Let A = \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}. For the determinant (λ-1)^(2)(λ+1) = 0. So, the eigen value λ = 1 with algebraic multiplicity of 2, and = -1 with algebraic multiplicity of 1. For λ = 1: Use Av = λv: \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=1\begin{pmatrix}x\\ y\\ z\end{pmatrix}. For λ = -1 : Use Av = λv: \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=-1\begin{pmatrix}x\\ y\\ z\end{pmatrix}. Then Find P, P^(-1), D and then e^(tA) = P e^(tD) P^(-1)
Let A = \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}. For the determinant (λ-1)^(2)(λ+1) = 0. So, the eigen value λ = 1 with algebraic multiplicity of 2, and = -1 with algebraic multiplicity of 1. For λ = 1: Use Av = λv: \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=1\begin{pmatrix}x\\ y\\ z\end{pmatrix}. For λ = -1 : Use Av = λv: \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=-1\begin{pmatrix}x\\ y\\ z\end{pmatrix}. Then Find P, P^(-1), D and then e^(tA) = P e^(tD) P^(-1)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Let A = \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}. For the determinant (λ-1)^(2)(λ+1) = 0. So, the eigen value λ = 1 with algebraic multiplicity of 2, and = -1 with algebraic multiplicity of 1. For λ = 1: Use Av = λv: \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=1\begin{pmatrix}x\\ y\\ z\end{pmatrix}. For λ = -1 : Use Av = λv: \begin{pmatrix}1&2&0\\ 0&1&-2\\ 2&2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=-1\begin{pmatrix}x\\ y\\ z\end{pmatrix}. Then Find P, P^(-1), D and then e^(tA) = P e^(tD) P^(-1)
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