/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Show that the system \(\mathbf{x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 32

Show that the system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\) has constant solutions other than \(\mathbf{x}(t)=0\) if and only if there exists a (constant) vector \(\mathbf{x} \neq \mathbf{0}\) with \(\mathbf{A x}=\mathbf{0}\). (It is shown in linear algebra that such a vector \(\mathbf{x}\) exists exactly when \(\operatorname{det}(\mathbf{A})=0 .\) )

Short Answer

Expert verified
Constant solutions exist if and only if \( \operatorname{det}(\mathbf{A}) = 0 \).

Step by step solution

01

System of Differential Equation

Consider the system \( \mathbf{x}^{\prime} = \mathbf{A} \mathbf{x} \). We are given that \( \mathbf{x}(t) = \mathbf{0} \) is a constant solution, and we need to find if there are other constant solutions.
02

Define Constant Solution

A constant solution \( \mathbf{x}(t) = \mathbf{c} \) implies its derivative \( \mathbf{x}^{\prime} = \mathbf{0} \), for some constant vector \( \mathbf{c} eq \mathbf{0} \). Substituting this into the differential equation gives \( \mathbf{A} \mathbf{c} = \mathbf{0} \).
03

Condition for Nontrivial Solution

To have a nontrivial constant solution \( \mathbf{c} eq \mathbf{0} \) such that \( \mathbf{A} \mathbf{c} = \mathbf{0} \), \( \mathbf{A} \) must be singular. This means the determinant of \( \mathbf{A} \) must be zero, i.e., \( \operatorname{det}(\mathbf{A}) = 0 \).
04

Conclusion from Linear Algebra

From linear algebra, a nonzero solution to \( \mathbf{A} \mathbf{x} = \mathbf{0} \) exists if and only if \( \operatorname{det}(\mathbf{A}) = 0 \). Therefore, the system has constant solutions other than \( \mathbf{x}(t) = \mathbf{0} \) if \( \operatorname{det}(\mathbf{A}) = 0 \).
05

Final Verification

Verify by direct computation: If \( \operatorname{det}(\mathbf{A}) = 0 \), then \( \mathbf{A} \) has a nontrivial null space, implying there exists a nonzero vector \( \mathbf{x} \) such that \( \mathbf{A} \mathbf{x} = \mathbf{0} \). Conversely, if \( \operatorname{det}(\mathbf{A}) eq 0 \), \( \mathbf{x} = \mathbf{0} \) is the only solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Algebra
Linear algebra is a branch of mathematics focused on vectors, vector spaces, and linear transformations. At its core, linear algebra helps us understand the behavior of systems of linear equations.
  • It provides tools for solving systems of equations using matrices and determinants.
  • Key concepts include eigenvalues, eigenvectors, and linear independence.
  • It's used in various fields like engineering, physics, and computer science for modeling and problem-solving.
Understanding linear algebra helps in grasping the workings of differential equations, especially when determining solutions to matrices involved in these equations. This foundational knowledge allows us to investigate and find solutions, such as the constant solutions explored in the given problem. A solid grasp of linear algebra is essential for tackling more complex systems that appear in theoretical and applied contexts.
Constant Solutions
Constant solutions refer to solutions of differential equations that do not change over time. For a system like \(\mathbf{x}^{\prime} = \mathbf{A} \mathbf{x}\), a constant solution means \(x(t)\) remains the same for all \(t\).
  • These solutions imply that the derivative, \(\mathbf{x}^{\prime}\), is zero.
  • This results in the equation \(\mathbf{A} \mathbf{c} = \mathbf{0}\), where \(\mathbf{c}\) is a constant vector.
  • The interest in constant solutions arises in determining stability and equilibrium points in systems.
In our specific case, finding nontrivial constant solutions leads to investigating whether a non-zero vector \(\mathbf{x}\) satisfies \(\mathbf{A} \mathbf{x} = \mathbf{0}\). This deepens our understanding of equilibrium within the system.
Singular Matrix
A singular matrix plays a crucial role in this exercise. In linear algebra, a matrix is termed singular if it does not have an inverse. This is indicated by a determinant of zero.
  • A singular matrix means there exist non-zero solutions to the equation \(\mathbf{A} \mathbf{x} = \mathbf{0}\).
  • Such matrices indicate a loss of full rank, meaning the columns (or rows) are not linearly independent.
  • This phenomenon reflects a kind of degeneracy in the system, often indicating multiple or infinite solutions.
In our problem, when \(\mathbf{A}\) is singular, it means we can find constant solutions other than \(\mathbf{x}(t)=0\), thus letting us explore a wider set of possible behaviors for the system defined by the matrix \(\mathbf{A}\).
Determinant
The determinant of a matrix is a special number calculated from a square matrix. It provides important information about the matrix and is crucial for understanding various properties of linear transformations.
  • A determinant of zero signifies that the matrix is singular, and there's no unique solution to \(\mathbf{A} \mathbf{x} = \mathbf{0}\).
  • If the determinant is non-zero, all solutions are trivial (like \(\mathbf{x} = \mathbf{0}\)), and the matrix is non-singular.
  • Calculating determinants involves specific operations that derive a scalar value providing insights into the matrix's properties.
In our exercise, checking the determinant of matrix \(\mathbf{A}\) helps determine the existence of nontrivial constant solutions for the system \(\mathbf{x}^{\prime} = \mathbf{A} \mathbf{x}\). When the determinant is zero, it shows that alternate solutions to \(\mathbf{x}(t)=0\) are possible, allowing for a deeper analysis of the system's behavior.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. \(\mathbf{x}^{\prime}=\left[\begin{array}{cc}4 & -3 \\ 6 & -7\end{array}\right] \mathbf{x} ; \mathbf{x}_{1}=\left[\begin{array}{c}3 e^{2 t} \\ 2 e^{2 t}\end{array}\right], \mathbf{x}_{2}=\left[\begin{array}{r}e^{-5 t} \\ 3 e^{-5 t}\end{array}\right]\)

