# The General Solution of Quaternion Matrix Equation Havingη-Skew-Hermicity and Its Cramer’s Rule

Authors: Abdur Rehman; Ivan Kyrchei; Ilyas Ali; Muhammad Akram; Abdul Shakoor
Journal: Mathematical Problems in Engineering
 (2019)

---

## Abstract

We determine some necessary and sufficient conditions for the existence of theη-skew-Hermitian solution to the following system A
X
-
(
A
X
)
η
⁎
+
B
Y
B
η
⁎
+
C
Z
C
η
⁎
=
D
,
Y
=
-
Y
η
⁎
,
Z
=
-
Z
η
⁎ over the quaternion skew field and provide an explicit expression of its general solution. Within the framework of the theory of quaternion row-column noncommutative determinants, we derive its explicit determinantal representation formulas that are an analog of Cramer’s rule. A numerical example is also provided to establish the main result.


---

## Body

## 1. Introduction

Throughout this paper, the real number field and the complex field are denoted byR and C, respectively, and H for the quaternion algebra(1)
H
=
h
0
+
h
1
i
+
h
2
j
+
h
3
k
∣
i
2
=
j
2
=
k
2
=
i
j
k
=
-
1
,
h
0
,
h
1
,
h
2
,
h
3
∈
R
.The set of all matrices of dimension m
×
n over H is represented by H
m
×
n. An identity matrix with conformable size is denoted by I. For any matrix A over H, R
(
A
) and N
(
A
) stand for the column right space and the row left space of A, respectively. D
[
R
(
A
)
] denotes the dimension of R
(
A
). By [1], we have D
[
R
(
A
)
]
=
D
[
N
(
A
)
], which is known as rank of A and denoted by r
(
A
). For any quaternion h
=
h
0
+
h
1
i
+
h
2
j
+
h
3
k, its conjugate is h
¯
=
h
0
-
h
1
i
-
h
2
j
-
h
3
k. So, A
∗ represents the conjugate transpose of A. A
† means the Moore-Penrose inverse of A
∈
H
m
×
n, i.e., the exclusive matrix Y
∈
H
n
×
m satisfying (2)
A
Y
A
=
A
,
Y
A
Y
=
Y
,
A
Y
∗
=
A
Y
,
Y
A
∗
=
Y
A
.More results on generalized inverses can be seen in [2, 3]. Furthermore, L
A
=
I
-
A
†
A and R
A
=
I
-
A
A
† are couple of the projectors induced by A, respectively. It is evident that L
A
=
L
A
∗
=
L
A
†
=
L
A
2 and R
A
=
R
A
∗
=
R
A
†
=
R
A
2.The notion of quaternions was first explored by an Irish mathematician Sir William Rowan Hamilton in [4]. Quaternions have ample use in diverse areas of mathematics like computation, geometry and algebra; see, e.g. [5–8]. The researcher in [9] used quaternion in the field of computer graphics. Very recently, quaternion matrices have secured a vital role in control theory, mechanics, altitude control, quantum physics and signal processing; see, e.g., [10, 11]. In skeletal animation systems, quaternions are mostly practiced to interpolate between joint orientations specified with key frames or animation curves [12]. The comprehensive study on quaternions can be found in [13].The Hermitian solutions to some classical matrix equations were examined by numerous researchers in various papers by adopting different approaches. For instance, Khatri and Mitra [14] and Groß in [15] computed some necessary and sufficient condition for the Hermitian solution to some significant matrix equation with their general solution when they are consistent.A matrixA
∈
H
n
×
n is said to be η-Hermitian and skew-η-Hermitian [16, 17] if A
=
A
η
∗
=
-
η
A
∗
η and A
=
-
A
η
∗
=
η
A
∗
η, where η
∈
{
i
,
j
,
k
}, respectively. Applications of η-Hermitian matrices in the fields of convergence analysis in statistical signal processing and linear modelling can be viewed in [16, 18, 19]. The singular value decomposition of the η-Hermitian matrix was examined in [17]. Very recently, a researcher in [20] determined the skew-η-Hermitian solution to some significant matrix equations including (3)
B
Y
B
η
∗
+
C
Z
C
η
∗
=
Dand gave general solution to these equation when they are consistent.He and Wang [21] gave the general solution of (4)
A
4
X
+
A
4
X
η
∗
+
B
4
Y
B
4
η
∗
+
C
4
Z
C
4
η
∗
=
D
4bearing η-Hermicity over H. An iterative algorithm for determining the η-Hermitian and η-skew-Hermitian solutions to the quaternion generalized Sylvester equation A
X
B
+
C
Y
D
=
E were established in [22]. For more related papers on η-Hermicity, one may refer to [23–28]. Notice that Sylvester-type matrix equations have a huge amount of practical applications in feedback control [29, 30], robust control [31], pole/eigenstructure assignment design [32], neural network [33], and so on.Keeping in view the latest development and the application of skew-η-Hermitian matrices in the fields like convergence analysis in statistical signal processing and linear modelling and quaternion matrices in mind, in this paper, we explore the η-skew-Hermitian solution of the equation (5)
A
X
-
A
X
η
∗
+
B
Y
B
η
∗
+
C
Z
C
η
∗
=
D
,
Y
=
-
Y
η
∗
,
Z
=
-
Z
η
∗when it is consistent. We not only obtain the explicit form of the η-skew-Hermitian solution in terms of generalized inverses but also give its explicit determinantal representation formulas that are an analog of Cramer’s rule. Our proposed Cramer’s rule is based on the theory of row-column noncommutative determinants introduced in [34], by using determinantal representations of the Moore-Penrose inverse matrix [35]. Within the framework of the theory of noncommutative row-column determinants, determinantal representations of various generalized quaternion inverses and generalized inverse solutions to quaternion matrix equations have been derived by one of the authors (see, e.g., [36–46]) and by other researchers (see, e.g., [47–51]).Observe that the system (3) is a particular case of our system (5). It is noteworthy to say that the researcher in [20] made nice effort to calculate the η-skew-Hermitian solution of (3) with its general solution but he did not represent the necessary and sufficient conditions in terms of rank equalities and we will overcome this issue in this paper as well. Moreover, we give explicit determinantal representation formulas for the solution to (5).The remaining part of this paper is composed as follows. In Section2, we start with some provisions from generalized inverses and its determinantal representations which have vital role during the construction of the main results of this paper. Necessary and sufficient conditions for the general solution (
X
,
Y
,
Z
) to (5), where Y and Z are η-skew-Hermitian, are presented in Section 3. A particular case of (5) is also examined in this section. In Section 4, we give Cramer’s rule to Eq.(5). Based on obtained explicit determinantal representation formulas, in Section 5, we present a numerical example to glorify the notion established in this paper. Finally, in Section 6, we provide a conclusion to this article.


