[A] = [L][U] Where: [L] = lower triangular matrix [U] = upper triangular matrix
How can this be used? Given [A][X] = [B] 1. Decompose [A] into [L] and [U] 2. Solve [L][Y] = [B] 3. Solve [U][X] = [Y]
a [ A] = [ L][U ] = b d
0 c e
0 1 0 0 f 0
g 1 0
h i 1
[U] is the same as the coefficient matrix at the end of the forward elimination step. [L] is obtained using the multipliers that were used in the forward elimination process
1 0 0 d [ A] = [ L][U ] = a 1 0 0 b c 1 0
e g 0
f h i
0 0 1 d [ A] = [ L][U ] = 0 1 a g 1 b c i
e h 0
f 0 0
OR
etc..
ag ah a b bg + c bh + ci d dg + e dh + ei +
f
Consider the system of equations below
2 x1 + x2 − 3 x3 = −9
− x1 + 2 x2 + 4 x3 = 27 3 x1 − 2 x2 + 5 x3 = 7
Ax=B
1 − 3 x1 − 9 2 − 1 2 x = 27 4 2 3 − 2 5 x3 7
ag ah a = b bg + c bh + ci d dg + e dh + ei + Solving
a=2 b = −1 d =3
for the variables:
ag = 1 2g = 1
ah = −3 2 h = −3
1 g= 2
−3 h= 2
f
bg + c = 2
dg + e = −2
1 (1)( ) + c = 2 2 5 c= 2
1 (3)( ) + e = −2 2 −7 e= 2
bh + ci = 4 −3 5 (−1)( ) + i = 4 2 2 i =1 dg + ei + f = 5 −3 −7 (3)( ) + (1) + f = 5 2 2 f = 13
By
2 −1 3
0 5 2 7 − 2
0 y1 −9 0 y 2 =27 y 7 3 5
forward substitution:
2 y1 = − 9 9 y1 = − 2 5 − y1 + y 2 = 27 2 y2 = 9
7 3 y1 − y 2 +13 y3 =7 2 y3 =4
1 0 0 By
1 2 1 0
3 9 − x1 − 2 2 1 x = 9 2 1 x3 4
backward substitution:
x3 = 4 x2 + x3 = 9 x2 = 5
1 3 9 x1 + x2 − x3 = − 2 2 2 x1 = −1 x1 −1 = 5 ∴ x 2 4 x3