Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The matrix \(\Psi(t)\) is a fundamental matrix of the given homogeneous linear system. Find a constant matrix \(C\) such that \(\hat{\Psi}(t)=\Psi(t) C\) is a fundamental matrix satisfying \(\hat{\Psi}(0)=I\), where \(I\) is the \((2 \times 2)\) identity matrix. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \mathbf{y}, \quad \Psi(t)=\left[\begin{array}{cc} e^{t} & e^{t} \\ e^{-t} & -e^{-t} \end{array}\right] $$

Short Answer

Expert verified
Question: Find the constant matrix \(C\) such that \(\hat{\Psi}(t)=\Psi(t) C\) is a fundamental matrix satisfying \(\hat{\Psi}(0)=I\), where \(\Psi(t)=\left[\begin{array}{cc} e^{t} & e^{t} \\\ e^{-t} & -e^{-t} \end{array}\right]\). Answer: The constant matrix \(C\) is given by \(C=\left[\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} \end{array}\right]\).

Step by step solution

01

Write the given equations

We are given that $$ \Psi(t)=\left[\begin{array}{cc} e^{t} & e^{t} \\\ e^{-t} & -e^{-t} \end{array}\right] $$ and we need to find a constant matrix \(C\) such that \(\hat{\Psi}(t)=\Psi(t) C\) with \(\hat{\Psi}(0)=I\).
02

Define the matrices and equation to solve for C

Let \(C\) be a constant \(2 \times 2\) matrix: $$ C=\left[\begin{array}{cc} a & b \\\ c & d \end{array}\right] $$ Then, the product \(\Psi(t) C\) is given by: $$ \Psi(t) C = \left[\begin{array}{cc} e^{t} & e^{t} \\\ e^{-t} & -e^{-t} \end{array}\right]\left[\begin{array}{cc} a & b \\\ c & d \end{array}\right] $$
03

Find the product of the matrices

To find the product of the matrices, we use the matrix multiplication method: $$ \hat{\Psi}(t)= \left[\begin{array}{cc} e^{t}a+e^{t}c & e^{t}b+e^{t}d \\\ e^{-t}a-e^{-t}c & e^{-t}b-e^{-t}d \end{array}\right] $$
04

Apply the condition \(\hat{\Psi}(0)=I\) to find the constant matrix C

According to the given condition, the identity matrix is: $$ I = \left[\begin{array}{cc} 1 & 0 \\\ 0 & 1 \end{array}\right] $$ So, we set \(\hat{\Psi}(0)\) equal to the identity matrix and solve for each element in the constant matrix C: $$ \left[\begin{array}{cc} e^{0}a+e^{0}c & e^{0}b+e^{0}d \\\ e^{-0}a-e^{-0}c & e^{-0}b-e^{-0}d \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\\ 0 & 1 \end{array}\right] $$ Which gives us the following system of equations: $$ \begin{cases} a+c=1 \\ b+d=0 \\ a-c=0 \\ b-d=1 \end{cases} $$
05

Solve the system of equations

Solving the system of equations for a, b, c, and d, we get: $$ \begin{cases} a=c \\ b=1-d \\ a+c=1 \\ b-d=1 \end{cases} $$ By substituting the values, we find: $$ a=\frac{1}{2}, \quad b=d=\frac{1}{2}, \quad c=\frac{1}{2} $$
06

Write the resulting constant matrix

Now that we have the values for a, b, c, and d, we can write the constant matrix \(C\) as: $$ C=\left[\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\\ \frac{1}{2} & \frac{1}{2} \end{array}\right] $$ Thus, we have found the required constant matrix \(C\) such that \(\hat{\Psi}(t)=\Psi(t)C\) is a fundamental matrix satisfying \(\hat{\Psi}(0)=I\).

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 Vaia!