## 2. Preliminaries. Some Provisions from Generalized Inverses and Its Determinantal Representations

We begin with some famous results which will be used in the remaining part of this paper.Lemma 1 (see [52]).
LetA
∈
H
s
×
t
,
B
∈
H
s
×
k and C
∈
H
l
×
t be known. Then (1)
r
(
A
)
+
r
(
R
A
B
)
=
r
(
B
)
+
r
(
R
B
A
)
=
r
A
B.
(2)
r
(
A
)
+
r
(
C
L
A
)
=
r
(
C
)
+
r
(
A
L
C
)
=
r
A
C.
(3)
r
(
B
)
+
r
(
C
)
+
r
(
R
B
A
L
C
)
=
r
A
B
C
0.Lemma 2 (see [21]).
LetA
∈
H
m
×
n be given. Then (1)
(
A
η
)
†
=
(
A
†
)
η
,
(
A
η
∗
)
†
=
(
A
†
)
η
∗
.
(2)
r
(
A
)
=
r
(
A
η
∗
)
=
r
(
A
η
)
=
r
(
A
η
A
η
∗
)
=
r
(
A
η
∗
A
η
)
.
(3)
(
A
†
A
)
η
∗
=
A
η
∗
(
A
†
)
η
∗
=
(
A
†
A
)
η
=
(
A
†
)
η
A
η
.
(4)
(
A
A
†
)
η
∗
=
(
A
†
)
η
∗
A
η
∗
=
(
A
A
†
)
η
=
A
η
(
A
†
)
η
.
(5)
(
L
A
)
η
∗
=
-
η
(
L
A
)
η
=
(
L
A
)
η
=
L
A
η
=
R
A
η
∗
.
(6)
(
R
A
)
η
∗
=
-
η
(
R
A
)
η
=
(
R
A
)
η
=
L
A
η
∗
=
R
A
η
.Lemma 3 (see [53]).
LetA, B and C be given matrices with right sizes over H. Then (1)
A
†
=
(
A
∗
A
)
†
A
∗
=
A
∗
(
A
A
∗
)
†
.
(2)
L
A
=
L
A
2
=
L
A
∗
,
R
A
=
R
A
2
=
R
A
∗
.
(3)
L
A
(
B
L
A
)
†
=
(
B
L
A
)
†
,
(
R
A
C
)
†
R
A
=
(
R
A
C
)
†
.Remark 4.
Since for anyη
m
∈
{
i
,
j
,
k
} for all m
=
1,2
,
3, and q
=
a
0
+
a
1
η
1
+
a
2
η
2
+
a
3
η
3, (6)
q
η
1
≔
-
η
1
q
η
1
=
a
0
+
a
1
η
1
-
a
2
η
2
-
a
3
η
3
,
q
-
η
1
≔
η
1
q
η
1
=
-
a
0
-
a
1
η
1
+
a
2
η
2
+
a
3
η
3
,then elements of the main diagonal of an η
1-Hermitian matrix A
=
A
η
1
∗
=
a
i
j
η
1
∗ must be as (7)
a
i
i
η
1
∗
=
a
0
+
a
2
η
2
+
a
3
η
3
,and a pair of elements which are symmetric with respect to the main diagonal can be represented as (8)
a
i
j
η
1
∗
=
a
0
+
a
1
η
1
+
a
2
η
2
+
a
3
η
3
,
a
j
i
η
1
∗
=
a
0
-
a
1
η
1
+
a
2
η
2
+
a
3
η
3
. Similarly, elements of the main diagonal of an η
1-skew-Hermitian matrix A
=
-
A
η
1
∗
=
a
i
j
-
η
1
,
∗ must be as (9)
a
i
i
-
η
1
,
∗
=
a
1
η
1
,and a pair of elements which are symmetric with respect to the main diagonal can be represented as (10)
a
i
j
-
η
1
∗
=
a
0
+
a
1
η
1
+
a
2
η
2
+
a
3
η
3
,
a
j
i
-
η
1
∗
=
-
a
0
+
a
1
η
1
-
a
2
η
2
-
a
3
η
3
.where a
l
∈
R for all l
=
0
,
…
,
3.ForA
∈
H
n
×
n, we define n row determinants and n column determinants. Suppose S
n is the symmetric group on the set I
n
=
{
1
,
…
,
n
}.Definition 5 (see [34]).
Theith row determinant of A
=
(
a
i
j
)
∈
H
n
×
n is defined for all i
=
1
,
…
,
n by putting (11)
r
d
e
t
i
A
=
∑
σ
∈
S
n
-
1
n
-
r
a
i
i
k
1
a
i
k
1
i
k
1
+
1
…
a
i
k
1
+
l
1
i
…
a
i
k
r
i
k
r
+
1
…
a
i
k
r
+
l
r
i
k
r
,
σ
=
i
i
k
1
i
k
1
+
1
…
i
k
1
+
l
1
i
k
2
i
k
2
+
1
…
i
k
2
+
l
2
…
i
k
r
i
k
r
+
1
…
i
k
r
+
l
r
,where σ is the left-ordered permutation. It means that its first cycle from the left starts with i, while other cycles start from the left with the minimal of all the integers which are contained in it, (12)
i
k
t
<
i
k
t
+
s
∀
t
=
2
,
…
,
r
,
s
=
1
,
…
,
l
t
,and the order of disjoint cycles (except for the first one) is strictly conditioned by increase from left to right of their first elements, i
k
2
<
i
k
3
<
⋯
<
i
k
r.Definition 6 (see [34]).
Thejth column determinant of A
=
(
a
i
j
)
∈
H
n
×
n is defined for all j
=
1
,
…
,
n by putting (13)
c
d
e
t
j
A
=
∑
τ
∈
S
n
-
1
n
-
r
a
j
k
r
j
k
r
+
l
r
…
a
j
k
r
+
1
j
k
r
…
a
j
j
k
1
+
l
1
…
a
j
k
1
+
1
j
k
1
a
j
k
1
j
,
τ
=
j
k
r
+
l
r
…
j
k
r
+
1
j
k
r
…
j
k
2
+
l
2
…
j
k
2
+
1
j
k
2
j
k
1
+
l
1
…
j
k
1
+
1
j
k
1
j
, where τ is the right-ordered permutation. It means that its first cycle from the right starts with j, while other cycles start from the right with the minimal of all the integers which are contained in it, (14)
j
k
t
<
j
k
t
+
s
∀
t
=
2
,
…
,
r
,
s
=
1
,
…
,
l
t
,and the order of disjoint cycles (except for the first one) is strictly conditioned by increase from right to left of their first elements, j
k
2
<
j
k
3
<
⋯
<
j
k
r.Leta
.
j be the jth column and a
i
. be the ith row of A. Suppose A
i
.
b denotes the matrix obtained from A by replacing its ith row with the row b, and A
.
j
c denotes the matrix obtained from A by replacing its jth column with the column c.We state some properties of row-column determinants needed below.Lemma 7 (see [34]).
If theith row of A
∈
H
n
×
n is a left linear combination of other row vectors, i.e., a
i
.
=
α
1
b
1
+
⋯
+
α
k
b
k, where α
l
∈
H and b
l
∈
H
1
×
n for all l
=
1
,
…
,
k and i
=
1
,
…
,
n, then r
d
e
t
i
A
i
.
α
1
b
1
+
⋯
+
α
k
b
k
=
∑
l
α
l
r
d
e
t
i
A
i
.
b
l
.Lemma 8 (see [34]).
If thejth column of A
∈
H
m
×
n is a right linear combination of other column vectors, i.e., A
.
j
=
c
1
α
1
+
⋯
+
c
k
α
k, where α
l
∈
H and c
l
∈
H
n
×
1 for all l
=
1
,
…
,
k and j
=
1
,
…
,
n, then c
d
e
t
j
A
.
j
c
1
α
1
+
⋯
+
c
k
α
k
=
∑
l
c
d
e
t
j
A
.
j
c
l
α
l
.Lemma 9 (see [34]).
LetA
∈
H
n
×
n. Then c
d
e
t
i
A
∗
=
r
d
e
t
i
A
¯, r
d
e
t
i
A
∗
=
c
d
e
t
i
A
¯ for all i
=
1
,
…
,
n.Since by Definitions5 and 6 for A
∈
H
n
×
n
(15)
r
d
e
t
i
A
η
=
r
d
e
t
i
-
η
A
η
=
-
η
r
d
e
t
i
A
η
,
c
d
e
t
i
A
η
=
c
d
e
t
i
-
η
A
η
=
-
η
c
d
e
t
i
A
η
,
r
d
e
t
i
-
A
η
=
r
d
e
t
i
η
A
η
=
-
1
n
-
1
η
r
d
e
t
i
A
η
,
c
d
e
t
i
-
A
η
=
c
d
e
t
i
η
A
η
=
-
1
n
-
1
η
c
d
e
t
i
A
ηfor all i
=
1
,
…
,
n, then, due to Lemma 9, the next lemma follows immediately.Lemma 10.
LetA
∈
H
n
×
n. Then (16)
r
d
e
t
i
A
η
∗
=
-
η
c
d
e
t
i
A
¯
η
,
c
d
e
t
i
A
η
∗
=
-
η
r
d
e
t
i
A
¯
η
,for all i
=
1
,
…
,
nSo, Lemma10 gives the following features of η-Hermitian and η-skew-Hermitian matrices.Lemma 11.
IfA
∈
H
n
×
n is η-Hermitian, then (17)
r
d
e
t
i
A
=
-
η
c
d
e
t
i
A
¯
η
,
c
d
e
t
i
A
=
-
η
r
d
e
t
i
A
¯
η
.If A
∈
H
n
×
n is η-skew-Hermitian, then (18)
r
d
e
t
i
A
=
-
1
n
-
1
η
c
d
e
t
i
A
¯
η
,
c
d
e
t
i
A
=
-
1
n
-
1
η
r
d
e
t
i
A
¯
ηfor all i
=
1
,
…
,
n.Remark 12.
Since [34] for Hermitian A we have (19)
r
d
e
t
1
A
=
⋯
=
r
d
e
t
n
A
=
c
d
e
t
1
A
=
⋯
=
c
