Answered: Compute the matrix exponential eAt for…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter2: Matrices

Section2.1: Operations With Matrices

Problem 71E

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Question

Compute the matrix exponential e^A^t for system x’= Ax given in the following Problem

x’₁ = 5x₁ - 4x₂, x’₂ = 3x₁ - 2x₂

Expert Solution

Step 1

To Determine: Compute the matrix exponential e^A^t for system x’= Ax given in the following Problem

Given: x’₁ = 5x₁ - 4x₂, x’₂ = 3x₁ - 2x₂

Explanation: we have a system of equations

$x_{1}' = 5 x_{1} - 4 x_{2} a n d x_{2}' = 3 x_{1} - 2 x_{2}$

this can be written as $[\begin{matrix} x_{1}' \\ x_{2}' \end{matrix}] = [\begin{matrix} 5 & - 4 \\ 3 & - 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}]$

now let us consider the matrix

$A = [\begin{matrix} 5 & - 4 \\ 3 & - 2 \end{matrix}]$ then the characteristic equation will be $|\begin{matrix} A - λ I \end{matrix}| = 0$

so we have

$\begin{array}{rcl} |\begin{matrix} A - λ I \end{matrix}| & = & |\begin{matrix} 5 - λ & - 4 \\ 3 & - 2 - λ \end{matrix}| \\ = & (5 - λ) (- 2 - λ) + 12 \\ = & λ^{2} - 3 λ + 2 \\ = & (λ - 1) (λ - 2) \\ |\begin{matrix} A - λ I \end{matrix}| & = & 0 \\ λ & = & 1, 2 \end{array}$

now we will find the eigenvector $λ = 1$

the characteristic vector corresponding to the characteristic root is given by $(A - I) X = 0$

THAT IS

$[\begin{matrix} 5 - 1 & - 4 \\ 3 & - 2 - 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] \Rightarrow [\begin{matrix} 4 & - 4 \\ 3 & - 3 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}]$

so we have $4 x_{1} - 4 x_{2} = 0$ ,now let $x_{1} = t t h e n x_{2} = x_{1} = t$

so we have ${(x_{1}, x_{2})}^{t} = {(t, t)}^{t} = t {(1, 1)}^{t}$

characteristic vectors corresponding to the characteristic root 1 are given by $[\begin{matrix} 1 \\ 1 \end{matrix}] e^{t}$

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