/*! 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 8 find a fundamental matrix for th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 8

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{1} & {-1} \\ {5} & {-3}\end{array}\right) \mathbf{x} $$

Short Answer

Expert verified
Question: Find a fundamental matrix for the given system of linear differential equations satisfying the initial condition \(\Phi(0)=\mathbf{1}\): $$ \frac{d \mathbf{X}(t)}{dt} = \left(\begin{array}{ll}{1} & {-1} \\\ {5} & {-3}\end{array}\right) \mathbf{X}(t) $$ Answer: The fundamental matrix satisfying \(\Phi(0)=\mathbf{1}\) is: $$ \mathbf{\Phi}(0) = \begin{pmatrix} 1 & 1 \\ 5 & -1 \end{pmatrix} $$

Step by step solution

01

Calculate the eigenvalues and eigenvectors of the coefficient matrix

First, we need to calculate the eigenvalues and corresponding eigenvectors of the coefficient matrix. The given matrix is: $$ A=\left(\begin{array}{ll}{1} & {-1} \\\ {5} & {-3}\end{array}\right) $$ To find the eigenvalues \(\lambda\) of the matrix, we must solve the characteristic equation for each eigenvalue: $$ \det(A-\lambda I) = 0 $$ where \(I\) is the identity matrix: $$ \left|\begin{array}{cc}{1-\lambda} & {-1} \\\ {5} & {-3-\lambda}\end{array}\right|=0 $$ Calculate the determinant: $$ (1-\lambda)((-3)-\lambda) - (-1)(5) = 0 $$ This result in: $$ \lambda^2 + 2\lambda -8 = 0 $$ Solving the quadratic equation produces the eigenvalues: \(\lambda_1 = -4\) and \(\lambda_2 = 2\). Now, we will find the corresponding eigenvectors for each eigenvalue. For \(\lambda_1 = -4\), \((A-\lambda_1 I)\mathbf{v}=\mathbf{0}\) becomes: $$ \left(\begin{array}{ll}{5} & {-1} \\\ {5} & {1}\end{array}\right)\mathbf{v}=\mathbf{0} $$ From the first row of the equations, we have: $$ 5v_1 - v_2 = 0 $$ Choosing the simplest eigenvector, we can set \(v_1 = 1\) and \(v_2 = 5\). Thus, the eigenvector for \(\lambda_1\) is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 5 \end{pmatrix}\). For \(\lambda_2 = 2\), \((A-\lambda_2 I)\mathbf{v}=\mathbf{0}\) becomes: $$ \left(\begin{array}{ll}{-1} & {-1} \\\ {5} & {-5}\end{array}\right)\mathbf{v}=\mathbf{0} $$ From the first row of the equations, we have: $$ -v_1 - v_2 = 0 $$ Choosing the simplest eigenvector, we can set \(v_1 = 1\) and \(v_2 = -1\). Thus, the eigenvector for \(\lambda_2\) is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
02

Form the fundamental matrix

Now, use the eigenvectors to create the fundamental matrix, which consists of linearly independent solutions in form of \(X(t) = \mathbf{v}e^{\lambda t}\). Solution \(1\): \(X_1(t) = \mathbf{v}_1e^{\lambda_1 t} = \begin{pmatrix}1 \\ 5\end{pmatrix}e^{-4t}\) Solution \(2\): \(X_2(t) = \mathbf{v}_2e^{\lambda_2 t} = \begin{pmatrix}1 \\ -1\end{pmatrix}e^{2t}\) We can then create the fundamental matrix (\(\mathbf{\Phi}(t)\)) from the solutions: $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{-4t} & e^{2t} \\ 5e^{-4t} & -e^{2t} \end{pmatrix} $$
03

Find the fundamental matrix \(\Phi(t)\) satisfying the initial condition \(\Phi(0)=\mathbf{1}\)

To find the fundamental matrix satisfying \(\Phi(0) = \mathbf{1}\), replace the time variable t with 0: $$ \mathbf{\Phi}(0) = \begin{pmatrix} e^{-4(0)} & e^{2(0)} \\ 5e^{-4(0)} & -e^{2(0)} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 5 & -1 \end{pmatrix} $$ The fundamental matrix satisfying \(\Phi(0)=\mathbf{1}\) is: $$ \mathbf{\Phi}(0) = \begin{pmatrix} 1 & 1 \\ 5 & -1 \end{pmatrix} $$

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.