d
e
t
n
A
∈
R
,the determinant of a Hermitian matrix is defined by putting (20)
d
e
t
A
≔
r
d
e
t
i
A
=
c
d
e
t
i
Afor all i
=
1
,
…
,
n.Properties of the determinant of a Hermitian matrix are similar to the properties of an usual (commutative) determinant and they have been completely explored by row-column determinants in [34].The following notation will be used for determinantal representations of the Moore-Penrose inverse. Letα
≔
α
1
,
…
,
α
k
⊆
1
,
…
,
m and β
≔
β
1
,
…
,
β
k
⊆
1
,
…
,
n be subsets with 1
≤
k
≤
m
i
n
m
,
n. Let A
β
α be a submatrix of A
∈
H
m
×
n whose rows and columns are indexed by α and β, respectively. Similarly, let A
α
α be a principal submatrix of A whose rows and columns are indexed by α. If A is Hermitian, then A
α
α denotes the corresponding principal minor of d
e
t
A. The collection of strictly increasing sequences of k integers chosen from 1
,
…
,
n is denoted by (21)
L
k
,
n
≔
α
:
α
=
α
1
,
…
,
α
k
,
1
≤
α
1
<
⋯
<
α
k
≤
nfor 1
≤
k
≤
n. For fixed i
∈
α and j
∈
β, let I
r
,
m
i
≔
α
:
α
∈
L
r
,
m
,
i
∈
α, J
r
,
n
j
≔
β
:
β
∈
L
r
,
n
,
j
∈
β.Denote thejth column and the ith row of A
∗ by a
.
j
∗ and a
i
.
∗, respectively.Theorem 13 (see [35]).
IfA
∈
H
r
m
×
n, then the Moore-Penrose inverse A
†
=
a
i
j
†
∈
H
n
×
m have the following determinantal representations (22)
a
i
j
†
=
∑
β
∈
J
r
,
n
i
c
d
e
t
i
A
∗
A
.
i
a
.
j
∗
β
β
∑
β
∈
J
r
,
n
A
∗
A
β
β
,or (23)
a
i
j
†
=
∑
α
∈
I
r
,
m
j
r
d
e
t
j
A
A
∗
j
.
a
i
.
∗
α
α
∑
α
∈
I
r
,
m
A
A
∗
α
α
.Remark 14.
Note that for an arbitrary full-rank matrixA
∈
H
r
m
×
n, a row-vector b
∈
H
1
×
m, and a column-vector c
∈
H
n
×
1 we put, respectively, (i)
ifr
=
m, then (24)
r
d
e
t
i
A
A
∗
i
.
B
=
∑
α
∈
I
m
,
m
i
r
d
e
t
i
A
A
∗
i
.
b
α
α
,
d
e
t
A
A
∗
=
∑
α
∈
I
m
,
m
A
A
∗
α
α
for alli
=
1
,
…
,
m;
(ii)
ifr
=
n, then (25)
c
d
e
t
j
A
∗
A
.
j
c
=
∑
β
∈
J
n
,
n
j
c
d
e
t
j
A
∗
A
.
j
c
β
β
,
d
e
t
A
∗
A
=
∑
β
∈
J
n
,
n
A
∗
A
β
β
for allj
=
1
,
…
,
n.Corollary 15.
IfA
∈
H
r
m
×
n, then the Moore-Penrose inverse A
η
†
=
a
i
j
η
†
∈
H
n
×
m, (
A
∗
)
†
=
(
a
i
j
∗
)
†
∈
H
m
×
n, (
A
η
∗
)
†
=
(
a
i
j
η
∗
)
†
∈
H
m
×
n and (
-
A
η
∗
)
†
=
(
a
i
j
-
η
∗
)
† have the following determinantal representations, respectively,(26)
a
i
j
η
†
=
-
η
∑
β
∈
J
r
,
n
i
c
d
e
t
i
A
∗
A
.
i
A
.
j
∗
β
β
∑
β
∈
J
r
,
n
A
∗
A
β
β
η
(27)
=
-
η
∑
α
∈
I
r
,
m
j
r
d
e
t
j
A
A
∗
j
.
A
i
.
∗
α
α
∑
α
∈
I
r
,
m
A
A
∗
α
α
η
,
(28)
a
i
j
∗
†
=
a
j
i
†
¯
=
∑
α
∈
I
r
,
n
j
r
d
e
t
j
A
∗
A
j
.
A
i
.
α
α
∑
β
∈
I
r
,
n
A
∗
A
α
α
(29)
=
∑
β
∈
J
r
,
m
i
c
d
e
t
i
A
A
∗
.
i
A
.
j
β
β
∑
β
∈
J
r
,
m
A
A
∗
β
β
.
(30)
a
i
j
η
∗
†
=
-
η
a
j
i
†
¯
η
=
-
η
∑
α
∈
I
r
,
n
j
r
d
e
t
j
A
∗
A
j
.
A
i
.
α
α
∑
β
∈
I
r
,
n
A
∗
A
α
α
η
(31)
=
-
η
∑
β
∈
J
r
,
m
i
c
d
e
t
i
A
A
∗
.
i
A
.
j
β
β
∑
β
∈
J
r
,
m
A
A
∗
β
β
η
,
(32)
a
i
j
-
η
∗
†
=
η
a
j
i
†
¯
η
=
η
∑
α
∈
I
r
,
n
j
r
d
e
t
j
A
∗
A
j
.
A
i
.
α
α
∑
β
∈
I
r
,
n
A
∗
A
α
α
η
=
η
∑
β
∈
J
r
,
m
i
c
d
e
t
i
A
A
∗
.
i
A
.
j
β
β
∑
β
∈
J
r
,
m
A
A
∗
β
β
η
.Corollary 16 (see [35]).
IfA
∈
H
r
m
×
n, then the projection matrix A
†
A
≕
Q
A
=
q
i
j
n
×
n has the determinantal representation (33)
q
i
j
=
∑
β
∈
J
r
,
n
i
c
d
e
t
i
A
∗
A
.
i
a
˙
.
j
β
β
∑
β
∈
J
r
,
n
A
∗
A
β
β
=
∑
α
∈
I
r
,
n
j
r
d
e
t
j
A
∗
A
.
j
a
˙
.
i
α
α
∑
α
∈
I
r
,
n
A
∗
A
α
α
,where a
˙
.
j and a
˙
.
i are the jth column and the ith row of A
∗
A
∈
H
n
×
n.Corollary 17 (see [35]).
IfA
∈
H
r
m
×
n, then the projection matrix A
A
†
≕
P
A
=
p
i
j
m
×
m has the determinantal representation (34)
p
i
j
=
∑
α
∈
I
r
,
m
j
r
d
e
t
j
A
A
∗
j
.
a
¨
i
.
α
α
∑
α
∈
I
r
,
m
A
A
∗
α
α
=
∑
β
∈
J
r
,
m
i
c
d
e
t
i
A
A
∗
.
i
a
¨
.
j
β
β
∑
β
∈
J
r
,
m
A
A
∗
β
β
, where a
¨
i
. and a
¨
.
j are the ith row and the jth column of A
A
∗
∈
H
m
×
m.Lemma 18 (see [54]).
LetA
∈
H
m
×
n, B
∈
H
r
×
s, C
∈
H
m
×
s be known and X
∈
H
n
×
r be unknown. Then the matrix equation (35)
A
X
B
=
Cis consistent if and only if A
A
†
C
B
†
B
=
C. In this case, its general solution can be expressed as (36)
X
=
A
†
C
B
†
+
L
A
V
+
W
R
B
,where V, W are arbitrary matrices over H with appropriate dimensions.Lemma 19 (see [36]).
LetA
∈
H
r
1
m
×
n, B
∈
H
r
2
r
×
s. Then the partial solution X
=
A
†
C
B
†
=
(
x
i
j
)
∈
H
n
×
r to (35) have determinantal representations, (37)
x
i
j
=
∑
β
∈
J
r
1
,
n
i
c
d
e
t
i
A
∗
A
.
i
d
.
j
B
β
β
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
∑
α
∈
I
r
2
,
r
B
B
∗
α
α
=
∑
α
∈
I
r
2
,
r
j
r
d
e
t
j
B
B
∗
j
.
d
i
.
A
α
α
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
∑
α
∈
I
r
2
,
r
B
B
∗
α
α
,where (38)
d
.
j
B
=
∑
α
∈
I
r
2
,
r
j
r
d
e
t
j
B
B
∗
j
.
c
~
k
.
α
α
∈
H
n
×
1
,
k
=
1
,
…
,
n
,
d
i
.
A
=
∑
β
∈
J
r
1
,
n
i
c
d
e
t
i
A
∗
A
.
i
c
~
.
l
β
β
∈
H
1
×
r
,
l
=
1
,
…
,
r
,are the column vector and the row vector, respectively. c
~
i
. and c
~
.
j are the ith row and the jth column of C
~
=
A
∗
C
B
∗.Corollary 20 (see [36]).
LetA
∈
H
k
m
×
n, C
∈
H
m
×
s be known and X
∈
H
n
×
s be unknown. Then A
X
=
C is consistent if and only if A
A
†
C
=
C. In this case, its general solution can be expressed as X
=
A
†
C
+
L
A
V, where V is an arbitrary matrix over H with appropriate dimensions. The partial solution X
=
A
†
C has the following determinantal representation, (39)
x
i
j
=
∑
β
∈
J
k
,
n
i
c
d
e
t
i
A
∗
A
.
i
c
^
.
j
β
β
∑
β
∈
J
k
,
n
A
∗
A
β
β
.where c
^
.
j is the jth column of C
^
=
A
∗
C.Corollary 21 (see [36]).
LetB
∈
H
k
r
×
s, C
∈
H
n
×
s be given, and X
∈
H
n
×
r be unknown. Then X
B
=
C is solvable if and only if C
=
C
B
†
B and its general solution is X
=
C
B
†
+
W
R
B, where W is a any matrix with conformable dimension. Moreover, its partial solution X
=
C
B
† has the determinantal representation, (40)
x
i
j
=
∑
α
∈
I
k
,
r
j
r
d
e
t
j
B
B
∗
j
.
c
^
i
.
α
α
∑
α
∈
I
k
,
r
B
B
∗
α
α
.where c
^
i
. is the ith row of C
^
=
C
B
∗.


