method if an nxn matrix a has an lu-factorization, then the solution of ax = b can be determined by...
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Method
If an nxn matrix A has an LU-factorization, then the solution of AX = b can be determined by a Forward substitution followed by a Back substitution
[A] = [L][U]
where
[L] = lower triangular matrix all of whose diagonal
entries are different from zero.
[U] = upper triangular matrix all of whose diagonal
entries are different from zero.
LU Decomposition
LU DecompositionHow can this be used?
Given [A][X] = [b]
1. Decompose [A] into [L] and [U]
(LU)X = b
L(UX) = b ; Let UX = z
then LZ = b
1. Use Forward substitution to Solve [L][Z] = [b] for [Z]
2. Use Back substitution to Solve
[U][X] = [Z] for [X]
Exercise 2.5 ; Page # 136; Qn # 1
3
721
322
082
A
5
8
14
b
411
032
002
L
200
120
041
U
Ax = b(LU)X = bL(UX) = b
LZ = b; Let UX = Z
4
411
032
002
3
2
1
z
z
z
5
8
14
[L][Z] = [b]
Solve for [Z] using Forward Substitution :
5
8
14
411
032
002
3
2
1
z
z
z
54
832
142
321
21
1
zzz
zz
z
Complete the forward substitution to solve for [Z]
0
2
7
3
2
1
z
z
z
Z
[U][X] = [Z]
Solve for [X] using Back Substitution :
0
2
7
200
120
041
3
2
1
x
x
x
704
22
02
321
32
3
xxx
xx
x
0
1
3
3
2
1
x
x
x
X
The Solution is :
Method: [A] Decompose to [L] and [U]Storage of Multipliers Scheme :
33
2322
131211
3231
21
00
0
1
01
001
u
uu
uuu
ULA
Exercise 2.5; Page #136; Qn # 5 Finding the [U] matrix
284
1054
432
A
284
21-0
432
;2;224
121211
21 URRRAofaa
620
210
432
;2;224
131311
31
URRRAof
aa
Step 1:
Finding the [U] Matrix
Step 2:
620
210
432
1
U
200
210
432
;2;21
223231
22
32
URRRUof
aa
Matrix after Step 1:
Finding the [L] matrix
http://numericalmethods.eng.usf.edu
1
01
001
3231
21
L
224
aa
11
21
21 Aof
224
aa
11
31
31 Aof
284
1054
432
A
Finding the [L] Matrix
122
012
001
2L
12
012
001
32
1
l
L 21
2aa
1
22
32
32
Uof
[L][Z] = [b]
Solve for [Z] using Forward Substitution :
2
16
6
122
012
001
3
2
1
z
z
z
222
162
6
321
21
1
zzz
zz
z
Complete the forward substitution to solve for [Z]
2
4
6
3
2
1
z
z
z
Z
[U][X] = [Z]
Solve for [X] using Back Substitution :
2
4
6
200
210
432
3
2
1
x
x
x
6432
42
22
321
32
3
xxx
xx
x
1
2
4
3
2
1
x
x
x
X
The Solution is :
Limitations to use this Procedure :
Interchanging of any two rows not allowed .
Only elementary row operation permitted is the one that subtract a multiple of one row to another.
In matrix A, if then this procedure fails.
In matrix if , or in if then this procedure fails.
011a
1U 0
22a 0
33a
2U
Test for you?Exercise 2.5 ; Page # 136;
Qn # 7
19
121
502
324
A
3
1
1
b
?L ?U