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Backward substitution:
B 11 B 12 B 13
0B 22 B 23
00B 33
A 11 A 12 A 13
0A 22 A 23
00A 33
=I 14 ;
B =I 14
(B.57)
A
(i)B 11 A 11 =I 2
B 11 =A 1
11 ;
(ii)B 12 A 12 +B 12 A 22 =0
11 A 12 A 1[2]
B 11 A 12 A 1
A 1
B 12 =
22 =
22 ;
(B.58)
(iii)B 11 A 13 +B 12 A 23 +B 13 A 33 =0
(B 11 A 13 +B 12 A 23 )A 1
B 13 =
33 =
11 A 13 A 1[3]
11 A 12 A 1[2]
11 (A 11 A 12 +A 12 A 11 )A 1[3]
= A 1
+A 1
11
11
=A 1
A 13 +A 12 (A 1
A 1
A 11 )](A 1
A 1
11 A 1
11 [
11 )(A 11
A 12 +A 12
11 ) .
11
11
First, we have used ( A⊗B ) 1 = A 1
⊗B 1 for two invertible square matrices A and B , secondly
we have used the standard solution of a system of upper triangular matrix equations. For the
inverse polynomial representation, only the elements of the first row of the matrix B :=
B are
of interest. An explicit write-up is
x 1
x 2
=A 1
y 1
y 2
11 ) y 1
[2]
A 1
11 A 12 (A 1
A 1
11
11
y 2
A 1
A 12 (A 1
A 1
11 [A 13
11 )(A 11
A 12 +A 12
A 11 )]
11
11 ) y 1
[3]
(A 1
A 1
A 1
.
(B.59)
11
11
y 2
End of Example.
B-3 Inversion of a Multivariate Homogeneous Polynomial of Degree
n
Assume a multivariate homogeneous polynomial of degree n ,namely y ( x ), to be given and find
the inverse multivariate homogeneous polynomial of degree n ,namely x ( y ), and
n
y ( x )=A 11 x +A 12 x [2] + ··· +A 1 n− 1 x [ n− 1] +A 1 n x [ n ] =
A 1 k x [ k ]
p ,
x R
k =1
(B.60)
n
x ( y )=B 11 y +B 12 y [2] + ··· +B 1 n− 1 y [ n− 1] +B 1 n y [ n ] =
B 1 k y [ k ]
p ,
y
R
k =1
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