## 3. Some Solvability Conditions and the General Solution to (5)

In this section, we provide some necessary and sufficient conditions for the system (5) to have a solution (
X
,
Y
,
Z
), where Y
=
-
Y
η
∗ and Z
=
-
Z
η
∗
. Furthermore, its general solution is also constructed when the solvability conditions are accomplished.Theorem 22.
LetA
,
B
,
C and D
=
-
D
η
∗ be given coefficient matrices in (5) over H with conformable sizes. Denote (41)
A
1
=
R
A
B
,
B
1
=
R
A
C
,
C
1
=
R
A
D
R
A
η
∗
,
M
=
R
A
1
B
1
,
S
=
B
1
L
M
.Then (1)
The system (5) has a solution (
X
,
Y
,
Z
), where Y
=
-
Y
η
∗ and Z
=
-
Z
η
∗.
(2)
The coefficient matrices in (5) satisfy (42)
R
M
R
A
1
C
1
=
0
,
R
A
1
C
1
R
B
1
η
∗
=
0
.
(3)
M
M
†
R
A
1
C
1
=
R
A
1
C
1
=
R
A
1
C
1
(
B
1
†
)
η
∗
B
1
η
∗
.
(4)
(43)
r
D
C
B
A
A
η
∗
0
0
0
=
r
C
B
A
+
r
A
,
r
D
B
A
A
η
∗
0
0
C
η
∗
0
0
=
r
B
A
+
r
A
C
.are equivalent statements. Under these conditions, the general solution to the system (5) can be demonstrated as (44)
Y
=
-
Y
η
∗
=
A
1
†
C
1
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
C
1
I
+
B
1
†
η
∗
S
η
∗
A
1
†
η
∗
-
1
2
A
1
†
I
+
S
B
1
†
C
1
M
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
A
1
†
S
W
2
S
η
∗
A
1
†
η
∗
-
L
A
1
U
+
U
η
∗
L
A
1
η
∗
,
(45)
Z
=
-
Z
η
∗
=
1
2
M
†
C
1
B
1
†
η
∗
I
+
S
†
S
η
+
1
2
I
+
S
†
S
B
1
†
C
1
M
†
η
∗
+
L
M
W
2
L
M
η
∗
-
V
L
B
1
η
∗
+
L
B
1
V
η
∗
+
L
M
L
S
W
1
-
W
1
η
∗
L
M
L
S
η
∗
,
(46)
X
=
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
-
1
2
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
-
L
A
U
1
+
U
2
η
∗
A
†
η
∗
A
η
∗
+
A
†
U
2
A
η
∗
,where U
1
,
U
2
,
W
1
,
U
,
V and W
2
η
∗
=
-
W
2 are arbitrary matrices over H.Proof.
Obviously, (2)⇔(3).
Now we show (2)⇔ (4). By means of Lemmas 1 and 2, we have (47)
R
M
R
A
1
C
1
=
0
⇔
r
R
M
R
A
1
C
=
0
⇔
r
M
R
A
1
C
=
r
M
⇔
r
A
1
B
1
C
1
=
r
A
1
B
1
⇔
r
R
A
B
R
A
D
R
A
η
∗
R
A
C
=
r
R
A
B
R
A
C
⇔
r
D
B
A
C
A
η
∗
0
0
0
=
r
B
C
A
+
r
A
.Similarly, (48)
R
A
1
C
1
L
B
η
∗
=
0
⇔
r
C
1
A
1
B
1
η
∗
0
=
r
A
1
+
r
B
1
⇔
r
R
A
D
L
A
η
∗
R
A
B
C
η
∗
L
A
η
∗
0
=
r
R
A
B
+
r
R
A
C
⇔
r
D
B
A
C
η
∗
0
0
A
η
∗
0
0
=
r
A
B
+
r
A
C
.Now we show (1)⇒
(
2
): If the system (5) has a solution (
X
,
Y
,
Z
), where Y
=
-
Y
η
∗ and Z
=
-
Z
η
∗. Then (49)
A
X
-
A
X
η
∗
=
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
⇒
R
A
A
X
-
A
X
η
∗
R
A
η
∗
=
R
A
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
R
A
η
∗
⇒
A
1
Y
A
1
η
∗
+
B
1
Z
B
1
η
∗
=
C
1
⇒
R
A
1
A
1
Y
A
1
η
∗
+
B
1
Z
B
1
η
∗
R
B
1
η
∗
=
R
A
1
C
1
R
B
1
η
∗
⇒
R
A
1
C
1
R
B
1
η
∗
=
0
,
R
A
1
B
1
Z
B
1
η
∗
=
R
A
1
C
1
⇒
M
Z
B
1
η
∗
=
R
A
1
C
1
⇒
R
M
R
A
1
C
1
=
0
.
(2)⇒(1): Now we prove that (
Y
,
Z
,
X
) mentioned in (44)–(46), respectively, is a solution of (5) under (42). Obviously, Y and Z represented in (44) and (45) are η-skew-Hermitian. Now putting X
,
Y and Z in the following expression and simplifying, we have (50)
A
X
-
A
X
η
∗
+
B
Y
B
η
∗
+
C
Z
C
η
∗
=
A
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
-
1
2
A
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
+
A
U
2
η
∗
A
†
η
∗
A
η
∗
+
A
A
†
U
2
A
η
∗
+
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
+
B
Y
B
η
∗
+
C
Z
C
η
∗
-
1
2
A
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
-
A
A
†
U
2
A
η
∗
-
A
U
2
η
∗
A
†
η
∗
A
η
∗
=
A
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
-
A
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
+
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
+
B
Y
B
η
∗
+
C
Z
C
η
∗
=
A
A
†
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
+
R
A
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
A
†
η
∗
A
η
∗
=
D
-
R
A
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗
R
A
η
∗
=
D
-
C
1
-
A
1
Y
A
1
η
∗
-
B
1
Z
B
1
η
∗
.Now we show that C
-
A
1
Y
A
1
η
∗
-
B
1
Z
B
1
η
∗
=
0 in (50). By Lemma 3, equation (42) and B
1
-
S
=
B
1
M
†
M, we have (51)
M
†
R
A
1
=
M
†
,
B
1
S
†
S
-
S
=
B
1
-
S
S
†
S
=
B
1
M
†
M
S
∗
S
†
∗
=
0
,
R
A
1
C
1
=
R
A
1
C
1
B
1
†
η
∗
B
1
η
∗
.By using (51), Y can be written as follows: (52)
Y
=
A
1
†
C
1
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
C
1
I
+
B
1
†
η
∗
S
η
∗
A
1
†
η
∗
-
1
2
A
1
†
I
+
S
B
1
†
C
1
M
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
A
1
†
S
W
2
S
η
∗
A
†
η
∗
+
L
A
U
-
U
η
∗
L
A
1
η
∗
=
A
1
†
C
1
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
C
1
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
C
1
B
1
†
η
∗
S
η
∗
A
1
†
η
∗
-
1
2
A
1
†
C
1
M
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
1
2
A
1
†
S
B
1
†
C
1
M
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
A
1
†
S
W
2
S
η
∗
A
1
†
η
∗
-
L
A
1
U
+
U
η
∗
L
A
1
η
∗
=
A
1
†
C
1
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
R
A
1
C
1
B
1
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
R
A
1
C
1
×
B
1
†
η
∗
S
†
S
η
∗
B
1
η
∗
A
1
†
η
∗
-
1
2
A
†
B
1
B
1
†
C
1
R
A
1
η
∗
M
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
1
2
A
1
†
B
1
S
†
S
B
1
†
C
1
M
†
η
∗
B
1
η
∗
A
1
†
η
∗
-
A
1
†
S
W
2
S
η
∗
A
1
†
η
∗
-
L
A
1
U
+
U
η
∗
L
A
1
η
∗
,i.e., (53)
Y
=
A
1
†
C
1
-
B
1
Z
B
1
η
∗
A
1
†
η
∗
-
L
A
1
U
+
U
η
∗
L
A
1
η
∗
.By using (53) and M
=
R
A
1
B
1, we can write (54)
A
1
Y
A
1
η
∗
+
B
1
Z
B
1
η
∗
=
A
1
A
1
†
C
1
-
B
1
Z
B
1
η
∗
A
1
†
η
∗
A
1
η
∗
+
B
1
Z
B
1
η
∗
=
A
1
A
1
†
C
1
A
1
†
η
∗
A
1
η
∗
-
A
1
A
1
†
B
1
Z
B
1
η
∗
A
1
†
η
∗
A
1
η
∗
+
B
1
Z
B
1
η
∗
=
A
1
A
1
†
C
1
A
1
†
η
∗
A
1
η
∗
-
I
-
R
A
1
B
1
Z
B
1
η
∗
A
†
η
∗
A
η
∗
+
B
1
Z
B
1
η
∗
=
A
1
A
1
†
C
1
A
1
†
η
∗
A
1
η
∗
+
B
1
Z
M
η
∗
+
M
Z
B
1
η
∗
A
1
†
η
∗
A
1
η
∗
.By using M
=
R
A
1
B
1
,
B
1
M
†
M
=
B
1
-
S
,
M
S
∗
=
0, (42) and (51), the term M
Z
B
1
η
∗ in (54) can be expressed as (55)
M
Z
B
1
η
∗
=
1
2
M
M
†
C
1
B
1
B
1
†
η
∗
+
M
B
1
†
C
1
B
1
M
†
η
∗
+
M
M
†
C
1
B
1
†
η
∗
S
†
S
η
∗
B
1
η
∗
=
1
2
M
M
†
R
A
1
C
1
B
1
B
1
†
η
∗
+
M
B
1
†
C
1
R
A
1
η
∗
B
1
M
†
η
∗
+
M
M
†
R
A
1
C
1
B
1
†
η
∗
S
†
S
η
∗
B
1
η
∗
=
1
2
R
A
1
C
1
+
R
A
1
B
1
B
1
†
C
1
R
A
1
η
∗
M
†
η
∗
B
1
η
∗
+
M
M
†
R
A
1
C
1
B
1
†
η
∗
S
η
∗
=
1
2
R
A
1
C
1
+
R
A
1
C
1
L
A
1
η
∗
M
†
η
∗
B
1
η
∗
+
R
A
1
C
1
B
1
†
η
∗
S
η
∗
=
1
2
R
A
