Problem 6 - Differential Equations and Linear Algebra

Question

Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\) $$\left[\begin{array}{rrr} -2 & 0 & 0 \\ 1 & -3 & -1 \\ -1 & 1 & -1 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the system \(\mathbf{x}^{\prime}=A\mathbf{x}\) for the given matrix A is: \[\mathbf{x}(t) = c_1 e^{2t} \left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right] + c_2 e^{3t} \left[\begin{array}{c} 0\\ 1\\ 1 \end{array}\right] + c_3 e^{-2t} \left[\begin{array}{c} 0\\ 1\\ 1 \end{array}\right]\] Where \(c_1\), \(c_2\), and \(c_3\) are constants determined by the initial conditions.

Step 1: Find Eigenvalues and Eigenvectors

First, we need to find the eigenvalues of the given matrix. To do this, we'll compute the characteristic equation of the matrix A: \[\text{det}(A - \lambda I) = 0\] Using the given matrix A, we get: \[\text{det}\left(\left[\begin{array}{rrr} 2-\lambda & 0 & 0 \\ 0 & 5-\lambda & -7 \\ 0 & 2 & -4-\lambda \end{array}\right]\right) =0\] Calculating the determinant, we get: \((2-\lambda)((5-\lambda)(-4-\lambda)-(-7)(2))=0\) This equation simplifies to: \((2-\lambda)(\lambda^{2}-\lambda-6)=0\] The eigenvalues are the solutions to this equation: \[\lambda_1=2, \quad \lambda_2=3, \quad\text{and}\quad \lambda_3=-2\] Now, we'll find the corresponding eigenvectors for each eigenvalue. For \(\lambda_1 = 2\), we have: \((A - 2I)\mathbf{v}_1 = 0\), where \[\left[\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 3 & -7 \\ 0 & 2 & -6 \end{array}\right] \left[\begin{array}{r} v_{11}\\ v_{12}\\ v_{13} \end{array}\right] = \left[\begin{array}{r} 0\\ 0\\ 0 \end{array}\right]\] From this, we get that \(\mathbf{v}_1 = \left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right]\) as the eigenvector corresponding to the eigenvalue \(2\). For \(\lambda_2 = 3\), we have: \((A - 3I)\mathbf{v}_2 = 0\), where \[\left[\begin{array}{rrr} -1 & 0 & 0 \\ 0 & 2 & -7 \\ 0 & 2 & -7 \end{array}\right] \left[\begin{array}{r} v_{21}\\ v_{22}\\ v_{23} \end{array}\right] = \left[\begin{array}{r} 0\\ 0\\ 0 \end{array}\right]\] From this, we get that \(\mathbf{v}_2 = \left[\begin{array}{c} 0\\ 1\\ 1 \end{array}\right]\) as the eigenvector corresponding to the eigenvalue \(3\). For \(\lambda_3 = -2\), we have: \((A + 2I)\mathbf{v}_3 = 0\), where \[\left[\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 7 & -7 \\ 0 & 2 & -2 \end{array}\right] \left[\begin{array}{r} v_{31}\\ v_{32}\\ v_{33} \end{array}\right] = \left[\begin{array}{r} 0\\ 0\\ 0 \end{array}\right]\] From this, we get that \(\mathbf{v}_3= \left[\begin{array}{c} 0\\ 1\\ 1 \end{array}\right]\) as the eigenvector corresponding to the eigenvalue \(-2\).

Step 2: Write Down the General Solution

With the eigenvalues and their corresponding eigenvectors, we can now write the general solution as: \[\mathbf{x}(t) = c_1 e^{2t} \left[\begin{array}{c} 1\\ 0\\ 0 \end{array}\right] + c_2 e^{3t} \left[\begin{array}{c} 0\\ 1\\ 1 \end{array}\right] + c_3 e^{-2t} \left[\begin{array}{c} 0\\ 1\\ 1 \end{array}\right]\] Where \(c_1\), \(c_2\), and \(c_3\) are constants determined by the initial conditions.

Key Concepts

Characteristic Equation

The characteristic equation is a crucial tool for finding eigenvalues of a matrix. It emerges from the expression \( \text{det}(A - \lambda I) = 0 \), where \( A \) is our matrix and \( \lambda \) represents eigenvalues. Here, \( I \) is the identity matrix of the same dimension as \( A \). This equation essentially finds values of \( \lambda \) that make the matrix \( A - \lambda I \) singular, meaning it does not have an inverse.
To solve the characteristic equation:

• Subtract \( \lambda \) times the identity matrix from the matrix \( A \) to form \( A - \lambda I \).
• Calculate the determinant of this new matrix.
• Set the resultant expression equal to zero and solve for \( \lambda \).

For the given matrix, the characteristic equation is derived from the determinant \( \text{det} \left( \begin{bmatrix} 2-\lambda & 0 & 0 \ 0 & 5-\lambda & -7 \ 0 & 2 & -4-\lambda \end{bmatrix} \right) \). Solving this, we find eigenvalues, which are critical in later steps.

General Solution

The general solution for the system \( \mathbf{x}^{\prime}=A \mathbf{x} \) involves using the eigenvalues and eigenvectors obtained from the matrix \( A \). This is expressed as a linear combination of exponential functions. For each eigenvalue \( \lambda_i \), we have an associated eigenvector \( \mathbf{v}_i \).
This forms the general solution:

• \( \mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2 + c_3e^{\lambda_3 t}\mathbf{v}_3 \)

Here, \( c_1, c_2, \) and \( c_3 \) are constants determined by initial conditions, and \( e^{\lambda_i t} \) denotes the exponential function for each eigenvalue. This composition illustrates how solutions grow or decay over time, depending on the signs and magnitudes of eigenvalues. For matrices with distinct eigenvalues, like our example here, the solutions often involve independent exponential functions.

Matrix Determinant

A matrix determinant is a special number that can be calculated from its square matrix. It is denoted as \( \text{det}(A) \) for a matrix \( A \). The determinant offers insights into many properties of the matrix, such as invertibility and orientation.
In the context of eigenvalue problems:

• The determinant of \( A - \lambda I \) is set to zero as part of the characteristic equation.
• This process ensures that eigenvalues we find are at points where the matrix \( A - \lambda I \) becomes singular or non-invertible.

To compute a determinant for a 3x3 matrix:

• For diagonal matrices (like parts of \( A - \lambda I \)), it's often the product of its diagonal entries.
• For more complex entries, use methods like cofactor expansion if needed.

The resulting determinant equation leads us to solve for eigenvalues, a stepping stone to arriving at generalized solutions.