Problem 28 - Advanced Engineering Mathematics

Question

We have det \((\mathbf{A}-\lambda \mathbf{I})=(\lambda-4)^{3}=0 .\) For \(\lambda_{1}=4\) we obtain $$\mathbf{K}=\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right)$$ Solutions of \(\left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{P}=\mathbf{K}\) and \(\left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{Q}=\mathbf{P}\) are $$\mathbf{P}=\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \quad \text { and } \quad \mathbf{Q}=\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right)$$ so that $$\mathbf{X}=c_{1}\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) e^{4 t}+c_{2}\left[\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) t e^{4 t}+\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) e^{4 t}\right]+c_{3}\left[\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right) \frac{t^{2}}{2} e^{4 t}+\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) t e^{4 t}+\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right) e^{4 t}\right]$$

Step-by-Step Solution

Answer

\(\mathbf{X}\) solutions are based on repeated eigenvalue \(\lambda=4\) and generalized eigenspaces.

Step 1: Determine Eigenvalue

Given the equation \( \det (\mathbf{A}-\lambda \mathbf{I})=(\lambda-4)^3=0 \). This equation indicates that the eigenvalue \(\lambda\) is 4, with algebraic multiplicity 3.

Step 2: Set Up Differential Equations

Write down the equations given for solutions \( \left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{P} = \mathbf{K} \) and \( \left(\mathbf{A}-\lambda_{1} \mathbf{I}\right) \mathbf{Q} = \mathbf{P} \). Here, \( \mathbf{K} \) is \( \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \), \( \mathbf{P} \) is \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \), and \( \mathbf{Q} \) is \( \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \).

Step 3: Construct the Solution for \(\mathbf{X}\)

For the system, the solution vector \(\mathbf{X}\) with \(\lambda_1=4\) is constructed by combining terms from the solution vectors and exponential functions. It is given by:\[\mathbf{X} = c_1 \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} e^{4t} + c_2 \left[ \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} t e^{4t} + \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} e^{4t} \right] + c_3 \left[ \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} \frac{t^2}{2} e^{4t} + \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} t e^{4t} + \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} e^{4t} \right]\]

Step 4: Understand Components of the Solution

The solution is a linear combination of vectors multiplied by exponential functions of time \(t\). Each part corresponds to the multiple eigenvalues and generalized eigenvectors. \(c_1, c_2, \) and \(c_3\) are arbitrary constants determined by initial conditions.

Key Concepts

Eigenvalues and Eigenvectors

To explore the heart of differential equations, one key concept you need to grasp is eigenvalues and eigenvectors. These are fundamental in simplifying complex systems. In this exercise, we see the eigenvalue \( \lambda = 4 \) with an algebraic multiplicity of 3.
**What Does This Mean?**
- **Eigenvalue**: A value, \( \lambda \), for which there exists a non-zero vector (eigenvector) \( \mathbf{v} \) such that \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \). Here, the matrix \( \mathbf{A} \) when multiplied by \( \mathbf{v} \), only scales the vector by \( \lambda \).
- **Algebraic Multiplicity**: This tells us how many times a certain eigenvalue appears as a root of the characteristic equation. It's related to the dimension of the eigenspace.
Eigenvectors \( \mathbf{K}, \mathbf{P}, \) and \( \mathbf{Q} \) serve as steps of the solution to the differential equation involving these eigenvalues. Understanding these components clarifies how systems transform, solving them using combinations of these vectors.

Matrix Algebra

Matrix algebra is an essential tool when working with eigenvalues and eigenvectors. Think of matrices as the structural blueprint for transformations. They simplify the process of handling sets of linear equations.
**Utilizing Matrices:**
- **Characteristic Equation**: This is where we calculate \( \det(\mathbf{A} - \lambda \mathbf{I}) \). It's a pivotal step, as this equation guides us in finding eigenvalues.
- **Identity Matrix \( \mathbf{I} \)**: Acts as the 'do-nothing' transformation in matrix algebra. Subtracting \( \lambda \mathbf{I} \) from \( \mathbf{A} \) prepares the characteristic equation for simplifying the transformation.
Matrix algebra gives us a systematic way to understand how changes in inputs affect outputs. It makes capturing the dynamics of complex systems more manageable. Keep practicing breaking down matrices and you'll better navigate through linear transformations.

Linear Algebra

Linear algebra provides the foundational language to analyze vectors, matrices, and their transformations. It's like learning the grammar of a mathematical language used to describe and solve systems.
**Core Concepts in Linear Algebra:**
- **Vectors and Spaces**: Vectors are objects that have both direction and magnitude. They are the building blocks of linear algebra, given that they represent points in space.
- **Linearity**: Describes systems that satisfy properties of superposition and homogeneity, meaning solutions can be represented as linear combinations of vectors.
- **Linear Combinations**: These allow expressing any vector in terms of other vectors, which is crucial for understanding solutions to linear systems.
These concepts come together in the differential equations as seen in the solution \( \mathbf{X} = c_1 \mathbf{K} e^{4t} + c_2 (\mathbf{K} te^{4t} + \mathbf{P} e^{4t}) + c_3 (\mathbf{K} \frac{t^2}{2} e^{4t} + \mathbf{P} te^{4t} + \mathbf{Q} e^{4t}) \). Understanding how to manipulate these with linear algebraic methods is the key to deeply understanding the solution provided.