1 0 %3D 3 2

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Please answer part 4 of this question.

Problem 5
Diagonalize the following matrices:
1 0
1. А —
3 2
1
2. В —
1
1
1
0 2
3. С —
2 0
2 0
4. D =
1
2 -1
1
1
Transcribed Image Text:Problem 5 Diagonalize the following matrices: 1 0 1. А — 3 2 1 2. В — 1 1 1 0 2 3. С — 2 0 2 0 4. D = 1 2 -1 1 1
Expert Solution
Step 1

as you asking too many question i only answered first one.

 

A can be diagonalized if there exists an invertible matrix P and diagonal matrix D such that A=PDP-1



Here A  = 
  1 0  
  3 2  





Find eigenvalues of the matrix A

|A-λI|=0

   (1-λ)   0   
   3   (2-λ)   
 = 0



(1-λ)×(2-λ)-0×3=0

(2-3λ+λ2)-0=0

(λ2-3λ+2)=0

(λ-1)(λ-2)=0

(λ-1)=0or(λ-2)=0

 The eigenvalues of the matrix A are given by λ=1,2,


 

1. Eigenvectors for λ=1


1. Eigenvectors for λ=1

A-λI=
  1 0  
  3 2  
 - 1 
  1 0  
  0 1  


 = 
  1 0  
  3 2  
 - 
  1 0  
  0 1  

 = 
  0 0  
  3 1  


Now, reduce this matrix
interchanging rows R1R2

 = 
  3 1  
  0 0  


R1R1÷3

 = 
  1 0.33333333  
  0 0  


The system associated with the eigenvalue λ=1

(A-1I)
  x1  
  x2  
 = 
  1 0.33333333  
  0 0  
 
  x1  
  x2  
 = 
  0  
  0  


x1+0.33333333x2=0

x1=-0.33333333x2

 eigenvectors corresponding to the eigenvalue λ=1 is

v=
  -0.33333333x2  
  x2  


Let x2=1

v1=

2. Eigenvectors for λ=2




2. Eigenvectors for λ=2

  -0.33333333  
  1
A-λI=
  1 0  
  3 2  
 - 2 
  1 0  
  0 1  


 = 
  1 0  
  3 2  
 - 
  2 0  
  0 2  

 = 
  -1 0  
  3 0  


Now, reduce this matrix
interchanging rows R1R2

 = 
  3 0  
  -1 0  


R1R1÷3

 = 
  1 0  
  -1 0  


R2R2+R1

 = 
  1 0  
  0 0  


The system associated with the eigenvalue λ=2

(A-2I)
  x1  
  x2  
 = 
  1 0  
  0 0  
 
  x1  
  x2  
 = 
  0  
  0  


x1=0

 eigenvectors corresponding to the eigenvalue λ=2 is

v=
  0  
  x2  


Let x2=1

v2=
  0  
  1
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