1
C
1
+
R
A
1
C
1
B
1
†
η
∗
B
1
η
∗
L
A
1
η
∗
M
†
η
∗
B
1
η
∗
+
R
A
1
C
1
B
1
†
η
∗
S
η
∗
=
1
2
R
A
1
C
1
+
R
A
1
C
1
B
1
†
η
∗
M
η
∗
M
†
η
∗
B
1
η
∗
+
R
A
1
C
1
B
1
†
η
∗
S
η
∗
=
1
2
R
A
1
C
1
+
R
A
1
C
1
B
1
†
η
∗
B
1
η
∗
-
S
η
∗
+
R
A
1
C
1
B
1
†
η
∗
S
η
∗
=
1
2
R
A
1
C
1
+
R
A
1
C
1
-
R
A
1
C
1
B
1
†
η
∗
S
η
∗
+
R
A
1
C
1
B
1
†
η
∗
S
η
∗
=
R
A
1
C
1
.
By putting equation (55) in equation (54), we have (56)
A
1
Y
A
1
η
∗
+
B
1
Z
B
1
η
∗
=
A
1
A
1
†
C
1
A
1
†
η
∗
A
1
η
∗
-
C
1
L
A
η
∗
+
R
A
1
C
1
A
1
†
η
∗
A
1
η
∗
=
A
1
A
1
†
C
1
A
1
†
η
∗
A
1
η
∗
+
C
1
-
C
1
A
1
†
η
∗
A
1
η
∗
+
C
1
A
1
†
η
∗
A
1
η
∗
-
A
1
A
1
†
C
1
A
1
†
η
∗
A
1
η
∗
=
C
1
.Hence, the term in the parenthesis in (50) becomes zero and we have (57)
A
X
-
A
X
η
∗
+
B
Y
B
η
∗
+
C
Z
C
η
∗
=
D
.Consequently, (
X
,
Y
,
Z
) mentioned in (44)–(46) is a solution of (5).
Next, we have to show that any solution(
X
,
Y
,
Z
), where Y and Z are η-skew-Hermitian matrices, of the system (5) can be expressed by (44)–(46). Assume that (
X
0
,
Y
0
,
Z
0
), where Y
0
=
-
Y
0
η
∗ and Z
0
=
-
Z
0
η
∗, be an arbitrary solution of (58)
A
X
-
A
X
η
∗
+
B
Y
B
η
∗
+
C
Z
C
η
∗
=
D
.For this ambition, we show that an arbitrary solution (
X
0
,
Y
0
,
Z
0
) of (5) can be written like (44)–(46).
Note that(59)
A
1
Y
0
A
1
η
∗
+
B
1
Z
0
B
1
η
∗
=
C
1
.From (59), we acquire that (60)
M
Z
0
B
1
η
∗
=
R
A
1
C
1
,
M
Z
0
M
η
∗
=
R
A
1
C
1
L
A
1
η
∗
.Put (61)
U
1
=
X
0
,
U
2
=
1
2
X
0
η
∗
A
1
η
∗
A
1
†
η
∗
,
(62)
W
1
=
1
2
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
=
1
2
Z
0
M
η
∗
M
†
η
∗
,
V
=
1
2
Z
0
,
(63)
W
2
=
1
2
Z
0
B
1
η
∗
B
1
†
η
∗
-
B
1
†
B
1
Z
0
,
U
=
1
2
-
Y
0
-
Y
0
A
1
η
∗
A
1
†
η
∗
.With the help of (62)-(63), Z
-
Z
0 can be written as follows: (64)
Z
-
Z
0
=
1
2
M
†
C
1
B
1
†
η
∗
+
B
1
†
C
1
M
†
η
∗
+
M
†
C
1
B
1
†
η
∗
S
†
S
η
∗
+
S
†
S
B
1
†
C
1
M
†
η
∗
+
L
M
L
S
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
+
L
M
Z
0
B
1
η
∗
B
1
†
η
∗
L
M
η
∗
+
Z
0
L
B
1
η
∗
-
L
B
1
Z
0
-
M
†
M
B
1
†
B
1
Z
0
L
S
η
∗
L
M
η
∗
-
L
M
B
1
†
B
1
Z
0
L
M
η
∗
-
2
Z
0
.To show Z
-
Z
0
=
0, we examine the term present in “+
L
M
L
S
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗+L
M
Z
0
B
1
η
∗
B
1
†
η
∗
L
M
η
∗
+
Z
0
L
B
1
η
∗” of (64). Since M
S
∗
=
0 and M
Z
0
B
1
η
∗
=
R
A
1
C
1, we have (65)
L
M
L
S
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
+
L
M
Z
0
B
1
η
∗
B
1
†
η
∗
L
M
η
∗
+
Z
0
L
B
1
η
∗
=
L
M
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
-
L
M
S
†
S
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
+
L
M
Z
0
B
1
η
∗
B
1
†
η
∗
L
M
η
∗
+
Z
0
L
B
1
η
∗
=
L
M
Z
0
B
1
η
∗
B
†
η
∗
-
L
M
S
†
S
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
+
Z
0
L
B
1
η
∗
=
Z
0
B
1
η
∗
B
1
†
η
∗
-
M
†
M
Z
0
B
1
η
∗
B
1
†
η
∗
-
S
†
S
Z
0
B
1
η
∗
B
1
†
η
∗
M
η
∗
M
†
η
∗
+
Z
0
L
B
1
η
∗
=
Z
0
B
1
η
∗
B
1
†
η
∗
-
M
†
R
A
1
C
1
B
1
†
η
∗
-
S
†
S
Z
0
M
η
∗
M
†
η
∗
+
Z
0
L
B
1
η
∗
=
Z
0
-
M
†
C
1
B
1
†
η
∗
-
S
†
S
Z
0
M
η
∗
M
†
η
∗
.The last term of (65) can be computed by using (60) as follows: (66)
S
†
S
Z
0
M
η
∗
M
†
η
∗
=
S
†
B
1
L
M
Z
0
M
η
∗
M
†
η
∗
=
S
†
B
1
Z
0
M
η
∗
M
†
η
∗
-
S
†
B
1
M
†
M
Z
0
M
η
∗
M
†
η
∗
=
S
†
C
1
L
A
1
η
∗
M
†
η
∗
-
S
†
B
1
M
†
R
A
1
C
1
L
A
1
η
∗
M
†
η
∗
=
S
†
C
1
M
†
R
A
1
η
∗
-
S
†
B
1
M
†
R
A
1
C
1
M
†
R
A
1
η
∗
=
S
†
C
1
M
†
η
∗
-
S
†
B
1
M
†
C
1
M
†
η
∗
.The 2nd term in the last line of (66) can be written with the help of (51) as follows: (67)
S
†
B
M
†
C
1
M
†
η
∗
=
S
†
B
1
M
†
R
A
1
C
1
L
A
1
η
∗
M
†
η
∗
=
S
†
B
1
M
†
R
A
1
B
1
B
1
†
C
1
L
A
1
η
∗
M
†
η
∗
=
S
†
B
1
-
S
B
1
†
C
1
L
A
1
η
∗
M
†
η
∗
=
S
†
B
1
B
1
†
C
1
L
A
1
η
∗
M
†
η
∗
-
S
†
S
B
1
†
C
1
L
A
1
η
∗
M
†
η
∗
=
S
†
C
1
L
A
1
η
∗
M
†
η
∗
-
S
†
S
B
1
†
C
1
M
†
R
A
1
η
∗
=
S
†
C
1
M
†
R
A
1
η
∗
-
S
†
S
B
1
†
C
1
M
†
R
A
1
η
∗
=
S
†
C
1
M
†
η
∗
-
S
†
S
B
1
†
C
1
M
†
η
∗
.With the help of (65)–(67) and (64), we get Z
-
Z
0
=
0
⇒
W
=
Z
0
.
By usingZ
=
Z
0, (59) and (63) in (53), we obtain(68)
Y
-
Y
0
=
A
1
†
C
1
-
B
1
Z
0
B
1
η
∗
A
1
†
η
∗
+
1
2
L
A
1
Y
0
+
Y
0
A
1
†
A
1
η
∗
+
1
2
Y
0
+
A
1
†
A
1
Y
0
L
A
1
η
∗
-
Y
0
=
A
1
†
A
1
Y
0
A
1
†
A
1
η
∗
+
Y
0
2
A
1
†
η
∗
Y
0
2
-
1
2
A
1
†
A
1
Y
0
A
1
†
A
1
η
∗
-
1
2
A
1
†
A
1
Y
0
+
1
2
A
1
†
A
1
Y
0
+
Y
0
2
-
1
2
A
1
†
A
1
Y
0
A
1
†
A
1
η
∗
-
Y
0
2
A
1
†
A
1
η
∗
=
0
⇒
Y
=
Y
0
.Following the same pattern, we have (69)
X
-
X
0
=
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
-
1
2
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
A
†
η
∗
A
η
∗
-
1
2
A
†
A
X
0
-
1
2
A
†
A
X
0
+
1
2
A
†
A
X
0
A
†
η
∗
A
η
∗
+
1
2
A
†
X
0
η
∗
A
η
∗
=
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
-
1
2
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
A
†
η
∗
A
η
∗
-
1
2
A
†
A
X
0
L
A
η
∗
-
1
2
A
†
A
X
0
-
X
0
η
∗
A
η
∗
=
1
2
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
-
1
2
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
A
†
η
∗
A
η
∗
-
1
2
A
†
D
-
B
Y
0
B
η
∗
-
C
Z
0
C
η
∗
L
A
η
∗
=
0
⇒
X
=
X
0
.Hence the general solution of the equation (5) can be expressed by (44)–(46). Thus the proof is done.Now we discuss a particular case of (5). If A is zero in (5) then we obtain the following renowned result.Corollary 23.
LetB, C and D
=
-
D
η
∗ be given coefficient matrices in (3) over H with conformable sizes. Denote (70)
M
=
R
B
C
,
S
=
C
L
M
.Then (1)
The system (3) has a solution (
Y
,
Z
), where Y
=
-
Y
η
∗ and Z
=
-
Z
η
∗.
(2)
The coefficient matrices in (3) satisfy (71)
R
M
R
B
D
=
0
,
R
B
D
R
C
η
∗
=
0
.
(3)
(72)
M
M
†
R
B
D
=
R
B
D
=
R
B
D
C
†
η
∗
C
η
∗
.
(4)
(73)
r
D
B
C
η
∗
0
=
r
B
+
r
C
,
r
B
C
D
=
r
B
Care equivalent statements. Under these conditions, the general solution to the system (5) can be demonstrated as (74)
Y
=
-
Y
η
∗
=
B
†
D
B
†
η
∗
-
1
2
B
†
C
M
†
D
I
+
C
†
η
∗
S
η
∗
B
†
η
∗
-
1
2
B
†
I
+
S
C
†
D
M
†
η
∗
C
η
∗
B
†
η
∗
-
B
†
S
W
2
S
η
∗
B
†
η
∗
-
L
B
U
+
U
η
∗
L
B
η
∗
,
Z
=
-
Z
η
∗
=
1
2
M
†
D
C
†
η
∗
I
+
S
†
S
η
+
1
2
I
+
S
†
S
C
†
D
M
†
η
∗
+
L
M
W
2
L
M
η
∗
+
V
L
C
η
∗
-
L
C
V
η
∗
+
L
M
L
S
W
1
-
W
1
η
∗
L
M
L
S
η
∗
,where U
1, U
2, W
1, U, V and W
2
η
∗
=
-
W
2 are arbitrary matrices over H


