Cholesky Decomposition Method

Cholesky Decomposition Method – Solved Problems

The Cholesky Decomposition Method is used to solve systems of linear equations when the coefficient matrix is symmetric and positive definite. The matrix is decomposed as

A = LL^T

where L is a lower triangular matrix. The solution is obtained in two steps:

• Forward substitution: LY = B
• Backward substitution: L^TX = Y

Problem 1

Solve using Cholesky decomposition

4x + 2y = 6
2x + 3y = 7

Matrix form

A = [4 2 2 3]

L = [2 0 1 √2]

Forward substitution:

2y₁ = 6 y₁ = 3
3 + √2 y₂ = 7
y₂ = 2√2

Backward substitution:

√2 y = 2√2 y = 2
2x + 2 = 3 x = 1/2

Solution: x = 1/2 y = 2

Problem 2

Solve
4x + 2y + 2z = 8
2x + 5y + z = 3
2x + y + 3z = 5

Matrix
A = [4 2 2 2 5 1 2 1 3]

Cholesky factor
L = [2 0 0 1 2 0 1 0 √2]

Forward substitution gives
Y = [4 -1/2 1/√2]

Backward substitution
x = 2 y = -1 z = 1

Problem 3

Solve
9x + 3y = 12
3x + 5y = 7

Cholesky factor
L = [3 0 1 2]

Forward substitution
3y₁ = 12 y₁ = 4
4 + 2y₂ = 7
y₂ = 3/2

Backward substitution
x = 1 y = 1

Problem 4

Solve
16x + 4y = 20
4x + 10y = 18

L = [4 0 1 3]

Forward substitution
y₁ = 5
5 + 3y₂ = 18
y₂ = 3

Backward substitution gives
x = 1 y = 2

Problem 5

Solve
25x + 5y = 30
5x + 6y = 11

L = [5 0 1 √5]

Solution
x = 1 y = 1

Problem 6

Solve
4x + 2y + 2z = 2
2x + 10y + 4z = 6
2x + 4y + 9z = 5

After Cholesky decomposition and substitution:

x = 0 y = 1/2 z = 1/2

Problem 7

Solve
6x + 3y = 9
3x + 2y = 5

Using Cholesky decomposition

x = 1 y = 1

Problem 8

Solve
9x + 3y + 3z = 12
3x + 5y + z = 8
3x + y + 4z = 7

Solution

x = 1 y = 1 z = 1

Problem 9

Solve
16x + 8y = 24
8x + 5y = 13

Solution

x = 1 y = 1

Problem 10

Solve
4x + 2y + 2z = 6
2x + 5y + z = 7
2x + y + 3z = 5

Solution

x = 1 y = 1 z = 1