LU factorization: 5x1+2x2+3x3

LU factorization (Doolittle): write A = L·U, where L is lower triangular with 1s on the diagonal and U is upper triangular. Then solve L·y = b by forward substitution and U·x = y by back substitution.

Step 1: System

5x1 + 2x2 + 3x3 = 31
3x1 + 3x2 + 2x3 = 25
x1 + 2x2 + 4x3 = 25

A = [5 2 3; 3 3 2; 1 2 4], b = [31, 25, 25]ᵀ

Step 2: Find L and U

From A = L·U (compare row by row):

Row 1 of U: u11 = 5.0000, u12 = 2.0000, u13 = 3.0000
Column 1 of L: l21 = a21/u11 = 3.0000/5.0000 = 0.6000, l31 = a31/u11 = 1.0000/5.0000 = 0.2000
Row 2 of U: u22 = a22 − l21·u12 = 3.0000 − (0.6000)(2.0000) = 1.8000; u23 = a23 − l21·u13 = 2.0000 − (0.6000)(3.0000) = 0.2000
l32 = (a32 − l31·u12)/u22 = (2.0000 − (0.2000)(2.0000))/1.8000 = 0.8889
u33 = a33 − l31·u13 − l32·u23 = 4.0000 − (0.2000)(3.0000) − (0.8889)(0.2000) = 3.2222

Result:

L =

1.0000  0.0000  0.0000
0.6000  1.0000  0.0000
0.2000  0.8889  1.0000

U =

5.0000  2.0000  3.0000
0.0000  1.8000  0.2000
0.0000  0.0000  3.2222

(Check: L·U reproduces A.)

Step 3: Solve L·y = b (forward substitution)

y1 = 31.0000
y2 = 25.0000 − (0.6000)(31.0000) = 6.4000
y3 = 25.0000 − (0.2000)(31.0000) − (0.8889)(6.4000) = 13.1111

Step 4: Solve U·x = y (back substitution)

x3 = (13.1111) / 3.2222 = 4.0690
x2 = (6.4000 − (0.2000)(4.0690)) / 1.8000 = 3.1034
x1 = (31.0000 − (2.0000)(3.1034) − (3.0000)(4.0690)) / 5.0000 = 2.5172

Answer: x1 = 2.5172, x2 = 3.1034, x3 = 4.0690