88 ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ −
− =
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− −
= 11111
. 32
5 .
53 27
351852 .
5 7
. 1
4 .
5 1
2 10
} {
3 2
1
x x
x x
6 351852
. 5
11111 .
32
3
− =
− =
x
8 4
. 5
6 7
. 1
4 .
53
2
= −
− −
− =
x
5 .
10 6
1 8
2 27
1
= −
− −
− =
x For the alternative right-hand-side vector, forward substitution is implemented as
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− =
6 18
12 1
148148 .
1 .
1 3
. 1
} {d
12
1
= d
6 .
21 12
3 .
18
2
= +
= d
4 18
148148 .
12 1
. 6
3
− =
− −
− −
= d
Back substitution:
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− =
4 6
. 21
12 351852
. 5
7 .
1 4
. 5
1 2
10 }
{x
747405 .
351852 .
5 4
3
− =
− =
x
235294 .
4 4
. 5
747405 .
7 .
1 6
. 21
2
− =
− −
− =
x
972318 .
1 10
747405 .
1 235294
. 4
2 12
1
= −
− −
− −
= x
9.5 The system can be written in matrix form as
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− −
− =
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− −
− −
− −
= 20
34 38
} {
2 1
8 7
1 3
1 6
2 ]
[ b
A Partial pivot:
89 ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ −
− −
= ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− −
− −
− =
38 34
20 }
{ 1
6 2
7 1
3 2
1 8
] [
b A
Forward eliminate f
21
= −3−8 = 0.375
f
31
= 2 −8 = −0.25
5 .
1 75
. 5
75 .
7 375
. 1
2 1
8 ]
[ ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− −
− −
= A
Pivot again
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− −
− =
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− −
− −
− =
34 38
20 }
{ 75
. 7
375 .
1 5
. 1
75 .
5 2
1 8
] [
b A
f
21
= −0.25 f
31
= 0.375 Forward eliminate
f
32
= −1.375−5.75 = 0.23913
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− −
− −
= 108696
. 8
5 .
1 75
. 5
2 1
8 ]
[ A Therefore, the LU decomposition is
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− −
− −
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− =
108696 .
8 5
. 1
75 .
5 2
1 8
1 23913
. 375
. 1
25 .
1 ]
]{ [
U L
Forward elimination
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
− −
− ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
= 34
38 20
1 23913
. 375
. 1
25 .
1 }
{d
20
1
− =
d
43 20
25 .
38
2
− =
− −
− −
= d
21739 .
16 43
23913 .
20 375
. 34
3
− =
− −
− −
− =
d Back substitution:
90 ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ −
− −
= ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− −
− 21739
. 16
43 20
108696 .
8 5
. 1
75 .
5 2
1 8
3 2
1
x x
x
2 108696
. 8
21739 .
16
3
− =
− =
x
8 75
. 5
2 5
. 1
43
2
= −
− −
− −
= x
4 8
2 2
8 1
20
1
= −
− −
− −
− =
x
9.6 Here is an M-file to generate the LU decomposition without pivoting
function [L, U] = LUNaiveA LUNaiveA:
LU decomposition without pivoting. input:
A = coefficient matrix output:
L = lower triangular matrix U = upper triangular matrix
[m,n] = sizeA; if m~=n, errorMatrix A must be square; end
L = eyen; U = A;
forward elimination for k = 1:n-1
for i = k+1:n Li,k = Ui,kUk,k;
Ui,k = 0; Ui,k+1:n = Ui,k+1:n-Li,kUk,k+1:n;
end end
Test with Prob. 9.3 A = [10 2 -1;-3 -6 2;1 1 5];
[L,U] = LUnaiveA L =
1.0000 0 0 -0.3000 1.0000 0
0.1000 -0.1481 1.0000 U =
10.0000 2.0000 -1.0000 0 -5.4000 1.7000
0 0 5.3519
91 Verification that [L][U] = [A].
LU ans =
10.0000 2.0000 -1.0000 -3.0000 -6.0000 2.0000
1.0000 1.0000 5.0000 Check using the lu function,
[L,U]=luA L =
1.0000 0 0 -0.3000 1.0000 0
0.1000 -0.1481 1.0000 U =
10.0000 2.0000 -1.0000 0 -5.4000 1.7000
0 0 5.3519
9.7 The result of Example 9.4 can be substituted into Eq. 9.14 to give
