Geography Reference
In-Depth Information
1st polynomial:
x
1
x
2
=B
11
A
11
x
1
+A
12
x
1
x
1
x
2
+
⊗
x
2
x
2
+B
12
A
11
x
1
+A
12
x
1
x
1
x
2
A
11
x
1
+A
12
x
1
x
1
x
2
=
⊗
⊗
⊗
(B.40)
x
2
x
2
x
2
x
2
=B
11
A
11
x
1
+(B
11
A
12
+B
12
A
11
⊗
A
11
)
x
1
[2]
+
β
13
.
x
2
x
2
According to
Grafarend and Schaffrin
(
1993
)or
Steeb
(
1991
). Note that we here have used (
AC
)
⊗
(
BD
)=(
A
⊗
B
)(
C
D
), i.e.
A
11
x
1
⊗
A
11
x
1
=(A
11
⊗
A
11
)
x
1
x
1
x
2
.
⊗
⊗
(B.41)
x
2
x
2
x
2
2nd polynomial:
x
1
x
2
x
1
x
2
=B
11
y
1
y
1
y
2
+
β
23
=B
22
(A
11
⊗
⊗
⊗
A
11
)
y
2
x
1
x
2
x
1
x
2
+
β
23
.
⊗
(B.42)
Forward substitution:
⎡
x
1
x
2
⎤
⎡
x
1
x
2
⎤
⎣
⎦
=
B
11
B
12
A
11
A
12
0A
22
⎣
⎦
+
β
13
,
(B.43)
x
1
x
2
x
1
x
2
x
1
x
2
x
1
x
2
β
23
0B
22
⊗
⊗
subject to A
22
=A
11
⊗
A
11
. Note that the matrices A :=
A
and B :=
B
are upper triangular
B
=I
2
⇔
−
1
such that
B
=
A
.
A
Backward substitution:
B
11
B
12
0B
22
A
11
A
12
0A
22
=I
6
;
B
=I
6
⇔
(B.44)
A
(i)B
11
A
11
=I
2
⇒
B
11
=A
−
1
11
,
B
11
A
12
A
−
1
A
−
1
11
A
12
(A
−
1
A
−
1
(ii)B
11
A
12
+B
12
A
22
=0
⇒
B
12
=
−
22
=
−
⊗
11
)
.
(B.45)
11
B
−
1
for two invertible square matrices
A
and
B
. Second,
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
)
−
1
=
A
−
1
First, we have used (
A
⊗
⊗
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
−
⊗
.
(B.46)
11
11
y
2
End of Example.
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