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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=15-3/5xx25R_2 larr R_2-3/5xx R_1 9 9=18-3/5xx15R_2 larr R_2-3/5xx R_1 3 3=0-3/5xx-5R_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+1/5xx25R_3 larr R_3+1/5xx R_1 3 3=0+1/5xx15R_3 larr R_3+1/5xx R_1 10 10=11+1/5xx-5R_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=0-1/3xx0R_3 larr R_3-1/3xx R_2 0 0=3-1/3xx9R_3 larr R_3-1/3xx R_2 9 9=10-1/3xx3R_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 =
 25 15 -5 15 18 0 -5 0 11

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)
=
 5 0 0 3 3 0 -1 1 3

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