Question1: Find all the eigenvalues for the matrix A below 2 -1 0 A=-1 2 -1 0 -1 2
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
![Chapter 7: Eigenvalues
Question1: Find all the eigenvalues for the matrix A below Solution:
-1
Power Method:
max
A =-1 2 -1
[A VjT
absolute
value
No
error
0 -1 2
0.0000
1.0000
0.0000
-1.0000
2.0000
-1.0000
2.0000
Solution:
The characteristic equation A-A1=0 is
2.
|2-2
-1
3
-1
2-2
-il=0
4
-1 2-a
5
6.
-0.7071
1.0000
0.7071-24141
3.4141
-2.4141
3.4141
0.0003
Cubic polynomial:
(2-4)[(2-4)(2-4)-(-19(-1)]
-(-1D[(-1)(2-2)-(-1)(0)]
+o[(-1)(-1)-(2-2)(0)]=0
-12' +62-_A+_=0
0.7071 1.0000
-0.7071
The dominant eigenvalue is 3.4142
And its corresponding eigenvector is (-0,7071 1 -0.7071)'
Solution:
Inverse Power Method:
"EQN - DEGREE → 3"
. Eigenvalue 1=3.4142
Eigenvalue 2=-2
Eigenvalue 3=
(0.75 0.5 0.25
A= 0.51
0.25 0.5 0.75
0.5
max
No
[A VT
absolute
value
error
Question 2:
Given the matrix
0.0000
1.0000
0.0000
0.5000
1.0000 0.5000
1.0000
2 -1 0
A= -1 2 -1
0 -1
2
3.
Use the power method to find the dominant eigenvalues,
A / the smallest eigenvalues, 2 and its corresponding
cigenvector, v, for the matrix A using V° = (0 1 0).
Stop the iteration until |4-A|s0.005
6.
0.7071
1.0000
0.7071
1.2071
L7071
1.2071
1.7071
0.0002
0.7071 L0000 0.7071
The dominant cigenvalue is 1/1.7071. And its corre sponding cigenvector is (0.7071 1 07071)
* * * * x>>](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b8e46f9-2a94-469b-8964-c3231bbf1505%2Faabb8b27-33db-46d7-8691-9caaef5f6ba7%2Fy14thy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Chapter 7: Eigenvalues
Question1: Find all the eigenvalues for the matrix A below Solution:
-1
Power Method:
max
A =-1 2 -1
[A VjT
absolute
value
No
error
0 -1 2
0.0000
1.0000
0.0000
-1.0000
2.0000
-1.0000
2.0000
Solution:
The characteristic equation A-A1=0 is
2.
|2-2
-1
3
-1
2-2
-il=0
4
-1 2-a
5
6.
-0.7071
1.0000
0.7071-24141
3.4141
-2.4141
3.4141
0.0003
Cubic polynomial:
(2-4)[(2-4)(2-4)-(-19(-1)]
-(-1D[(-1)(2-2)-(-1)(0)]
+o[(-1)(-1)-(2-2)(0)]=0
-12' +62-_A+_=0
0.7071 1.0000
-0.7071
The dominant eigenvalue is 3.4142
And its corresponding eigenvector is (-0,7071 1 -0.7071)'
Solution:
Inverse Power Method:
"EQN - DEGREE → 3"
. Eigenvalue 1=3.4142
Eigenvalue 2=-2
Eigenvalue 3=
(0.75 0.5 0.25
A= 0.51
0.25 0.5 0.75
0.5
max
No
[A VT
absolute
value
error
Question 2:
Given the matrix
0.0000
1.0000
0.0000
0.5000
1.0000 0.5000
1.0000
2 -1 0
A= -1 2 -1
0 -1
2
3.
Use the power method to find the dominant eigenvalues,
A / the smallest eigenvalues, 2 and its corresponding
cigenvector, v, for the matrix A using V° = (0 1 0).
Stop the iteration until |4-A|s0.005
6.
0.7071
1.0000
0.7071
1.2071
L7071
1.2071
1.7071
0.0002
0.7071 L0000 0.7071
The dominant cigenvalue is 1/1.7071. And its corre sponding cigenvector is (0.7071 1 07071)
* * * * x>>
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