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Find cholesky decomposition 25x+15y-5z=35,15x+18y+0z=33,-5x+0y+11z=6 [ Calculator, Method and examples ]

Solution:
Your problem `->` cholesky decomposition 25x+15y-5z=35,15x+18y+0z=33,-5x+0y+11z=6


Total Equations are `3`

`25x+15y-5z=35 -> (1)`

`15x+18y+0z=33 -> (2)`

`-5x+0y+11z=6 -> (3)`

Now converting given equations into matrix form
`[[25,15,-5],[15,18,0],[-5,0,11]] [[x],[y],[z]]=[[35],[33],[6]]`

Now, A = `[[25,15,-5],[15,18,0],[-5,0,11]]`, X = `[[x],[y],[z]]` and B = `[[35],[33],[6]]`

Cholesky decomposition : `A=L*L^T`, Every symmetric positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose.


Here matrix is symmetric positive definate, so Cholesky decomposition is possible.

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` `0=15-3/5xx25`
`R_2 larr R_2-3/5xx R_1`
 `9` `9=18-3/5xx15`
`R_2 larr R_2-3/5xx R_1`
 `3` `3=0-3/5xx-5`
`R_2 larr R_2-3/5xx R_1`
`-5``0``11`


`R_3 larr R_3+1/5xx R_1`

 = 
`25``15``-5`
`0``9``3`
 `0` `0=-5+1/5xx25`
`R_3 larr R_3+1/5xx R_1`
 `3` `3=0+1/5xx15`
`R_3 larr R_3+1/5xx R_1`
 `10` `10=11+1/5xx-5`
`R_3 larr R_3+1/5xx R_1`


`R_3 larr R_3-1/3xx R_2`

 = 
`25``15``-5`
`0``9``3`
 `0` `0=0-1/3xx0`
`R_3 larr R_3-1/3xx R_2`
 `0` `0=3-1/3xx9`
`R_3 larr R_3-1/3xx R_2`
 `9` `9=10-1/3xx3`
`R_3 larr R_3-1/3xx R_2`


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 definate.


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`


Dets are `25,225,2025`

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



Formula
`l_(ki)=(a_(ki) - sum_{j=1}^{i-1} l_(ij) * l_(kj))/(l_(ii))`

`l_(kk)=sqrt(a_(kk)-sum_{j=1}^{k-1} l_(kj)^2)`

Here `A` = 
2515-5
15180
-5011


`l_(11)=sqrt(a_(11))=sqrt(25)=5`

`l_(21)=(a_(21))/l_(11)=(15)/(5)=3`

`l_(22)=sqrt(a_(22)-l_(21)^2)=sqrt(18-(3)^2)=sqrt(18-9)=3`

`l_(31)=(a_(31))/l_(11)=(-5)/(5)=-1`

`l_(32)=(a_(32)-l_(31) xx l_(21))/l_(22)=(0-(-1)xx(3))/(3)=(0-(-3))/(3)=1`

`l_(33)=sqrt(a_(33)-l_(31)^2-l_(32)^2)=sqrt(11-(-1)^2-(1)^2)=sqrt(11-2)=3`

So `L` = 
`l_(11)``0``0`
`l_(21)``l_(22)``0`
`l_(31)``l_(32)``l_(33)`
 = 
500
330
-113







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