0.29 -0.01 0.38 0.1 Let A = 0.25 0.75 0.5 The vector v₁ = 0.5 is an eigenvector for A, and two eigenvalues are 0.5 and 0.1. 0.23 0.13 0.56 0.2 Construct the solution of the dynamical system XK+1=Axk that satisfies x₁ = (-0.3,-0.3,0.8). What happens to xx as k→∞o? Find the solution of the equation XK+ 1 = Axk. Choose the correct answer below. 0.1 OA. X 0.5 +0.2(0.5) 0.2 2 -4 +0.4(0.1) 1 -2 1
0.29 -0.01 0.38 0.1 Let A = 0.25 0.75 0.5 The vector v₁ = 0.5 is an eigenvector for A, and two eigenvalues are 0.5 and 0.1. 0.23 0.13 0.56 0.2 Construct the solution of the dynamical system XK+1=Axk that satisfies x₁ = (-0.3,-0.3,0.8). What happens to xx as k→∞o? Find the solution of the equation XK+ 1 = Axk. Choose the correct answer below. 0.1 OA. X 0.5 +0.2(0.5) 0.2 2 -4 +0.4(0.1) 1 -2 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Matrix Algebra and Dynamical Systems
#### Eigenvectors and Eigenvalues
Consider the matrix \( A \) and its associated eigenvector \( \mathbf{v}_1 \):
\[ A = \begin{bmatrix} 0.29 & -0.01 & 0.38 \\ 0.25 & 0.75 & 0.5 \\ 0.23 & 0.13 & 0.56 \end{bmatrix} \]
\[ \mathbf{v}_1 = \begin{bmatrix} 0.1 \\ 0.5 \\ 0.2 \end{bmatrix} \]
This eigenvector is associated with \( A \), and the eigenvalues of \( A \) are 0.5 and 0.1.
#### Dynamical System
We are given a dynamical system described by:
\[ \mathbf{x}_{k+1} = A \mathbf{x}_k \]
with the initial condition:
\[ \mathbf{x}_0 = \begin{bmatrix} -0.3 \\ -0.3 \\ 0.8 \end{bmatrix} \]
Our goal is to determine the behavior of \( \mathbf{x}_k \) as \( k \to \infty \).
#### Numerical Solution
To find the numerical solution for the equation \( \mathbf{x}_{k+1} = A \mathbf{x}_k \), we must explore the given options. The correct answer must be determined based on how \( \mathbf{x}_k \) evolves over iterations.
The options provided are:
- **Option A:**
\[ \mathbf{x}_k = \begin{bmatrix} 0.1 \\ 0.5 \\ 0.2 \end{bmatrix} + 0.2(0.5)^k \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix} + 0.4(0.1)^k \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \]
- **Option B:**
\[ \mathbf{x}_k = \begin{bmatrix} 2 & -2 & 0 \\ -4 & 4 & 1 \\ 1 & 0 & 1 \end{bmatrix} + 0.2(0.5)^k \begin{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5483afab-7850-40f6-959f-e698dea1419a%2F4b987cae-3b75-4a74-aabb-6d97aafdac42%2F6kfshcs_processed.png&w=3840&q=75)
Transcribed Image Text:### Matrix Algebra and Dynamical Systems
#### Eigenvectors and Eigenvalues
Consider the matrix \( A \) and its associated eigenvector \( \mathbf{v}_1 \):
\[ A = \begin{bmatrix} 0.29 & -0.01 & 0.38 \\ 0.25 & 0.75 & 0.5 \\ 0.23 & 0.13 & 0.56 \end{bmatrix} \]
\[ \mathbf{v}_1 = \begin{bmatrix} 0.1 \\ 0.5 \\ 0.2 \end{bmatrix} \]
This eigenvector is associated with \( A \), and the eigenvalues of \( A \) are 0.5 and 0.1.
#### Dynamical System
We are given a dynamical system described by:
\[ \mathbf{x}_{k+1} = A \mathbf{x}_k \]
with the initial condition:
\[ \mathbf{x}_0 = \begin{bmatrix} -0.3 \\ -0.3 \\ 0.8 \end{bmatrix} \]
Our goal is to determine the behavior of \( \mathbf{x}_k \) as \( k \to \infty \).
#### Numerical Solution
To find the numerical solution for the equation \( \mathbf{x}_{k+1} = A \mathbf{x}_k \), we must explore the given options. The correct answer must be determined based on how \( \mathbf{x}_k \) evolves over iterations.
The options provided are:
- **Option A:**
\[ \mathbf{x}_k = \begin{bmatrix} 0.1 \\ 0.5 \\ 0.2 \end{bmatrix} + 0.2(0.5)^k \begin{bmatrix} 2 \\ -4 \\ 1 \end{bmatrix} + 0.4(0.1)^k \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \]
- **Option B:**
\[ \mathbf{x}_k = \begin{bmatrix} 2 & -2 & 0 \\ -4 & 4 & 1 \\ 1 & 0 & 1 \end{bmatrix} + 0.2(0.5)^k \begin{
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