Key Concepts

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

Homogeneous Linear Systems
Homogeneous linear systems are sets of linear equations in which all the constant terms are zero. As a result, these systems always have at least one solution: the trivial solution, where all variables are zero. For the equation \( \mathbf{y}^{\textprime} = A\mathbf{y} \), where \( A \) is a matrix of coefficients and \( \mathbf{y} \) is a vector of variables, we seek solutions that describe how \( \mathbf{y} \) behaves over time. When dealing with such systems, the fundamental matrix \( \Psi(t) \) comes into play. It's a solution to the matrix differential equation that helps us understand the behavior of the system over time. When constructing a fundamental matrix, we ensure it satisfies certain conditions, such as \( \hat{\Psi}(0) = I \), to help simplify initial value problems, where the system's status is known at \( t = 0 \).
Identity Matrix
The identity matrix, denoted as \( I \), is a special square matrix with ones on the main diagonal and zeros elsewhere. Its dimensions (\( n \times n \) for an \( n \)th order identity matrix) determine the size of the matrix. One of the key properties of the identity matrix is that when it is multiplied by any compatible matrix, the original matrix remains unchanged. This property comes into prominence when solving for the fundamental matrix in differential equations. Setting the value of \( \hat{\Psi}(t) \) equal to identity matrix at time \( t = 0 \) ensures that the resulting fundamental matrix satisfies initial value conditions—which is crucial in understanding the system's behavior at the initial state.
Matrix Multiplication
Matrix multiplication is a binary operation that takes a pair of matrices and produces another matrix. This operation is fundamental in linear algebra and appears frequently in solving systems of linear equations, transforming geometric figures, and many applications in engineering and science. Unlike scalar multiplication, matrix multiplication is not commutative—meaning the order of multiplication matters (\( AB \) does not necessarily equal \( BA \) ). To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. The elements of the resulting matrix are calculated by summing the products of the corresponding entries from the rows of the first matrix and the columns of the second matrix. As seen in the step-by-step solution presented, matrix multiplication is a crucial step in finding the constant matrix \( C \) that transforms the given fundamental matrix \( \Psi(t) \) to satisfy initial conditions.
System of Equations
A system of equations is a set of two or more equations with a common set of variables. In linear algebra, these systems can be solved using various methods such as substitution, elimination, and matrix operations. The goal is to find the values of the variables that satisfy all equations simultaneously. In the context of differential equations, systems often arise when applying conditions to find specific constants, like matrix \( C \) in our example. By equating \( \hat{\Psi}(0) \) to the identity matrix, we create a system of equations that can be solved to determine the values of the elements in matrix \( C \) that will fulfill the required initial conditions. Solving these systems is fundamental in finding specific solutions to differential equations that model various physical, economical, or engineering processes.

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

We know that similar matrices have the same eigenvalues (in fact, they have the same characteristic polynomial). There are many examples that show the converse is not true; that is, there are examples of matrices \(A\) and \(B\) that have the same characteristic polynomial but are not similar. Show that the following matrices \(A\) and \(B\) cannot be similar: $$ A=\left[\begin{array}{ll} 1 & 0 \\ 3 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] $$

Consider the homogeneous linear system \(\mathbf{y}^{\prime}=A \mathbf{y} .\) Recall that any associated fundamental matrix satisfies the matrix differential equation \(\Psi^{\prime}=A \Psi\). In each exercise, construct a fundamental matrix that solves the matrix initial value problem \(\Psi^{\prime}=A \Psi, \Psi\left(t_{0}\right)=\Psi_{0}\).\(\Psi^{\prime}=\left[\begin{array}{ll}3 & -4 \\\ 2 & -3\end{array}\right] \Psi, \quad \Psi(0)=\left[\begin{array}{ll}1 & 0 \\\ 0 & 1\end{array}\right]\)

Let \(A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]\) be a \((2 \times 2)\) real symmetric matrix. In Exercise 28 of Section \(4.4\), it was shown that such a matrix has only real eigenvalues. We now show that \(A\) has a full set of eigenvectors. Note, by Exercise 30 of Section \(4.4\), that if \(A\) has distinct eigenvalues, then \(A\) has a full set of eigenvectors. Thus, the only case to consider is the case where \(A\) has repeated eigenvalues, \(\lambda_{1}=\lambda_{2}\). (a) If \(\lambda_{1}=\lambda_{2}\), show that \(a=c, b=0\), and therefore \(A=a I\). (b) Exhibit a pair of linearly independent eigenvectors in this case.

Let \(A(t)\) be an ( \(n \times n\) ) matrix function that is both differentiable and invertible on some \(t\)-interval of interest. It can be shown that \(A^{-1}(t)\) is likewise differentiable on this interval. Differentiate the matrix identity \(A^{-1}(t) A(t)=I\) to obtain the following formula: $$ \frac{d}{d t}\left[A^{-1}(t)\right]=-A^{-1}(t) A^{\prime}(t) A^{-1}(t) $$ [Hint: Recall the product rule, equation (9). Notice that the formula you derive is not the same as the power rule of single-variable calculus.]

We consider systems of second order linear equations. Such systems arise, for instance, when Newton's laws are used to model the motion of coupled spring- mass systems, such as those in Exercises 31-32. In each of Exercises \(25-30\), let \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right] .\) Note that the eigenpairs of \(A\) are \(\lambda_{1}=3, \mathbf{x}_{1}=\left[\begin{array}{l}1 \\ 1\end{array}\right]\) and \(\lambda_{2}=1, \mathbf{x}_{2}=\left[\begin{array}{r}1 \\\ -1\end{array}\right] .\) (a) Let \(T=\left[\mathbf{x}_{1}, \mathbf{x}_{2}\right]\) denote the matrix of eigenvectors that diagonalizes \(A\). Make the change of variable \(\mathbf{z}(t)=T^{-1} \mathbf{y}(t)\), and reformulate the given problem as a set of uncoupled second order linear problems. (b) Solve the uncoupled problem for \(\mathbf{z}(t)\), and then form \(\mathbf{y}(t)=T \mathbf{z}(t)\) to solve the original problem.\(\mathbf{y}^{\prime \prime}+A \mathbf{y}=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{y}^{\prime}(0)=\left[\begin{array}{l}0 \\\ 1\end{array}\right]\)

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free