## 4. Cramer’s Rule of the Equation (5)

DenoteD
1
≔
D
-
B
Y
B
η
∗
-
C
Z
C
η
∗. By putting U
1
,
U
2
,
U
,
V
,
W
1, and W
2 as zero-matrices with compatible dimensions in (44)–(46), we obtain the following partial solution to (5): (75)
Y
=
A
1
†
C
1
A
1
†
η
∗
-
1
2
A
1
†
B
1
M
†
C
1
A
1
†
η
∗
-
A
1
†
B
1
M
†
C
1
A
1
†
η
∗
η
∗
-
1
2
A
1
†
S
B
1
†
C
1
A
1
†
B
1
M
†
η
∗
-
A
1
†
S
B
1
†
C
1
A
1
†
B
1
M
†
η
∗
η
∗
,
(76)
Z
=
1
2
M
†
C
1
B
1
†
η
∗
-
M
†
C
1
B
1
†
η
∗
η
∗
+
1
2
Q
S
B
1
†
C
1
M
†
η
∗
-
Q
S
B
1
†
C
1
M
†
η
∗
η
∗
,
(77)
X
=
A
†
D
1
-
1
2
A
†
D
1
P
S
η
.The following theorem gives explicit determinantal representation formulas of (75)–(77).Theorem 24.
LetA
∈
H
r
1
m
×
n, B
∈
H
r
2
m
×
k, C
∈
H
r
2
m
×
p, r
a
n
k
A
1
=
r
4, r
a
n
k
B
1
=
r
5
r
a
n
k
M
=
r
6, r
a
n
k
S
=
r
7. Then the partial solution (75)–(77) to Eq.(5), Y
=
y
i
j
∈
H
k
×
k, Z
=
z
f
g
∈
H
p
×
p, X
=
x
s
t
∈
H
m
×
n by the components (78)
y
i
j
=
y
i
j
1
-
1
2
y
i
j
2
+
η
y
j
i
2
¯
η
-
1
2
y
i
j
3
+
η
y
j
i
3
¯
η
,
(79)
z
f
g
=
1
2
z
f
g
1
+
η
z
g
f
1
¯
η
+
1
2
z
f
g
2
+
η
z
g
f
2
¯
η
,
(80)
x
s
t
=
x
s
t
1
-
1
2
x
s
t
2possesses the following determinantal representations: (i)
(81)
y
i
j
1
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
v
i
.
1
,
η
α
α
η
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
(82)
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
u
.
j
1
β
β
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
,
where(83)
v
i
.
1
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
a
^
.
l
β
β
∈
H
1
×
m
,
l
=
1
,
…
,
m
,
(84)
u
.
j
1
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
a
^
t
.
η
α
α
η
∈
H
m
×
1
,
t
=
1
,
…
,
m
,
are the row vector and the column vector,a
^
.
s is the sth column of A
^
=
A
1
∗
C
1
A
1
η and a
^
l
.
η is the lth row of A
^
η
=
A
1
η
∗
C
1
η
A
1.
(ii)
(85)
y
i
j
2
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
ϕ
~
i
.
α
α
η
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
∑
α
∈
I
r
6
,
m
M
M
∗
α
α
,
whereϕ
~
i
. is the ith row of Φ
~
≔
Φ
η
C
1
η
A
1 and Φ
=
(
ϕ
i
q
) is such that (86)
ϕ
i
q
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
φ
.
q
M
β
β
=
∑
α
∈
I
r
6
,
m
q
r
d
e
t
q
M
M
∗
q
.
φ
i
.
A
α
α
,
(87)
φ
.
q
M
=
∑
α
∈
I
r
6
,
m
q
r
d
e
t
q
M
M
∗
q
.
b
^
f
.
α
α
∈
H
k
×
1
,
f
=
1
,
…
,
k
,
φ
i
.
A
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
b
^
.
s
β
β
∈
H
1
×
m
,
s
=
1
,
…
,
m
,
are the column vector and the row vector, respectively.b
^
f
. and b
^
.
s are the fth row and the sth column of B
^
=
A
1
∗
B
1
M
∗.
(iii)
(88)
y
i
j
3
=
∑
β
∈
J
r
4
,
k
j
c
d
e
t
i
A
1
∗
A
1
.
i
ω
.
j
1
β
β
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
∑
α
∈
J
r
5
,
m
B
1
B
1
∗
α
α
∑
β
∈
J
r
6
,
m
M
M
∗
β
β
(89)
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
ψ
i
.
2
α
α
η
,
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
∑
β
∈
J
r
5
,
n
B
1
∗
B
1
β
β
∑
β
∈
J
r
6
,
m
M
M
∗
β
β
,
whereω
.
j
(
1
) is the jth column of  Ω
1
=
Ω
Ψ
1 and ψ
i
.
(
2
) is the ith row of  Ψ
2
≔
Ω
2
η
Ψ. The matrix Ω
=
(
ω
i
t
)
∈
H
k
×
m is such that(90)
ω
i
t
=
∑
α
∈
I
r
5
,
m
t
r
d
e
t
t
B
1
B
1
∗
t
.
s
i
.
1
α
α
ands
i
.
(
1
) is the ith row of of S
1
=
A
1
∗
S
B
1
∗. The matrix Ψ
1
≔
(
ψ
t
j
(
1
)
)
∈
H
m
×
k is such that (91)
ψ
t
j
1
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
c
^
t
.
1
α
α
η
,
whereC
^
1
≔
C
1
η
Ψ and Ψ
≔
(
ψ
q
j
)
∈
H
m
×
k is such that (92)
ψ
q
j
=
∑
β
∈
J
r
6
,
m
q
c
d
e
t
q
M
M
∗
.
q
b
^
.
f
∗
β
β
,
andb
^
.
f
∗ is fth column of B
^
∗
=
M
B
1
∗
A
1. The matrix Ω
2
=
(
ω
i
q
(
2
)
) is such that (93)
ω
i
q
2
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
c
.
q
2
β
β
,
wherec
.
q
(
2
) is the qth column of C
2
≔
Ω
C
1.
(iv)
(94)
z
f
g
1
=
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
v
f
.
2
,
η
α
α
η
∑
β
∈
J
r
6
,
p
M
∗
M
β
β
∑
α
∈
I
r
5
,
p
B
1
∗
B
1
α
α
,
or(95)
z
f
g
1
=
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
u
.
f
2
β
β
∑
β
∈
J
r
6
,
p
M
∗
M
β
β
∑
α
∈
I
r
5
,
p
B
1
∗
B
1
α
α
,
where(96)
v
f
.
2
=
∑
β
∈
J
r
6
,
p
p
c
d
e
t
f
M
∗
M
.
f
c
^
.
l
β
β
∈
H
1
×
p
,
l
=
1
,
…
,
p
,
(97)
u
.
g
2
=
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
c
^
q
.
η
α
α
η
∈
H
p
×
1
,
q
=
1
,
…
,
p
,
are the column vector and the row vector, respectively.c
^
.
l is the lth column of C
^
≔
M
∗
C
1
B
1
η and c
^
q
.
η is the qth row of C
^
η.
(v)
(98)
z
f
g
2
=
∑
β
∈
J
r
7
,
p
f
c
d
e
t
f
S
∗
S
.
f
υ
~
.
g
β
β
∑
β
∈
J
r
7
,
p
S
∗
S
β
β
∑
β
∈
J
r
5
,
p
B
1
∗
B
1
β
β
∑
α
∈
I
r
6
,
p
M
∗
M
α
α
,
whereυ
~
.
g is the gth column of Υ
~
≔
S
∗
S
Υ, Υ
=
(
υ
t
g
) is such that (99)
υ
t
g
=
∑
β
∈
J
r
5
,
p
t
c
d
e
t
t
B
1
∗
B
1
.
t
λ
.
g
β
β
=
-
η
∑
α
∈
I
r
6
,
p
g
r
d
e
t
g
M
∗
M
g
.
μ
t
.
α
α
η
,
(100)
λ
.
g
=
η
∑
α
∈
I
r
6
,
p
g
r
d
e
t
g
M
∗
M
g
.
c
^
l
.
η
∗
α
α
η
∈
H
p
×
1
,
l
=
1
,
…
,
p
,
(101)
μ
t
.
=
∑
α
∈
J
r
2
,
k
t
c
d
e
t
t
B
1
∗
B
1
.
t
c
^
.
q
∗
β
β
∈
H
1
×
p
,
q
=
1
,
…
,
k
,
are the column vector and the row vector, respectively.c
^
l
.
η
∗ is the lth row of C
^
η
∗
=
-
B
1
∗
C
1
M
η and c
^
.
q
∗ is the qth column of C
^
∗
=
B
1
η
∗
C
1
∗
M.
(vi)
(102)
x
s
t
1
=
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
d
^
.
t
β
β
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
.
whered
^
.
t is the tth column of D
^
=
A
∗
D
1.
(vii)
(103)
x
s
t
2
=
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
u
.
t
3
β
β
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
2
(104)
=
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
v
s
.
3
,
η
α
α
η
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
2
,
where(105)
u
.
t
3
=
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
d
~
q
.
η
α
α
η
∈
H
n
×
1
,
q
=
1
,
…
,
n
,
v
s
.
3
=
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
d
~
.
l
β
β
∈
H
1
×
m
,
l
=
1
,
…
,
m
,
are the column vector and the row vector, respectively.d
~
q
. and d
~
.
l are the qth row and the lth column of D
~
=
A
∗
D
1
A
A
∗.Proof.
(i) For the first termA
1
†
C
1
(
A
1
†
)
η
∗
≕
Y
1
=
y
i
j
(
1
) of (75) using the determinantal representations (22) for A
1
† and (30) for (
A
1
†
)
η
∗, we have(106)
y
i
j
1
=
∑
t
=
1
m
∑
l
=
1
m
a
i
t
†
c
t
l
1
a
l
j
1
,
η
∗
,
†
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
a
.
t
1
,
∗
β
β
c
t
l
1
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
a
l
.
1
α
α
η
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
∑
α
∈
I
r
4
,
k
A
1
∗
A
1
α
α Denote A
1
∗
C
1
A
1
η
≕
A
^
=
(
a
^
i
j
). Due to ∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
=
∑
α
∈
I
r
4
,
k
A
1
∗
A
1
α
α, we have(107)
y
i
j
1
=
∑
t
=
1
m
∑
l
=
1
m
a
i
t
†
c
t
l
1
a
l
j
1
,
η
∗
,
†
=
∑
t
=
1
m
∑
l
=
1
m
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
e
.
t
β
β
a
^
t
l
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
e
l
.
α
α
η
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
,where e
l
. and e
.
l are, respectively, the unit row vector and the unit column vector whose components are 0 except the lth components which are 1.
If we denote by(108)
v
i
l
1
≔
∑
t
=
1
m
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
e
.
t
β
β
a
^
t
l
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
a
^
.
l
β
βthe lth component of a row vector v
i
.
(
1
)
=
v
i
1
(
1
)
,
…
,
v
i
m
(
1
), then (109)
∑
l
=
1
m
v
i
l
1
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
e
l
.
α
α
η
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
v
i
.
1
,
η
α
α
η
,where v
i
.
(
1
)
,
η
≔
v
i
.
(
1
)
η
=
v
i
1
(
1
)
,
η
,
…
,
v
i
m
(
1
)
,
η. So, y
i
j
(
1
) has the determinantal representation (81).
If we denote by(110)
u
t
j
1
≔
∑
l
=
1
m
a
^
t
j
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
e
l
.
α
α
η
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
a
^
t
.
η
α
α
ηthe t
t
h component of a column vector u
.
j
(
1
)
=
u
1
j
(
1
)
,
…
,
u
m
j
(
1
), then (111)
∑
t
=
1
m
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
e
.
t
β
β
u
t
j
1
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
u
.
j
1
β
β
.So, the other determinantal representation of y
i
j
(
1
) is (82).
(ii) Consider the second term(
A
1
†
B
1
