Let's now find the corresponding eigenvectors: Plugging in a₁: Plugging in a2: 0 1 (9) 0 - 0 - λ 1 0 α β 0 2 λβ = 2λα = β = 2a λα = 2λβ = a = 2β 2λγ = y = 1 2λγ = λβ = λα = β = a λα = λβ = a =β 2λγ = λγ =1=0, = |αι) = = 102) λβ = αα λα = αβ 2λγ = = αγ -= ( (0)
Let's now find the corresponding eigenvectors: Plugging in a₁: Plugging in a2: 0 1 (9) 0 - 0 - λ 1 0 α β 0 2 λβ = 2λα = β = 2a λα = 2λβ = a = 2β 2λγ = y = 1 2λγ = λβ = λα = β = a λα = λβ = a =β 2λγ = λγ =1=0, = |αι) = = 102) λβ = αα λα = αβ 2λγ = = αγ -= ( (0)
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
I'm having a hard time understanding how they're getting these
When I solve and plug in beta = 2*alpha = 2 * (2*beta) -> beta = 4beta, so is it zero because thats the only number that makes this equality?
So like it's not so much algebra after solving as looking at the equality and making sense of it?
![Let's now find the corresponding eigenvectors:
Plugging in a₁:
Plugging in ag:
6 : 90--0---
λ 1
α
0 0 2
Ο
-
λβ = 2λα → β = 2a
λα = 2λβ → a = 2β } → |ai)
2λγ = 2λγ = y=1
λβ = λα = β = a
-
λα
λβ → a = β
2λγ = λγ = y = 0,
=
→ 102)
=
λβ = αα
λα αβ
αγ
=
-)
(3)
12](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb1a6ad76-e2c4-4fd8-a398-9f8475c442d4%2F43ef8e03-d3ec-421f-9e94-0a00a022c695%2Fwi8e9vs_processed.png&w=3840&q=75)
Transcribed Image Text:Let's now find the corresponding eigenvectors:
Plugging in a₁:
Plugging in ag:
6 : 90--0---
λ 1
α
0 0 2
Ο
-
λβ = 2λα → β = 2a
λα = 2λβ → a = 2β } → |ai)
2λγ = 2λγ = y=1
λβ = λα = β = a
-
λα
λβ → a = β
2λγ = λγ = y = 0,
=
→ 102)
=
λβ = αα
λα αβ
αγ
=
-)
(3)
12
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