This preview shows pages 1–3. Sign up to view the full content.
1
LECTURE 3.2
Dr. Vyas
Overview: (covers section 3.5)
•
Triangular Factorization : Direct and Indirect Factorization
•
Examples/Demos
TRIANGULAR FACTORIZATION:
DIRECT METHOD
1.
The purpose of triangular factorization is to speed up the computations for the
calculations involving same coefficient matrix
A
but different vector
B
.
Straightforward Gaussian elimination will have to carry out upper triangulation in
its entirety every time
B
is changed. To eliminate such unnecessary repetition,
triangular factorization technique is used.
2.
A
nonsingular
matrix
A
has a triangular factorization if it can be expressed as a
product of lower triangular matrix
L
(with unit diagonal elements) and an upper
triangular matrix
U
. Mathematically,
=
A
LU
,
0
kk
u
≠
.
3.
In matrix form, for example, this may be written as
11
12
13
11
12
13
21
22
23
21
22
23
31
32
31
32
33
33
1
0
0
1
0
0
1
0
0
a
a
a
u
u
u
a
a
a
m
u
u
m
m
a
a
a
u
=
(aka
Doolittle Factorization
)
4.
Suppose
that the coefficient matrix
A
for the linear system
=
AX
B
has a
triangular factorization
=
A
LU
, then
=
⇒
=
AX
B
LUX
B
5.
Solution to
=
LUX
B
can be obtained by first defining
Y = UX
and then solving
two systems:
a)
LY = B
to determine
Y
;
b)
UX = Y
to determine
X
. If no
row interchanges are needed, the method is called
direct factorization
.
Example 1:
Solve
LY = B
,
UX = Y
and verify that
B = AX
, given
[ 4 10
5]
′ = 
B
and
=
A
LU
is provided as,
1
2
1
1
2
3
1
0
0
2
4
6
2
4
6
1
5
3
1
0
0
3
6
1
3
2
0
0
3
1


=
Solution:
First,
use the
forward substitution algorithm
to determine
Y
via
LY = B
.
Because finding the algorithm/formulae is part of your homework, it is not shown here.
Instead a hand calculation procedure is clearly illustrated that should help you generate
the algorithm and modify the back substitution program.
1
1
11
/
y
b l
=
1
1
2
2
1
1
3
2
3
1
0
0
4
1
0
10
5
1
y
y
y

=
. Start with the first equation to find,
1
4
y
= 
1
2
1
2
(10
) /1 10 ( 2)
12
y
y
=

=
 
=
;
1
1
3
1
2
2
3
(5
)/1 5 2 4
3
y
y
y
=


= +  =
Check 1:
4 0 0
4
 + + = 
;
1
2
4( ) 12 0
10

+
+ =
;
1
1
2
3
4( ) 12( ) 3
5

+
+ =
Now solve the system of equations given by,
UX = Y
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
1
2
3
2
4
6
4
0
3
6
12
0
0
3
3
x
x
x


=
. Using the back substitution algorithm (method),
3
3/3 1
x
=
=
,
2
(12 6) / 3
2
x
=

=
,
1
( 4 ( 6) 4(2)) / 2
3
x
=   

= 
Check 2:
2( 3)
4(2) 6
4

+
 = 
; 3(2) 6(1)
12
+
=
; 3(1)
3
=
Check 3:
2( 3)
4(2) 6(1)
4

+

= 
; 1( 3) 5(2) 3(1)
10

+
+
=
; 1( 3) 3(2)
2(1)
5

+
+
=
Note that if we wanted to solve a different system such that only the
B
vector is
changed, then we just have to solve
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Vyas
 Math

Click to edit the document details