Eigenvalues
Eigenvalues are a fascinating concept in linear algebra, playing a critical role in understanding the behavior of linear transformations. They are numbers associated with a matrix that reveal significant properties about the matrix, especially in the context of differential equations.
To find eigenvalues, you solve the characteristic equation, which is obtained from the matrix equation \( \det(A - \lambda I) = 0 \), where \(A\) is your matrix, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix of the same size as \(A\).
This equation results in a polynomial usually known as the characteristic polynomial, and its roots are the eigenvalues.- **Purpose**: They help us understand the stability and dynamics of a system of differential equations.
- **Calculation**: In our problem, the matrix \(A\) has eigenvalues \(\lambda_1 = -4\) and \(\lambda_2 = 2\). These tell us how the solutions of the differential system behave over time.
Eigenvectors
Once we determine the eigenvalues, the next step is finding the associated eigenvectors. These are vector solutions corresponding to each eigenvalue. They provide direction in which the transformation applied by the matrix \(A\) stretches the vector by a factor equal to the eigenvalue.
For an eigenvalue \(\lambda\), its eigenvector \(\mathbf{v}\) is found by solving \((A - \lambda I) \mathbf{v} = \mathbf{0}\).
In our worked example:
  • For \(\lambda_1 = -4\), the corresponding eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \ 5 \end{pmatrix}\).
  • For \(\lambda_2 = 2\), the eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix}\).
Eigenvectors are vital since they span the solution space and define the direction of each part of the system's motion.
Fundamental Matrix
The fundamental matrix is an essential concept in solving linear systems of differential equations. It is a matrix formed by linearly independent solutions to the differential equation.
To form a fundamental matrix \(\mathbf{\Phi}(t)\), you take the general solutions of the differential equations, expressed as \(X(t) = \mathbf{v} e^{\lambda t}\) for each eigenvector and eigenvalue pair.
From our system:
  • \(X_1(t) = \begin{pmatrix} 1 \ 5 \end{pmatrix} e^{-4t}\)
  • \(X_2(t) = \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{2t}\)
These solutions populate the columns of the fundamental matrix:
\[\mathbf{\Phi}(t) = \begin{pmatrix} e^{-4t} & e^{2t} \ 5e^{-4t} & -e^{2t} \end{pmatrix}\]
Additionally, it must satisfy the condition \(\Phi(0) = \mathbf{1}\), the identity matrix. Solving this results in the identity, indicating that the fundamental matrix is accurately representing the system's solutions.
System of Equations
A system of linear equations forms the backbone of many mathematical and engineering problems. When dealing with multiple linear equations simultaneously, as in the given differential system, we essentially manage a system of equations.
In our context, the system is described by the matrix equation \(\mathbf{x}' = A\mathbf{x}\), where \(A\) is a matrix, and \(\mathbf{x}\) represents a vector of unknown functions.
Working with a system of equations involves:
  • Finding eigenvalues and eigenvectors to understand the geometric nature of the solution space.
  • Constructing the fundamental matrix to represent all solutions to the system.
  • Using initial conditions, if any, to ensure the solution fits specific scenario requirements.
This matrix-based approach is crucial because it allows for the systematic solution of systems that are too complicated to manage by direct integration.

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

Find the solution of the given initial value problem. Draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) - space and also draw the graph of \(x_{1}\) versus \(t .\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{-\frac{5}{2}} & {1} & {1} \\ {1} & {-\frac{5}{2}} & {1} \\ {1} & {1} & {-\frac{5}{2}}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{r}{2} \\ {3} \\\ {-1}\end{array}\right) $$

Let $$ \mathbf{J}=\left(\begin{array}{ccc}{\lambda} & {1} & {0} \\ {0} & {\lambda} & {1} \\ {0} & {0} & {\lambda}\end{array}\right) $$ where \(\lambda\) is an arbitrary real number. (a) Find \(\mathbf{J}^{2}, \mathbf{J}^{3},\) and \(\mathbf{J}^{4}\). (b) Use an inductive argument to show that $$ \mathbf{J}^{n}=\left(\begin{array}{ccc}{\lambda^{n}} & {n \lambda^{n-1}} & {[n(n-1) / 2] \lambda^{n-2}} \\ {0} & {\lambda^{n}} & {n \lambda^{n-1}} \\\ {0} & {0} & {\lambda^{n}}\end{array}\right) $$ (c) Determine exp(Jt). (d) Observe that if you choose \(\lambda=2\), then the matrix \(\mathbf{J}\) in this problem is the same as the matrix \(\mathbf{J}\) in Problem \(17(f)\). Using the matrix T from Problem \(17(f),\) form the product Texp(Jt) with \(\lambda=2\). Observe that the resulting matrix is the same as the fundamental matrix \(\Psi(t)\) in Problem \(17(e) .\)

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{cc}{-3} & {3 / 4} \\ {-5} & {1}\end{array}\right) $$

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

In this problem we show that the eigenvalues of a Hermitian matrix \(\Lambda\) are real. Let \(x\) be an eigenvector corresponding to the eigenvalue \(\lambda\). (a) Show that \((A x, x)=(x, A x)\). Hint: See Problem 31 . (b) Show that \(\lambda(x, x)=\lambda(x, x)\), Hint: Recall that \(A x=\lambda x\). (c) Show that \(\lambda=\lambda\); that is, the cigenvalue \(\lambda\) is real.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

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.