1. LU Decomposition
1

2. Introduction
• LU Decomposition is another method to solve a set of simultaneous linear
equations.
• For most non-singular matrix [A] that one could conduct Naïve Gauss
Elimination forward elimination steps, one can always write it as
[A] = [L][U]
where
[L] = lower triangular matrix
[U] = upper triangular matrix
2

3. Method: [A] Decomposes to [L] and [U]
3
    






















33
23
22
13
12
11
32
31
21
0
0
0
1
0
1
0
0
1
u
u
u
u
u
u
U
L
A



• [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

4. Finding the [U] matrix
Using the Forward Elimination Procedure of Gauss Elimination
1
12
144
1
8
64
1
5
25










 
1
12
144
56
.
1
8
.
4
0
1
5
25
56
.
2
1
2
;
56
.
2
25
64














 Row
Row
 
76
.
4
8
.
16
0
56
.
1
8
.
4
0
1
5
25
76
.
5
1
3
;
76
.
5
25
144
















 Row
Row
Step 1:

5. Finding the [U] Matrix
Step 2:














76
.
4
8
.
16
0
56
.
1
8
.
4
0
1
5
25
 
7
.
0
0
0
56
.
1
8
.
4
0
1
5
25
5
.
3
2
3
;
5
.
3
8
.
4
8
.
16

















Row
Row
 













7
.
0
0
0
56
.
1
8
.
4
0
1
5
25
U
Matrix after Step 1:

6. Finding the [L] matrix
Using the multipliers used during the Forward Elimination Procedure










1
0
1
0
0
1
32
31
21



56
.
2
25
64
11
21
21 


a
a

76
.
5
25
144
11
31
31 


a
a

From the first step
of forward
elimination
1
12
144
1
8
64
1
5
25











7. Finding the [L] Matrix
 











1
5
.
3
76
.
5
0
1
56
.
2
0
0
1
L
From the second
step of forward
elimination














76
.
4
8
.
16
0
56
.
1
8
.
4
0
1
5
25
5
.
3
8
.
4
8
.
16
22
32
32 




a
a


8. Does [L][U] = [A]?
   























7
.
0
0
0
56
.
1
8
.
4
0
1
5
25
1
5
.
3
76
.
5
0
1
56
.
2
0
0
1
U
L
?

9. Using LU Decomposition to solve SLEs
Solve the following set of
linear equations using LU
Decomposition































2
279
2
177
8
106
1
12
144
1
8
64
1
5
25
3
2
1
.
.
.
x
x
x
Using the procedure for finding the [L] and [U] matrices
    
























7
.
0
0
0
56
.
1
8
.
4
0
1
5
25
1
5
.
3
76
.
5
0
1
56
.
2
0
0
1
U
L
A

10. Example
Set [L][D] = [B]
Solve for [D]































2
.
279
2
.
177
8
.
106
1
5
.
3
76
.
5
0
1
56
.
2
0
0
1
3
2
1
d
d
d
2
.
279
5
.
3
76
.
5
2
.
177
56
.
2
8
.
106
3
2
1
2
1
1






d
d
d
d
d
d

11. Example
Complete the forward substitution to solve for [D]
 
   
735
.
0
21
.
96
5
.
3
8
.
106
76
.
5
2
.
279
5
.
3
76
.
5
2
.
279
2
.
96
8
.
106
56
.
2
2
.
177
56
.
2
2
.
177
8
.
106
2
1
3
1
2
1















d
d
d
d
d
d
 























735
.
0
21
.
96
8
.
106
3
2
1
d
d
d
D

12. Example
Set [U][X] = [D]
Solve for [X] The 3 equations become


































735
0
21
96
8
106
7
.
0
0
0
56
.
1
8
.
4
0
1
5
25
3
2
1
.
.
.
x
x
x
735
.
0
7
.
0
21
.
96
56
.
1
8
.
4
8
.
106
5
25
3
3
2
3
2
1








x
x
x
x
x
x

13. Example
From the 3rd equation
050
1
7
0
735
0
735
0
7
0
3
3
3
.
x
.
.
x
.
x
.



Substituting in x3 and using the
second equation
21
96
56
1
8
4 3
2 .
x
.
x
. 



 
70
19
8
4
050
1
56
1
21
96
8
4
56
1
21
96
2
2
3
2
.
x
.
.
.
.
x
.
a
.
.
x










14. Example
Substituting in x3 and x2 using
the first equation
8
106
5
25 3
2
1 .
x
x
x 


Hence the Solution Vector is:





















050
.
1
70
.
19
2900
.
0
3
2
1
x
x
x
 
2900
0
25
050
1
70
19
5
8
106
25
5
8
106 3
2
1
.
.
.
.
x
x
.
x








15. Overview of LU Decomposition
15

16. Example
Solve the following set of
linear equations using LU
Decomposition




















3
2
1
9
8
7
6
5
4
3
2
1
x
x
x

17. Using LU Decomposition to solve SLEs
    
























0
0
0
6
3
0
3
2
1
1
2
7
0
1
4
0
0
1
U
L
A

18. Example
• Find an LU decomposition for the following example:
y1+2y2=2
3y1+6y2-y3=8
y1+2y2+y3=0
18

19. Solution
Solve the following set of
linear equations using LU
Decomposition
Using the procedure for finding the [L] and [U] matrices
    
























0
0
0
1
0
0
0
2
1
1
1
1
0
1
3
0
0
1
U
L
A

20. Thank You!
20