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Problem: Is Positive Definite matrix [[25,15,-5],[15,18,0],[-5,0,11]] [ Calculator, Method and examples ]

Solution:
Your problem `->` Is Positive Definite matrix [[25,15,-5],[15,18,0],[-5,0,11]]


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

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


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_2-3/5xx R_1`

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


`R_3 larr R_3+1/5xx R_1`

 = 
`25``15``-5`
`0``9``3`
`0``3``10`


`R_3 larr R_3-1/3xx 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.


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 = 0`


`1^(st)` iteration :

`f(x_0)=f(0)=1 xx 0^3-54 xx 0^2+673 xx 0-2025=-2025`

`f'(x_0)=f'(0)=3 xx 0^2-108 xx 0+673=673`

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

`x_1 = 0 - (-2025)/(673)`

`x_1 = 3.00891530460624`


`2^(nd)` iteration :

`f(x_1)=f(3.00891530460624)=1 xx 3.00891530460624^3-54 xx 3.00891530460624^2+673 xx 3.00891530460624-2025=-461.651421478971`

`f'(x_1)=f'(3.00891530460624)=3 xx 3.00891530460624^2-108 xx 3.00891530460624+673=375.197861033407`

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

`x_2 = 3.00891530460624 - (-461.651421478971)/(375.197861033407)`

`x_2 = 4.23933655542959`


`3^(rd)` iteration :

`f(x_2)=f(4.23933655542959)=1 xx 4.23933655542959^3-54 xx 4.23933655542959^2+673 xx 4.23933655542959-2025=-66.223869251573`

`f'(x_2)=f'(4.23933655542959)=3 xx 4.23933655542959^2-108 xx 4.23933655542959+673=269.067575304209`

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

`x_3 = 4.23933655542959 - (-66.223869251573)/(269.067575304209)`

`x_3 = 4.48546011445264`


`4^(th)` iteration :

`f(x_3)=f(4.48546011445264)=1 xx 4.48546011445264^3-54 xx 4.48546011445264^2+673 xx 4.48546011445264-2025=-2.48582175321596`

`f'(x_3)=f'(4.48546011445264)=3 xx 4.48546011445264^2-108 xx 4.48546011445264+673=248.928364954152`

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

`x_4 = 4.48546011445264 - (-2.48582175321596)/(248.928364954152)`

`x_4 = 4.49544620725357`


`5^(th)` iteration :

`f(x_4)=f(4.49544620725357)=1 xx 4.49544620725357^3-54 xx 4.49544620725357^2+673 xx 4.49544620725357-2025=-0.004042097009886`

`f'(x_4)=f'(4.49544620725357)=3 xx 4.49544620725357^2-108 xx 4.49544620725357+673=248.118919423546`

`x_5 = x_4 - f(x_4)/(f'(x_4))`

`x_5 = 4.49544620725357 - (-0.004042097009886)/(248.118919423546)`

`x_5 = 4.49546249822009`


`6^(th)` iteration :

`f(x_5)=f(4.49546249822009)=1 xx 4.49546249822009^3-54 xx 4.49546249822009^2+673 xx 4.49546249822009-2025=-0.000000010752174`

`f'(x_5)=f'(4.49546249822009)=3 xx 4.49546249822009^2-108 xx 4.49546249822009+673=248.11759941094`

`x_6 = x_5 - f(x_5)/(f'(x_5))`

`x_6 = 4.49546249822009 - (-0.000000010752174)/(248.11759941094)`

`x_6 = 4.49546249826343`


Approximate root of the equation `x^3-54x^2+673x-2025=0` using Newton Raphson mehtod is `4.49546249826343`

`n``x_0``f(x_0)``f'(x_0)``x_1`Update
10-20256733.00891530460624`x_0 = x_1`
23.00891530460624-461.651421478971375.1978610334074.23933655542959`x_0 = x_1`
34.23933655542959-66.223869251573269.0675753042094.48546011445264`x_0 = x_1`
44.48546011445264-2.48582175321596248.9283649541524.49544620725357`x_0 = x_1`
54.49544620725357-0.004042097009886248.1189194235464.49546249822009`x_0 = x_1`
64.49546249822009-0.000000010752174248.117599410944.49546249826343`x_0 = x_1`


`:. `x=4.49546


Now, using long division `(x^3-54x^2+673x-2025)/(x-4.49546)=x^2-49.50454x+450.45421`


Now, `x^2-49.50454x+450.45421=0`

`:. x=12.01568` and `x=37.48885`


`:.` The eigenvalues of the matrix `A` are given by `lamda=4.49546,12.01568,37.48885`

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







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