Chapter 5: Problem 9
Compute the matrix exponential \(e^{\mathbf{A} t}\) for each system \(\mathbf{x}^{\prime}=\mathbf{A x}\) given. $$ x_{1}^{\prime}=5 x_{1}-4 x_{2}, x_{2}^{\prime}=2 x_{1}-x_{2} $$
Short Answer
Expert verified
The matrix exponential is \(e^{\mathbf{A}t} = \begin{bmatrix} 0.5 e^{3t} + 0.5 e^t & -0.5 e^{3t} + 0.5 e^t \\ 0.25 e^{3t} + 0.5 e^t & -0.25 e^{3t} + 0.5 e^t \end{bmatrix}.\)
Step by step solution
01
Write Matrix Representation
We start by writing the given system of differential equations in matrix form. This can be expressed as: \[\begin{bmatrix} x_1' \ x_2' \\end{bmatrix} = \begin{bmatrix} 5 & -4 \ 2 & -1 \\end{bmatrix} \begin{bmatrix} x_1 \ x_2 \\end{bmatrix}.\]Here, the matrix \(\mathbf{A}\) is \[\mathbf{A} = \begin{bmatrix} 5 & -4 \ 2 & -1 \end{bmatrix}.\]
02
Compute Eigenvalues of \(\mathbf{A}\)
To compute the matrix exponential, we first find the eigenvalues of \(\mathbf{A}\). The eigenvalues \(\lambda\) are found by solving the characteristic equation:\[det(\mathbf{A} - \lambda \mathbf{I}) = 0,\]where \(\mathbf{I}\) is the identity matrix. For \(\mathbf{A} = \begin{bmatrix} 5 & -4 \ 2 & -1 \end{bmatrix}\), the equation is:\[det\begin{bmatrix} 5 - \lambda & -4 \ 2 & -1 - \lambda \end{bmatrix} = (5 - \lambda)(-1 - \lambda) - (-4)(2) = 0.\]Simplifying gives us the characteristic polynomial:\[(5 - \lambda)(-1 - \lambda) - (-8) = \lambda^2 - 4\lambda + 3 = 0.\]The solutions are \(\lambda = 3\) and \(\lambda = 1\).
03
Find Eigenvectors
Next, we find the eigenvectors corresponding to each eigenvalue. For \(\lambda = 3\):\[\begin{bmatrix} 5 - 3 & -4 \ 2 & -1 - 3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 2 & -4 \ 2 & -4 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]This yields the equation \(x_1 - 2x_2 = 0\), so \(x_2 = \frac{x_1}{2}\). An eigenvector is \(\begin{bmatrix} 1 \ 0.5 \end{bmatrix}\).For \(\lambda = 1\):\[\begin{bmatrix} 4 & -4 \ 2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}.\]This yields the equation \(x_1 - x_2 = 0\), so \(x_2 = x_1\). An eigenvector is \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\).
04
Matrix Exponential using Eigenvalues and Eigenvectors
Using the eigenvalues and eigenvectors, we can write the matrix \(\mathbf{A}\) in terms of its eigenvectors \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\) containing the eigenvalues. \[\mathbf{A} = \mathbf{P} \mathbf{D} \mathbf{P}^{-1},\]where\[\mathbf{P} = \begin{bmatrix} 1 & 1 \ 0.5 & 1 \end{bmatrix}, \quad \mathbf{D} = \begin{bmatrix} 3 & 0 \ 0 & 1 \end{bmatrix}.\]The matrix exponential \(e^{\mathbf{A}t}\) is given by:\[e^{\mathbf{A} t} = \mathbf{P} e^{\mathbf{D} t} \mathbf{P}^{-1}.\]Since \(e^{\mathbf{D} t}\) = \(\begin{bmatrix} e^{3t} & 0 \ 0 & e^t \end{bmatrix}\), we have:\[e^{\mathbf{A} t} = \begin{bmatrix} 1 & 1 \ 0.5 & 1 \end{bmatrix} \begin{bmatrix} e^{3t} & 0 \ 0 & e^t \end{bmatrix} \begin{bmatrix} 1 & -1 \ -0.5 & 1 \end{bmatrix}.\]
05
Compute the Final Expression for Matrix Exponential
Finish by calculating the product:\[\mathbf{P} e^{\mathbf{D} t} = \begin{bmatrix} 1 & 1 \ 0.5 & 1 \end{bmatrix} \begin{bmatrix} e^{3t} & 0 \ 0 & e^t \end{bmatrix} = \begin{bmatrix} e^{3t} & e^t \ 0.5e^{3t} & e^t \end{bmatrix},\]and\[e^{\mathbf{A} t} = \begin{bmatrix} e^{3t} & e^t \ 0.5e^{3t} & e^t \end{bmatrix} \begin{bmatrix} 1 & -1 \ -0.5 & 1 \end{bmatrix}.\]Calculating the final product results in:\[e^{\mathbf{A} t} = \begin{bmatrix} 0.5 e^{3t} + 0.5 e^t & -0.5 e^{3t} + 0.5 e^t \ 0.25 e^{3t} + 0.5 e^t & -0.25 e^{3t} + 0.5 e^t \end{bmatrix}.\]
06
Conclusion