Show that the matrix \(A\) is nilpotent and then use this fact to find (as in Example 3) the matrix exponential \(e^{\mathbf{A} t} .\) $$ \mathbf{A}=\left[\begin{array}{rrr} 1 & -1 & -1 \\ 1 & -1 & 1 \\ 0 & 0 & 0 \end{array}\right] $$

$$ x_{1}^{\prime}=5 x_{1}-9 x_{2}, \quad x_{2}^{\prime}=2 x_{1}-x_{2} $$

Compute the matrix exponential \(e^{\mathbf{A} t}\) for each system \(\mathbf{x}^{\prime}=\mathbf{A x}\) given. $$ x_{1}^{\prime}=9 x_{1}-8 x_{2}, x_{2}^{\prime}=6 x_{1}-5 x_{2} $$

The characteristic equation of the coefficient matrix \(\mathbf{A}\) of the system $$ \mathbf{x}^{\prime}=\left[\begin{array}{rrrr} 3 & -4 & 1 & 0 \\ 4 & 3 & 0 & 1 \\ 0 & 0 & 3 & -4 \\ 0 & 0 & 4 & 3 \end{array}\right] \mathbf{x} $$ is $$ \phi(\lambda)=\left(\lambda^{2}-6 \lambda+25\right)^{2}=0 $$ Therefore, \(\mathbf{A}\) has the repeated complex conjugate pair \(3 \pm 4 i\) of eigenvalues. First show that the complex vectors $$ \mathbf{v}_{1}=\left[\begin{array}{llll} 1 & i & 0 & 0 \end{array}\right]^{T} \quad \text { and } \quad \mathbf{v}_{2}=\left[\begin{array}{llll} 0 & 0 & 1 & i \end{array}\right]^{T} $$ form a length 2 chain \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) associated with the eigenvalue \(\lambda=3-4 i\). Then calculate the real and imaginary parts of the complex-valued solutions $$ \mathbf{v}_{1} e^{\lambda t} \quad \text { and } \quad\left(\mathbf{v}_{1} t+\mathbf{v}_{2}\right) e^{\lambda t} $$ to find four independent real-valued solutions of \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.