Graphics Programs Reference
In-Depth Information
Solution We use Gauss elimination, storing the multipliers in the upper triangular
portion of A .At the completion of elimination, the matrix will have the form of U
in
Eq. (2.25).
The termstobeeliminatedinthe first pass are A 21 and A 31 using the elementary
operations
row2
row2
(
1)
×
row1
row3
row3
(1)
×
row1
Storing the multipliers(
1 and 1) in the locations occupiedby A 12 and A 13 , we get
3
11
0 2 4
047
A =
The second pass is the operation
row3
row3
2
×
row2
which yields after overwriting A 23 with the multiplier2
3
11
0 22
00
A = 0
L T =
\
D
\
1
Hence
1 00
3 00
0 2
L
=
110
121
D
=
0
00
1
EXAMPLE 2.11
Solve Ax
=
b , where
6
4
1
00
···
x 1
x 2
x 3
.
x 9
x 10
3
0
0
.
0
4
4
6
4
1
0
···
1
4
6
4
1
···
A
=
=
. . .
. . .
. . .
. . .
···
0
1
4
6
4
···
001
4
7
Solution As the coefficient matrix issymmetric and pentadiagonal, we utilize the
functions LUdec5 and LUsol5 :
% Example 2.11 (Solution of pentadiagonal eqs.)
n=10;
Search WWH ::

Custom Search