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
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