Therefore, A has the repeated complex conjugate pair 3+ 4i of eigenvalues. First show that the complex vec- tors vi = [1 i 0 0]" and v2 = [0 0 1 i]" form a length 2 chain {v1, V2} associated with the eigen- value A = 3 – 4i. Then calculate the real and imaginary parts of the complex-valued solutions Vjedt and (vịt +v2)edr to find four independent real-valued solutions of x' = Ax.
Therefore, A has the repeated complex conjugate pair 3+ 4i of eigenvalues. First show that the complex vec- tors vi = [1 i 0 0]" and v2 = [0 0 1 i]" form a length 2 chain {v1, V2} associated with the eigen- value A = 3 – 4i. Then calculate the real and imaginary parts of the complex-valued solutions Vjedt and (vịt +v2)edr to find four independent real-valued solutions of x' = Ax.
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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Help please with 33
![**Problem 33: Characteristic Equation of a Matrix**
Consider the coefficient matrix \( \mathbf{A} \) of the system given by:
\[
\mathbf{x}' = \begin{bmatrix} 3 & -4 & 1 & 0 \\ 4 & 3 & 0 & 1 \\ 0 & 0 & 3 & -4 \\ 0 & 0 & 4 & 3 \end{bmatrix} \mathbf{x}
\]
The characteristic equation of this matrix \( \mathbf{A} \) is:
\[
\phi(\lambda) = (\lambda^2 - 6\lambda + 25)^2 = 0.
\]
This equation is derived from finding the eigenvalues of matrix \( \mathbf{A} \), which are the solutions to \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The expression represents a polynomial equation whose roots correspond to these eigenvalues.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03106fa3-330c-422f-ad56-a6718a5c3a13%2F887f6379-908d-4326-ae35-ab2512c6c8cf%2Frr6l04k_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 33: Characteristic Equation of a Matrix**
Consider the coefficient matrix \( \mathbf{A} \) of the system given by:
\[
\mathbf{x}' = \begin{bmatrix} 3 & -4 & 1 & 0 \\ 4 & 3 & 0 & 1 \\ 0 & 0 & 3 & -4 \\ 0 & 0 & 4 & 3 \end{bmatrix} \mathbf{x}
\]
The characteristic equation of this matrix \( \mathbf{A} \) is:
\[
\phi(\lambda) = (\lambda^2 - 6\lambda + 25)^2 = 0.
\]
This equation is derived from finding the eigenvalues of matrix \( \mathbf{A} \), which are the solutions to \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The expression represents a polynomial equation whose roots correspond to these eigenvalues.
![Therefore, A has the repeated complex conjugate pair \(3 \pm 4i\) of eigenvalues. First, show that the complex vectors
\[
\mathbf{v_1} = \begin{bmatrix} 1 \\ i \\ 0 \\ 0 \end{bmatrix}^T \quad \text{and} \quad \mathbf{v_2} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ i \end{bmatrix}^T
\]
form a length 2 chain \(\{\mathbf{v_1}, \mathbf{v_2}\}\) associated with the eigenvalue \(\lambda = 3 - 4i\). Then calculate the real and imaginary parts of the complex-valued solutions
\[
\mathbf{v_1} e^{\lambda t} \quad \text{and} \quad (\mathbf{v_1}t + \mathbf{v_2}) e^{\lambda t}
\]
to find four independent real-valued solutions of \(x' = Ax\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F03106fa3-330c-422f-ad56-a6718a5c3a13%2F887f6379-908d-4326-ae35-ab2512c6c8cf%2F326ap96_processed.png&w=3840&q=75)
Transcribed Image Text:Therefore, A has the repeated complex conjugate pair \(3 \pm 4i\) of eigenvalues. First, show that the complex vectors
\[
\mathbf{v_1} = \begin{bmatrix} 1 \\ i \\ 0 \\ 0 \end{bmatrix}^T \quad \text{and} \quad \mathbf{v_2} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ i \end{bmatrix}^T
\]
form a length 2 chain \(\{\mathbf{v_1}, \mathbf{v_2}\}\) associated with the eigenvalue \(\lambda = 3 - 4i\). Then calculate the real and imaginary parts of the complex-valued solutions
\[
\mathbf{v_1} e^{\lambda t} \quad \text{and} \quad (\mathbf{v_1}t + \mathbf{v_2}) e^{\lambda t}
\]
to find four independent real-valued solutions of \(x' = Ax\).
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