The matrix exponential \(e^{\mathbf{A}t}\) for the system can be expressed as:\[\begin{bmatrix} 0.5 e^{3t} + 0.5 e^t & -0.5 e^{3t} + 0.5 e^t \ 0.25 e^{3t} + 0.5 e^t & -0.25 e^{3t} + 0.5 e^t \end{bmatrix}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Differential Equations
Differential equations play a crucial role in many scientific and engineering fields. They represent the relationship between a function and its derivatives. In simpler terms, they explain how a quantity changes over time. In this problem, the differential equation is represented as \( \mathbf{x}^{\prime} = \mathbf{A x} \), which describes a system where the change of vector \( \mathbf{x} \) is determined by the multiplication of a matrix \( \mathbf{A} \) with \( \mathbf{x} \). This expression is common when dealing with systems of linear differential equations.
The specific system presented can be conceptually grasped by breaking it down into two parts:
The specific system presented can be conceptually grasped by breaking it down into two parts:
- \(x_{1}^{\prime} = 5x_{1} - 4x_{2}\)
- \(x_{2}^{\prime} = 2x_{1} - x_{2}\)
Eigenvalues and Eigenvectors: The Key to Solving Systems
When working with differential equations in matrix form, eigenvalues and eigenvectors provide us with a powerful way to solve them. Eigenvalues are coefficients that provide important information about the characteristics of a system.
A matrix like \( \mathbf{A} \) from our problem can be analyzed to find its eigenvalues. These are found by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), which gives us the eigenvalues \( \lambda \). These eigenvalues reveal the tendencies of the system's dynamics over time. For our matrix \( \mathbf{A} \), the eigenvalues are 3 and 1, which imply different growth rates when each solution component is expressed as an exponential function.
A matrix like \( \mathbf{A} \) from our problem can be analyzed to find its eigenvalues. These are found by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), which gives us the eigenvalues \( \lambda \). These eigenvalues reveal the tendencies of the system's dynamics over time. For our matrix \( \mathbf{A} \), the eigenvalues are 3 and 1, which imply different growth rates when each solution component is expressed as an exponential function.
- Eigenvectors provide directions of these characteristic rates, helping us find a basis in which the matrix \( \mathbf{A} \) acts as simple scaling.
- Coupled with eigenvalues, they enable us to express the matrix exponential \( e^{\mathbf{A}t} \) conveniently.
Using Linear Algebra to Compute Matrix Exponential
Linear algebra provides the tools to dissolve complex matrix operations into manageable steps, especially when computing the matrix exponential \( e^{\mathbf{A} t} \). With eigenvalues and eigenvectors, we utilize the diagonalization technique, where a matrix \( \mathbf{A} \) is expressed in terms of a diagonal matrix \( \mathbf{D} \) and a basis of eigenvectors in matrix \( \mathbf{P} \). This can be shown through the equation \( \mathbf{A} = \mathbf{P} \mathbf{D} \mathbf{P}^{-1} \).
This relationship simplifies the exponentiation process:
This relationship simplifies the exponentiation process:
- First, express \( e^{\mathbf{A}t} = \mathbf{P} e^{\mathbf{D}t} \mathbf{P}^{-1} \).
- Multiply each term sequentially, starting with \( e^{\mathbf{D}t} \), which involves exponentiating the diagonal matrix. This step reduces to exponentiating each eigenvalue.
- Finally, multiply through with the matrices \( \mathbf{P} \) and \( \mathbf{P}^{-1} \) to obtain the full expression for \( e^{\mathbf{A}t} \).