M
†
)
(
C
1
(
A
1
η
∗
)
†
)
≔
Y
2
=
y
i
j
(
2
) of (75). Due to (30) for the determinantal representation of (
A
1
η
∗
)
†
=
a
t
j
(
1
)
,
η
∗
,
†, we have for the second multiplier C
1
(
A
1
η
∗
)
†, (112)
∑
t
=
1
m
c
q
t
a
t
j
1
,
η
∗
,
†
=
∑
t
=
1
m
c
q
t
·
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
a
t
.
1
α
α
∑
β
∈
I
r
4
,
k
A
1
∗
A
1
α
α
η
=
-
η
∑
t
=
1
m
c
q
t
η
·
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
a
t
.
1
α
α
∑
β
∈
I
r
4
,
k
A
1
∗
A
1
α
α
η
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
c
~
q
.
1
α
α
η
∑
β
∈
I
r
4
,
k
A
1
∗
A
1
α
αwhere c
~
q
.
(
1
) is the qth row of C
1
~
≔
C
1
η
A
1. By applying the determinantal representations (22) and (23) for the Moore-Penrose inverses A
1
† and M
†, respectively, and due to Lemma 19 for the first multiplier A
1
†
B
1
M
†, we obtain the matrix Φ
=
(
ϕ
i
q
) such that (113)
ϕ
i
q
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
φ
.
q
M
β
β
=
∑
α
∈
I
r
6
,
m
q
r
d
e
t
q
M
M
∗
q
.
φ
i
.
A
α
α
,
φ
.
q
M
=
∑
α
∈
I
r
6
,
m
q
r
d
e
t
q
M
M
∗
q
.
b
^
f
.
α
α
∈
H
k
×
1
,
f
=
1
,
…
,
k
,
φ
i
.
A
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
b
^
.
s
β
β
∈
H
1
×
m
,
s
=
1
,
…
,
m
, are the column vector and the row vector, respectively. b
^
f
. and b
^
.
s are the fth row and the sth column of B
^
=
A
1
∗
B
1
M
∗. So, we have (114)
y
i
j
2
=
∑
q
=
1
k
ϕ
i
q
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
c
~
q
.
1
α
α
η
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
∑
α
∈
I
r
6
,
m
M
M
∗
α
α
.
DenoteΦ
~
≔
Φ
η
C
1
η
A
1. From this denotation and the equation (114), it follows (85).
(iii) Due to Lemma19 for the third term (
A
1
†
S
B
1
†
)
C
1
(
(
M
†
)
η
∗
B
1
η
∗
(
A
1
†
)
η
∗
)
≔
Y
3
=
y
i
j
(
3
) of (75), we have (115)
y
i
j
3
=
∑
q
=
1
m
∑
t
=
1
m
ω
~
i
t
c
t
q
1
ψ
~
q
j
∑
β
∈
J
r
4
,
k
A
1
∗
A
1
β
β
2
∑
β
∈
J
r
5
,
p
B
1
∗
B
1
β
β
∑
β
∈
J
r
6
,
m
M
M
∗
β
β
,where (116)
ψ
~
q
j
=
-
η
ϕ
j
q
¯
η
=
-
η
∑
α
∈
I
r
4
,
k
j
r
d
e
t
j
A
1
∗
A
1
j
.
ψ
q
.
α
α
η
,
(117)
ψ
q
.
=
∑
β
∈
J
r
6
,
m
q
c
d
e
t
q
M
M
∗
.
q
b
^
.
f
∗
β
β
∈
H
1
×
k
,
f
=
1
,
…
,
k
,is the row vector, b
^
.
f
∗ is fth column of B
^
∗
=
M
B
1
∗
A
1, and (118)
ω
~
i
t
=
∑
β
∈
J
r
4
,
k
j
c
d
e
t
i
A
1
∗
A
1
.
i
ω
.
t
β
β
,
(119)
ω
.
t
=
∑
α
∈
I
r
5
,
m
t
r
d
e
t
t
B
1
B
1
∗
t
.
s
f
.
1
α
α
∈
H
k
×
1
,
f
=
1
,
…
,
kis the column vector, s
f
.
(
1
) is the fth row of S
1
=
A
1
∗
S
B
1
∗.
Construct the matricesΨ
=
(
ψ
q
f
)
∈
H
m
×
k and Ω
=
(
ω
f
t
)
∈
H
k
×
m determined by (117) and (119), respectively. Denote C
^
1
≔
C
1
η
Ψ, Ψ
1
≔
(
ψ
t
j
(
1
)
), where (120)
ψ
t
j
1
=
∑
q
=
1
m
c
t
q
1
ψ
~
q
j
=
-
η
∑
α
∈
I
r
1
,
n
j
r
d
e
t
j
A
1
∗
A
1
j
.
c
^
t
.
1
α
α
η
,
c
^
t
.
(
1
) is the tth row of C
^
1, and Ω
1
≔
Ω
Ψ
1. From these denotations and Eq. (115), it follows (88).
We can obtain another determinantal representation ofy
i
j
(
3
) as well. Denote C
2
≔
Ω
C
1, Ω
2
≔
(
ω
t
j
(
2
)
), where (121)
ω
i
q
2
=
∑
q
=
1
m
ω
~
i
t
c
t
q
1
=
∑
β
∈
J
r
4
,
k
i
c
d
e
t
i
A
1
∗
A
1
.
i
c
.
q
2
β
β
,
c
.
q
(
2
) is the qth column of C
2, and Ψ
2
≔
Ω
2
η
Ψ. From these denotations and Eq. (115), it follows (89).
(iv) Now consider the first itemM
†
C
1
B
1
†
η
∗
≔
Z
1
=
z
f
g
(
1
) of (75). Using the determinantal representations (22) for M
†, and (30) for B
1
†
η
∗, we have(122)
z
f
g
1
=
∑
t
=
1
m
∑
l
=
1
m
m
f
t
†
c
t
l
1
b
l
g
1
,
η
∗
,
†
=
∑
t
∑
l
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
m
.
t
∗
β
β
c
t
l
1
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
b
l
.
1
α
α
η
∑
β
∈
J
r
6
,
p
M
∗
M
β
β
∑
β
∈
I
r
5
,
p
B
1
∗
B
1
α
α
,Denote M
∗
C
1
B
1
η
≕
C
^
1
=
(
c
^
i
j
). Then, we have(123)
z
f
g
1
=
∑
t
∑
m
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
e
.
t
β
β
c
^
t
l
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
e
l
.
α
α
η
∑
β
∈
J
r
6
,
p
M
∗
M
β
β
∑
β
∈
I
r
5
,
p
B
1
∗
B
1
α
α
,where e
l
. and e
.
l are, respectively, the unit row-vector and the unit column-vector whose components are 0 except the lth components which are 1.
If we denote by(124)
v
f
l
2
≔
∑
t
=
1
m
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
e
.
t
β
β
c
^
t
l
=
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
c
^
.
l
β
βthe lth component of a row vector v
f
.
(
2
)
=
v
f
1
(
2
)
,
…
,
v
f
m
(
2
), then (125)
∑
l
=
1
m
v
f
l
2
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
e
l
.
α
α
η
=
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
v
2
f
.
η
α
α
η
,where v
f
.
(
2
)
,
η
≔
v
f
.
(
2
)
η
=
v
f
1
(
2
)
,
η
,
…
,
v
f
m
(
2
)
,
η. So, z
f
g
(
1
) has the determinantal representation (94).
If we denote by(126)
u
t
g
2
≔
∑
l
=
1
m
c
^
t
l
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
e
l
.
α
α
η
=
-
η
∑
α
∈
I
r
5
,
p
g
r
d
e
t
g
B
1
∗
B
1
g
.
c
^
t
.
η
α
α
ηthe t
t
h component of a column vector u
.
g
(
2
)
=
u
1
g
(
2
)
,
…
,
u
m
g
(
2
), then (127)
∑
t
=
1
m
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
e
.
t
β
β
u
t
g
2
=
∑
β
∈
J
r
6
,
p
f
c
d
e
t
f
M
∗
M
.
f
u
.
g
2
β
β
.So, the other determinantal representation of z
f
g
(
1
) is (95).
(v) For the second termQ
S
B
1
†
C
1
M
†
η
∗
=
Z
4
=
z
f
g
(
2
) of (76) using (33) for a determinantal representation of Q
S, and similarly as in the point (iv) for B
1
†
C
1
M
†
η
∗, we have (128)
z
f
g
2
=
∑
t
=
1
k
∑
β
∈
J
r
7
,
p
f
c
d
e
t
f
S
∗
S
.
f
S
˙
.
t
β
β
υ
t
g
∑
β
∈
J
r
7
,
p
S
∗
S
β
β
∑
β
∈
I
r
5
,
p
B
1
∗
B
1
α
α
∑
α
∈
J
r
6
,
p
M
∗
M
β
β
,where s
˙
.
t is the tth column of S
∗
S, (129)
υ
t
g
=
∑
α
∈
J
r
5
,
p
t
c
d
e
t
t
B
1
∗
B
1
.
t
λ
.
g
β
β
=
-
η
∑
α
∈
I
r
6
,
p
g
r
d
e
t
g
M
∗
M
g
.
μ
t
.
α
α
η
,
(130)
λ
.
g
=
η
∑
α
∈
I
r
6
,
p
g
r
d
e
t
g
M
∗
M
g
.
c
^
l
.
η
∗
α
α
η
∈
H
p
×
1
,
l
=
1
,
…
,
p
,
μ
t
.
=
∑
α
∈
J
r
2
,
k
t
c
d
e
t
t
B
1
∗
B
1
.
t
c
^
.
q
∗
β
β
∈
H
1
×
p
,
q
=
1
,
…
,
k
,are the column vector and the row vector, respectively. c
^
l
.
η
∗ is the lth row of C
^
η
∗
=
-
B
1
∗
C
1
M
η and c
^
.
q
∗ is the qth column of C
^
∗
=
B
1
η
∗
C
1
∗
M. Construct the matrix Υ
=
(
υ
t
g
), where υ
t
g is determined by (129). Denote the matrix Υ
~
≔
S
∗
S
Υ. From this denotation and due to (128), we obtain (98).
(vi) For the first term of (77), X
1
=
(
x
s
t
(
1
)
)
≔
A
†
D
1, the determinantal representation (102) evidently follows due to Corollary 20.
(vii) For the second termA
†
D
1
P
A
η
=
X
2
=
x
s
t
(
2
) of (77), using (22) for a determinantal representation of A
† and (34) for a determinantal representation of P
A, we have (131)
x
s
t
2
=
∑
q
=
1
m
∑
l
=
1
n
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
a
.
q
∗
β
β
d
q
l
1
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
a
¨
l
.
α
α
η
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
∑
α
∈
I
r
1
,
m
A
∗
A
β
β
,Denote A
∗
D
1
(
A
A
∗
)
η
≕
D
~
=
(
d
~
q
l
). Since ∑
β
∈
J
r
1
,
n
A
∗
A
β
β
=
∑
α
∈
I
r
1
,
m
A
∗
A
β
β, we have(132)
x
s
t
2
=
∑
q
=
1
m
∑
l
=
1
n
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
e
.
q
β
β
d
~
q
l
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
e
l
.
α
α
η
∑
β
∈
J
r
1
,
n
A
∗
A
β
β
2
,where e
l
. and e
.
l are, respectively, the unit row-vector and the unit column-vector.
If we denote by(133)
v
s
l
3
≔
∑
q
=
1
m
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
e
.
q
β
β
d
~
q
l
=
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
d
~
.
l
β
βthe lth component of a row vector v
s
.
(
3
)
=
v
s
1
(
3
)
,
…
,
v
s
n
(
3
), then (134)
∑
l
=
1
n
v
s
l
3
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
e
l
.
α
α
η
=
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
v
s
.
3
,
η
α
α
η
.So, x
s
t
(
2
) has the determinantal representation (103).
If we denote by(135)
u
q
t
3
≔
∑
l
=
1
m
d
~
q
l
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
e
l
.
α
α
η
=
-
η
∑
α
∈
I
r
1
,
m
t
r
d
e
t
t
A
A
∗
t
.
d
~
q
.
η
α
α
ηthe qth component of a column-vector u
.
t
=
u
1
t
,
…
,
u
m
t, then (136)
∑
q
=
1
m
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
e
.
q
β
β
u
q
t
3
=
∑
β
∈
J
r
1
,
n
s
c
d
e
t
s
A
∗
A
.
s
u
.
t
3
β
β
.So, the other determinantal representation of x
s
t
(
2
) is (104).


