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21. is Positive Definite Matrix example ( Enter your problem )
  1. Definition & Examples
  2. Example-2
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  20. is Periodic Matrix
  21. is Positive Definite Matrix
  22. is Negative Definite Matrix
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  25. is Strictly Diagonally Dominant Matrix
  26. Auto detect the matrix type

20. is Periodic Matrix
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2. Example-2
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1. Definition & Examples





1. is Positive Definite Matrix ?
[[25,15,-5],[15,18,0],[-5,0,11]]


Solution:
A = 
2515-5
15180
-5011


A matrix is positive definite if it's symmetric and all its pivots are positive.

Test method 1: Existence of all positive Pivots.
First apply Gaussian Elimination method to find Pivots
A = 
2515-5
15180
-5011


R_2 larr R_2-0.6xx R_1

 = 
2515-5
093
-5011


R_3 larr R_3+0.2xx R_1

 = 
2515-5
093
0310


R_3 larr R_3-0.3333xx R_2

 = 
2515-5
093
009


Pivots are the first non-zero element in each row of this eliminated matrix.

:. Pivots are 25,9,9

Here all pivots are positive, so matrix is positive definite.


A matrix is positive definite if Determinants of all upper-left sub-matrices are positive.

Test method 2: Determinants of all upper-left sub-matrices are positive.
A = 
2515-5
15180
-5011


 25 
=25


 25  15 
 15  18 
=225


 25  15  -5 
 15  18  0 
 -5  0  11 
=2025


Determinants are 25,225,2025

Here all determinants are positive, so matrix is positive definite.


A matrix is positive definite if it's symmetric and all its eigenvalues are positive.

Test method 3: All positive eigen values.
|A-lamdaI|=0

 (25-lamda)  15  -5 
 15  (18-lamda)  0 
 -5  0  (11-lamda) 
 = 0


:.(25-lamda)((18-lamda) × (11-lamda) - 0 × 0)-15(15 × (11-lamda) - 0 × (-5))+(-5)(15 × 0 - (18-lamda) × (-5))=0

:.(25-lamda)((198-29lamda+lamda^2)-0)-15((165-15lamda)-0)-5(0-(-90+5lamda))=0

:.(25-lamda)(198-29lamda+lamda^2)-15(165-15lamda)-5(90-5lamda)=0

:. (4950-923lamda+54lamda^2-lamda^3)-(2475-225lamda)-(450-25lamda)=0

:.(-lamda^3+54lamda^2-673lamda+2025)=0

:.-(lamda^3-54lamda^2+673lamda-2025)=0

:.(lamda^3-54lamda^2+673lamda-2025)=0

Roots can be found using newton raphson method
Newton Raphson method for x^3-54x^2+673x-2025=0


Here x^3-54x^2+673x-2025=0

Let f(x) = x^3-54x^2+673x-2025

:. f'(x) = 3x^2-108x+673

x_0 = 4


1^(st) iteration :

f(x_0)=f(4)=4^3-54 xx 4^2+673 xx 4-2025=-133

f'(x_0)=f'(4)=3 xx 4^2-108 xx 4+673=289

x_1 = x_0 - f(x_0)/(f'(x_0))

x_1=4 - (-133)/(289)

x_1=4.46020761


2^(nd) iteration :

f(x_1)=f(4.46020761)=4.46020761^3-54 xx 4.46020761^2+673 xx 4.46020761-2025=-8.7977561

f'(x_1)=f'(4.46020761)=3 xx 4.46020761^2-108 xx 4.46020761+673=250.97793369

x_2 = x_1 - f(x_1)/(f'(x_1))

x_2=4.46020761 - (-8.7977561)/(250.97793369)

x_2=4.49526152


3^(rd) iteration :

f(x_2)=f(4.49526152)=4.49526152^3-54 xx 4.49526152^2+673 xx 4.49526152-2025=-0.04986905

f'(x_2)=f'(4.49526152)=3 xx 4.49526152^2-108 xx 4.49526152+673=248.13388462

x_3 = x_2 - f(x_2)/(f'(x_2))

x_3=4.49526152 - (-0.04986905)/(248.13388462)

x_3=4.49546249


4^(th) iteration :

f(x_3)=f(4.49546249)=4.49546249^3-54 xx 4.49546249^2+673 xx 4.49546249-2025=-0.00000164

f'(x_3)=f'(4.49546249)=3 xx 4.49546249^2-108 xx 4.49546249+673=248.11759994

x_4 = x_3 - f(x_3)/(f'(x_3))

x_4=4.49546249 - (-0.00000164)/(248.11759994)

x_4=4.4954625


Approximate root of the equation x^3-54x^2+673x-2025=0 using Newton Raphson method is 4.4954625 (After 4 iterations)

nx_0f(x_0)f'(x_0)x_1Update
14-1332894.46020761x_0 = x_1
24.46020761-8.7977561250.977933694.49526152x_0 = x_1
34.49526152-0.04986905248.133884624.49546249x_0 = x_1
44.49546249-0.00000164248.117599944.4954625x_0 = x_1


:. x=4.4954625


Now, using long division (x^3-54x^2+673x-2025)/(x-4.4954625)=x^2-49.5045375x+450.45420817


Now, x^2-49.5045375x+450.45420817=0

:. x=12.01568365 and x=37.48885386


:. The eigenvalues of the matrix A are given by lamda=4.4954625,12.01568365,37.48885386

Here all determinants are positive, so matrix is positive definite.


This material is intended as a summary. Use your textbook for detail explanation.
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20. is Periodic Matrix
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2. Example-2
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