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;
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