise, row reduction is used to simplify matrices like \( A - 4I \) and \( A + 3I \), which are interactions involving the identity matrix \( I \) with the scalar \( \lambda \). Row reduction helps uncover eigenvectors by allowing us to focus on nontrivial solutions of the equation \( (A - \lambda I) \mathbf{v} = 0 \). Simplifying these matrices isolates the free variables, leading directly to expressions defining the null space, hence the eigenvectors.

General Solution of Systems of Equations

The general solution to a system of linear differential equations ties together our learned concepts of eigenvalues and eigenvectors. When we derive eigenvectors as solutions from equations involving matrix transformations like \( A - \lambda I \), we compose them into solutions expressed in terms of time \( t \). The general solution is usually formed by combining these eigenvectors with exponential terms \( e^{\lambda t} \), each associated with an eigenvalue. In this problem, the solution takes the form \( \mathbf{x}(t) = c_1 e^{4t} \mathbf{v}_1 + c_2 e^{-3t} \mathbf{v}_2 \), where \( c_1 \) and \( c_2 \) are constants determined by initial conditions if provided. Each term involves one of the eigenvectors we found earlier, multiplied by an exponential term that introduces dynamics over time. This methodology provides a complete characterization of how the system evolves, and is applicable in many situations where determining behavior over time is needed.