Graphics Reference
In-Depth Information
Next, consider questions (3) and (4). To see the possible role of polynomial divi-
sion and bases in answers to those questions, we look at the special case of linear
equations and parameterizations. The standard solution to the problem in question
(3) when we have linear equations is to turn it into a matrix problem and apply the
Gauss-Jordan elimination method and standard row reductions. This approach can
be thought of as a method for finding certain bases.
10.10.1. Example.
Consider two linear equations
(
)
=- +=
fXYZ
,,
,,
X
2 0
2340
Y Z
(
)
=-+=
(10.57)
gXYZ
X
Y
Z
.
Applying the standard row reduction method to the matrix which represents the
system of equations (10.57) leads to the row echelon form of that matrix:
121
234
-
-
121
012
-
105
012
Ê
Ë
ˆ
¯
Ê
Ë
ˆ
¯
Æ
Ê
Ë
ˆ
¯
Æ
It follows that solving (10.57) is equivalent to solving
XZ
YZ
+=
+=
50
2 .
(10.58)
Equations (10.58) are easily solved for X and Y in terms of Z and turned into a para-
metric solution of the form
Xt
Yt
Zt
=-
=-
=
5
2
.
A similar approach works for question (4) in the linear case.
10.10.2. Example.
Consider the parameterization
xt
=+ -
=-+
=
t
7
11
2
x
2
t
1
35
t
2
1
2
x
t
- ,
(10.59)
3
2
which we rewrite as
tt x
tt
+-
-=
70
12 1
2
-
-
x
+ =
--=.
1
0
12
2
(10.60)
3
t
x
5
0
2
3
Again, the matrix associated to the system in (10.60) can brought into standard form
using row reductions:
