Let {xn} be a sequence such that xn = of a suitable matrix to solve for xn. 5xn-1+6xn-2, x1 = -2, xo = 4. Use power
Let {xn} be a sequence such that xn = of a suitable matrix to solve for xn. 5xn-1+6xn-2, x1 = -2, xo = 4. Use power
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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![Transcription for Educational Website:
---
**Problem Statement:**
Let \(\{x_n\}\) be a sequence such that
\[ x_n = 5x_{n-1} + 6x_{n-2}, \]
with initial conditions \( x_1 = -2 \) and \( x_0 = 4 \).
**Task:**
Use the power of a suitable matrix to solve for \( x_n \).
---
**Detailed Explanation:**
The problem involves finding a sequence \(\{x_n\}\) defined by a linear recurrence relation with constant coefficients. The goal is to express \(x_n\) in terms of matrix operations.
1. **Define a Matrix Representation:**
To use matrices, set up a vector that can express the relation using matrix multiplication. Define the vector:
\[
\mathbf{v}_n = \begin{bmatrix} x_n \\ x_{n-1} \end{bmatrix}
\]
The recurrence relation can be rewritten in matrix form:
\[
\mathbf{v}_{n} = \begin{bmatrix} 5 & 6 \\ 1 & 0 \end{bmatrix} \mathbf{v}_{n-1}
\]
Let \(\mathbf{A} = \begin{bmatrix} 5 & 6 \\ 1 & 0 \end{bmatrix}\).
2. **Initial Conditions:**
Given:
\[
\mathbf{v}_1 = \begin{bmatrix} x_1 \\ x_0 \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \end{bmatrix}
\]
3. **Solution Using Matrix Powers:**
The general solution for \(\mathbf{v}_n\) can be expressed as:
\[
\mathbf{v}_n = \mathbf{A}^{n-1} \cdot \mathbf{v}_1
\]
Calculate \(\mathbf{A}^{n-1}\) and multiply it by \(\mathbf{v}_1\) to find \(x_n\).
This matrix approach allows for efficient computation of terms in the sequence by leveraging the power of matrices and linear algebra.
---
This allows students to explore the interplay between linear algebra and recurrence relations, highlighting the practical applications of matrices in solving complex](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0780f626-a981-4f24-912b-c0e4a815959e%2F672efef9-8b50-4f92-82f8-5e80ef4f2ab9%2Faxnsir9_processed.png&w=3840&q=75)
Transcribed Image Text:Transcription for Educational Website:
---
**Problem Statement:**
Let \(\{x_n\}\) be a sequence such that
\[ x_n = 5x_{n-1} + 6x_{n-2}, \]
with initial conditions \( x_1 = -2 \) and \( x_0 = 4 \).
**Task:**
Use the power of a suitable matrix to solve for \( x_n \).
---
**Detailed Explanation:**
The problem involves finding a sequence \(\{x_n\}\) defined by a linear recurrence relation with constant coefficients. The goal is to express \(x_n\) in terms of matrix operations.
1. **Define a Matrix Representation:**
To use matrices, set up a vector that can express the relation using matrix multiplication. Define the vector:
\[
\mathbf{v}_n = \begin{bmatrix} x_n \\ x_{n-1} \end{bmatrix}
\]
The recurrence relation can be rewritten in matrix form:
\[
\mathbf{v}_{n} = \begin{bmatrix} 5 & 6 \\ 1 & 0 \end{bmatrix} \mathbf{v}_{n-1}
\]
Let \(\mathbf{A} = \begin{bmatrix} 5 & 6 \\ 1 & 0 \end{bmatrix}\).
2. **Initial Conditions:**
Given:
\[
\mathbf{v}_1 = \begin{bmatrix} x_1 \\ x_0 \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \end{bmatrix}
\]
3. **Solution Using Matrix Powers:**
The general solution for \(\mathbf{v}_n\) can be expressed as:
\[
\mathbf{v}_n = \mathbf{A}^{n-1} \cdot \mathbf{v}_1
\]
Calculate \(\mathbf{A}^{n-1}\) and multiply it by \(\mathbf{v}_1\) to find \(x_n\).
This matrix approach allows for efficient computation of terms in the sequence by leveraging the power of matrices and linear algebra.
---
This allows students to explore the interplay between linear algebra and recurrence relations, highlighting the practical applications of matrices in solving complex
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