The zeros in the given matrix make its characteristic polynomial easy to calculate. Find the general solution of x' = Ax. 2 0 0 0 - 14 -5 12 -6 A = 0 0 5 0 0 -14 - -2 ... x(t) = 0
The zeros in the given matrix make its characteristic polynomial easy to calculate. Find the general solution of x' = Ax. 2 0 0 0 - 14 -5 12 -6 A = 0 0 5 0 0 -14 - -2 ... x(t) = 0
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Transcription for Educational Website:
**Title: Solving a System of Differential Equations Using the Characteristic Polynomial**
**Problem Statement:**
The zeros in the given matrix make its characteristic polynomial easy to calculate. Find the general solution of \( \mathbf{x}' = A\mathbf{x} \).
\[
A = \begin{pmatrix}
2 & 0 & 0 & 0 \\
-14 & -5 & -12 & -6 \\
0 & 0 & 5 & 0 \\
0 & 0 & -14 & -2
\end{pmatrix}
\]
---
**Solution:**
\[
x(t) = \_\_\_\_
\]
(Note: The box at the end appears to be for further calculations, possibly a placeholder for the solution. The red text "vvv" seems to be an annotation or part of a handwritten note, not relevant to the actual problem.)
---
**Explanation:**
Understanding the given matrix and its structure can help in simplifying the calculation of the characteristic polynomial, which is essential in determining the eigenvalues. These eigenvalues, in turn, are used to find the general solution of the system of differential equations represented by \( \mathbf{x}' = A\mathbf{x} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06bbd155-c9d1-4bdc-8881-0cd893ac016a%2Fd1a14b41-edde-4f5d-8602-9bdf890d79bf%2Ffx9p9q_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription for Educational Website:
**Title: Solving a System of Differential Equations Using the Characteristic Polynomial**
**Problem Statement:**
The zeros in the given matrix make its characteristic polynomial easy to calculate. Find the general solution of \( \mathbf{x}' = A\mathbf{x} \).
\[
A = \begin{pmatrix}
2 & 0 & 0 & 0 \\
-14 & -5 & -12 & -6 \\
0 & 0 & 5 & 0 \\
0 & 0 & -14 & -2
\end{pmatrix}
\]
---
**Solution:**
\[
x(t) = \_\_\_\_
\]
(Note: The box at the end appears to be for further calculations, possibly a placeholder for the solution. The red text "vvv" seems to be an annotation or part of a handwritten note, not relevant to the actual problem.)
---
**Explanation:**
Understanding the given matrix and its structure can help in simplifying the calculation of the characteristic polynomial, which is essential in determining the eigenvalues. These eigenvalues, in turn, are used to find the general solution of the system of differential equations represented by \( \mathbf{x}' = A\mathbf{x} \).
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