Graphics Programs Reference
In-Depth Information
Solution
Laplace's development (see Appendix A2)ofthedeterminant about the first
row of
A
yields
1
+
6
+
1
4
.
7
−
0
.
8
3
.
2
−
0
.
8
3
.
2
4
.
7
|
A
|=
2
.
0
.
1
.
−
6
.
5
4
.
1
3
.
1
4
.
1
3
.
1
−
6
.
5
=
2
.
1(14
.
07)
+
0
.
6(15
.
60)
+
1
.
1(
−
35
.
37)
=
0
Since the determinant iszero, the matrix issingular. It can be verified that the singu-
larityis due to the following rowdependency: (row3)
=
(3
×
row1)
−
(row2).
EXAMPLE 2.2
Solve the equations
Ax
=
b
, where
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
8
−
62
28
A
=
−
4
11
−
7
b
=
−
40
33
4
−
76
knowing that the LU decomposition of the coefficientmatrix is(you shouldverify this)
⎡
⎣
⎤
⎦
⎡
⎣
⎤
⎦
2
00
4
31
04
−
A
=
LU
=
−
120
1
3
002
−
−
11
Solution
We first solve the equations
Ly
=
b
by forward substitution:
2
y
1
=
28
y
1
=
28
/
2
=
14
−
y
1
+
2
y
2
=−
40
y
2
=
(
−
40
+
y
1
)
/
2
=
(
−
40
+
14)
/
2
=−
13
y
1
−
y
2
+
y
3
=
33
y
3
=
33
−
y
1
+
y
2
=
33
−
14
−
13
=
6
The solution
x
is then obtained from
Ux
=
y
by back substitution:
2
x
3
=
y
3
x
3
=
y
3
/
2
=
6
/
2
=
3
4
x
2
−
3
x
3
=
y
2
x
2
=
(
y
2
+
3
x
3
)
/
4
=
−
13
+
3(3)
]
/
4
=−
1
[
4
x
1
−
3
x
2
+
x
3
=
y
1
x
1
=
(
y
1
+
3
x
2
−
x
3
)
/
4
=
[14
+
3(
−
1)
−
3]
/
4
=
2
13
]
T
Hence the solutionis
x
=
[
2
−
2.2
Gauss Elimination Method
Introduction
Gauss eliminationis the most familiar method for solving simultaneousequations. It
consists of two parts: the eliminationphase and the solutionphase.As indicatedin
Table 2.1, the function of the eliminationphase istotransform the equations into the
form
Ux
=
c
. The equations are then solvedbyback substitution. In order to illustrate
Search WWH ::
Custom Search