Graphics Programs Reference
In-Depth Information
writtenas
⎡
⎣
⎤
⎦
4
−
2111
−
2
4
−
2
−
16
1
−
2
4
17
and the equivalentequations producedbythe first and the second passes of Gauss
eliminationwould appear as
⎡
⎣
⎤
⎦
4
−
2
1
11
.
00
0
3
−
1
.
5
−
10
.
50
0
−
1
.
53
.
75
14
.
25
⎡
⎣
⎤
⎦
4
−
21 11
.
0
0
3
−
1
.
5
−
10
.
5
00 3
9
.
0
It is importanttonote that the elementary row operationinEq. (2.6)leaves
the determinant of the coefficient matrix unchanged. This is rather fortunate, since
the determinantofa triangular matrix is very easy to compute—it is the product
of the diagonal elements (you can verify thisquite easily). In otherwords,
|
A
| = |
U
| =
U
11
×
U
22
×···×
U
nn
(2.7)
Back substitution phase
The unknownscan nowbecomputedbyback substitu-
tioninthe mannerdescribedinthe previous article.Solving Eqs. (c), (b) and (a) in
thatorder, we get
x
3
=
/
=
9
3
3
x
2
=
(
−
10
.
5
+
1
.
5
x
3
)
/
3
=
[
−
10
.
5
+
1
.
5(3)]
/
3
=−
2
x
1
=
(11
+
2
x
2
−
x
3
)
/
4
=
[11
+
2(
−
2)
−
3]
/
4
=
1
Algorithm for Gauss Elimination Method
Elimination phase
Let us look at the equations atsome instant during the elim-
inationphase.Assumethat the first
k
rowsof
A
have already been transformed to
upper triangular form. Therefore, the current pivot equationis the
k
th equation, and
all the equations belowit arestill to betransformed. Thissituationis depictedbythe
augmented coefficient matrix shown below. Note that the components of
A
are not
the coefficients of the originalequations(exceptfor the first row), since they have
beenalteredbythe eliminationprocedure. The same applies to the components of
the constant vector
b
.
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