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Home > Matrix & Vector calculators > is Positive Definite Matrix example
21. is Positive Definite Matrix example ( Enter your problem )
( Enter your problem )
  1. Definition & Examples
  2. Example-2
 		Other related methods
  1. is Row Matrix
  2. is Column Matrix
  3. is Square Matrix
  4. is Horizontal Matrix
  5. is Vertical Matrix
  6. is Diagonal Matrix
  7. is Identity Matrix
  8. is Scalar Matrix
  9. is Null Matrix
  10. is Lower Triangle Matrix
  11. is Upper Triangle Matrix
  12. is Orthogonal Matrix
  13. is Singular Matrix
  14. is Nonsingular Matrix
  15. is Symmetric Matrix
  16. is Skew Symmetric Matrix
  17. is Nilpotent Matrix
  18. is Involutary Matrix
  19. is Idempotent Matrix
  20. is Periodic Matrix
  21. is Positive Definite Matrix
  22. is Negative Definite Matrix
  23. is Derogatory Matrix
  24. is Diagonally Dominant Matrix
  25. is Strictly Diagonally Dominant Matrix
  26. Auto detect the matrix type
20. is Periodic Matrix
(Previous method)		2. Example-2
(Next example)

1. Definition & Examples


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

Solution:
`A` = 
`25``15``-5`
`15``18``0`
`-5``0``11`


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


`R_2 larr R_2-0.6xx R_1`

 = 
`25``15``-5`
`0``9``3`
`-5``0``11`


`R_3 larr R_3+0.2xx R_1`

 = 
`25``15``-5`
`0``9``3`
`0``3``10`


`R_3 larr R_3-0.3333xx R_2`

 = 
`25``15``-5`
`0``9``3`
`0``0``9`


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


 `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)

`n``x_0``f(x_0)``f'(x_0)``x_1`Update
14-1332894.46020761`x_0 = x_1`
24.46020761-8.7977561250.977933694.49526152`x_0 = x_1`
34.49526152-0.04986905248.133884624.49546249`x_0 = x_1`
44.49546249-0.00000164248.117599944.4954625`x_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
(Previous method)		2. Example-2
(Next example)


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