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x
x 2
x 3
b 11 b 12 b 13
0 b 22 b 23
00 b 33
a 11 a 12 a 13
0 a 22 a 23
00 a 33
x
x 2
x 3
β 14
β 24
β 34
=
+
,
(B.16)
subject to a 22 = a 11 ,a 23 =2 a 11 a 12 , and a 33 = a 11 .
Both the matrices A :=
A and B :=
B are upper triangular such that
1
B =I 3
A .
B =
(B.17)
A
3rd step (backward substitution):
a 1
a 3
11 a 12 a 4
11 (2 a 1
11 a 12
a 11 a 12 a 13
0 a 11 2 a 11 a 12
00 a 11
a 13 )
11
=
1
a 2
2 a 4
0
11 a 12
B =
=
(B.18)
11
A
a 3
0
0
11
b 11 = a 1
a 3
11 a 12 ,b 13 = a 4
11 (2 a 1
11 a 12
11 ,b 12 =
a 13 ) ,
or
b 33 a 33 = b 33 a 11 =1
b 33 = a 1
33 = a 3
11 ,b 22 a 22 = b 22 a 11 =1
b 22
= a 1
22 = a 2
11 ,b 22 a 23 + b 23 a 33 =2 a 1
11 a 12 + b 23 a 11 =0 ⇒ b 23
2 a 4
=
11 a 12 ,
(B.19)
b 11 = a 1
11 ,b 11 a 12 + b 12 a 22 = a 1
11 a 12 + b 12 a 11 =0
b 11 a 11 =1
b 12
a 3
11 a 12 ,b 11 a 13 + b 12 a 23 + b 13 a 33 = a 1
a 3
11 a 12 a 23 + b 13 a 11 =0
=
11 a 13
b 13 = a 4
11 (2 a 1
11 a 12 − a 13 ) ,
x ( y )= a 1
a 3
11 a 12 y 2 + a 4
11 (2 a 1
11 a 12
a 13 ) y 3 .
11 y
(B.20)
End of Example.
B-2 Inversion of a Bivariate Homogeneous Polynomial of Degree
n
}
of Box B.2 , to be given and find the inverse bivariate homogeneous polynomial of degree n ,namely
x ( y )or
Assume the bivariate homogeneous polynomial of degree n ,namely y ( x )or
{
y 1 ( x 1 ,x 2 ) ,y 2 ( x 1 ,x 2 )
{
x 1 ( y 1 ,y 2 ) ,x 2 ( y 1 ,y 2 )
}
, i.e. from the set of coecients
{
A 11 , A 12 , ..., A 1 n− 1 , A 1 n
}
,by
the algorithm that is outlined in Box B.2 , find the set of coecients
{
B 11 , B 12 , ..., B 1 n− 1 , B 1 n
}
.
Box B.2 (Algorithm for the construction of an inverse bivariate homogeneous polynomial of
degree n ).
y ( x ):= y 1
=
y 2
=A 11 x 1
+A 12 x 1
x 1
x 2
+
+A 1 n− 1 x 1
x 1
x 2
···
⊗···⊗
n
x 2
x 2
x 2
1times
+A 1 n x 1
x 1
x 2
=
⊗···⊗
(B.21)
n times
x 2
 
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