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11. Cholesky decomposition method example ( Enter your problem )
  1. Example `6x+15y+55z=76,15x+55y+225z=295,55x+225y+979z=1259`
  2. Example `25x+15y-5z=35,15x+18y+0z=33,-5x+0y+11z=6`
Other related methods
  1. Inverse Matrix method
  2. Cramer's Rule method
  3. Gauss-Jordan Elimination method
  4. Gauss Elimination Back Substitution method
  5. Gauss Seidel method
  6. Gauss Jacobi method
  7. Elimination method
  8. LU decomposition using Gauss Elimination method
  9. LU decomposition using Doolittle's method
  10. LU decomposition using Crout's method
  11. Cholesky decomposition method
  12. SOR (Successive over-relaxation) method
  13. Relaxation method

1. Example `6x+15y+55z=76,15x+55y+225z=295,55x+225y+979z=1259`
(Previous example)
12. SOR (Successive over-relaxation) method
(Next method)

2. Example `25x+15y-5z=35,15x+18y+0z=33,-5x+0y+11z=6`





Solve Equations 25x+15y-5z=35,15x+18y+0z=33,-5x+0y+11z=6 using Cholesky decomposition method

Solution:
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


`L xx L^T` = 
500
330
-113
 `xx` 
53-1
031
003
 = 
2515-5
15180
-5011


and `A` = 
2515-5
15180
-5011


Now, `Ax=B`, and `A=LL^T => LL^Tx=B`

let `L^Tx=y`, then `Ly=B =>`

`5``0``0`
`3``3``0`
`-1``1``3`
 `xx` 
`y_1`
`y_2`
`y_3`
 = 
`35`
`33`
`6`


```5y_1``=``35```
```3y_1``+``3y_2``=``33```
`-``y_1``+``y_2``+``3y_3``=``6```


Now use forward substitution method
From (1)
`5y_1=35`

`=>5y_1=35`

`=>y_1=(35)/(5)=7`

From (2)
`3y_1+3y_2=33`

`=>3(7)+3y_2=33`

`=>21+3y_2=33`

`=>3y_2=33-21=12`

`=>y_2=(12)/(3)=4`

From (3)
`-y_1+y_2+3y_3=6`

`=>-(7)+(4)+3y_3=6`

`=>-3+3y_3=6`

`=>3y_3=6+3=9`

`=>y_3=(9)/(3)=3`

Now, `L^Tx=y`

`5``3``-1`
`0``3``1`
`0``0``3`
 `xx` 
`x`
`y`
`z`
 = 
`7`
`4`
`3`


```5x``+``3y``-``z``=``7```
```3y``+``z``=``4```
```3z``=``3```


Now use back substitution method
From (3)
`3z=3`

`=>z=(3)/(3)=1`

From (2)
`3y+z=4`

`=>3y+(1)=4`

`=>3y+1=4`

`=>3y=4-1=3`

`=>y=(3)/(3)=1`

From (1)
`5x+3y-z=7`

`=>5x+3(1)-(1)=7`

`=>5x+2=7`

`=>5x=7-2=5`

`=>x=(5)/(5)=1`

Solution by Cholesky Decomposition method is
`x=1,y=1,z=1`


This material is intended as a summary. Use your textbook for detail explanation.
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1. Example `6x+15y+55z=76,15x+55y+225z=295,55x+225y+979z=1259`
(Previous example)
12. SOR (Successive over-relaxation) method
(Next method)





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