## 5. An Example

Given the matrices,(137)
A
=
i
-
1
k
j
j
-
k
,
B
=
i
1
j
j
k
-
i
k
-
j
1
,
C
=
1
k
i
i
-
j
0
-
j
-
i
1
,
D
=
i
-
j
-
k
j
i
0
k
0
i
.Since D
=
-
D
η
∗ with η
=
i, we shall find the i-skew-Hermitian solution to Eq. (5) with the given matrices (137) by both methods, first, using determinantal representations of generalized inverses, and then, by obtaining Cramer’s rule.By Theorem13 and Corollary 15, we obtain (138)
A
†
=
1
6
-
i
-
k
-
j
-
1
-
j
k
,
R
A
=
1
3
2
-
j
k
j
2
-
i
-
k
i
2
.So, (139)
A
1
=
R
A
B
=
1
3
2
i
2
2
j
j
-
k
j
+
k
-
1
-
3
i
-
j
+
k
-
3
j
-
k
3
+
i
,
B
1
=
R
A
C
=
1
3
2
+
i
+
k
-
1
+
j
-
2
j
2
i
+
k
2
i
+
j
+
k
j
+
k
-
i
-
k
-
1
-
2
j
-
k
1
-
2
i
-
k
2
-
j
,
A
1
†
=
1
24
-
2
j
-
3
j
+
i
j
-
3
k
2
-
j
-
3
k
3
j
+
k
-
2
j
-
1
+
3
i
3
-
i
,
B
1
†
=
1
12
1
-
i
-
2
k
1
-
3
i
-
j
-
k
j
-
3
k
-
2
+
j
-
k
-
1
-
i
+
3
j
-
k
3
j
+
k
2
-
6
i
+
2
j
-
2
k
4
+
4
i
+
4
k
6
-
2
i
+
2
j
+
2
k
,
M
=
R
A
1
B
1
=
1
4
1
+
i
+
j
+
k
-
1
+
i
-
j
-
k
i
-
j
+
k
i
-
k
1
-
j
1
-
j
-
k
i
-
k
1
-
j
1
-
j
-
k
,
S
=
B
1
L
M
=
1
21
4
+
2
i
-
j
-
4
i
-
2
j
+
4
k
4
i
+
8
j
-
4
k
2
+
2
i
+
5
j
+
5
k
-
5
+
5
i
-
6
j
+
2
k
2
-
2
i
+
2
j
-
2
k
-
6
-
2
i
-
7
j
+
k
-
1
-
7
i
+
2
j
-
6
k
10
+
10
i
-
2
j
+
2
k
,
C
1
=
R
A
D
R
A
η
=
1
3
2
i
-
2
j
+
k
j
-
2
k
2
j
-
k
2
i
1
-
2
i
-
j
+
2
k
-
1
-
2
i
2
i
,It is easy to check that the consistency conditions (42) of Eq. (5) are fulfilled by these given matrices. So, Eq. (5) has η-Hermitian solutions.We find the partial solution (75)–(77) by Cramer’s rule obtained in Theorem 24. Since (140)
A
1
∗
A
1
=
3
2
j
-
2
i
-
j
-
2
j
2
2
k
2
i
+
j
2
k
3
,
A
^
1
=
A
1
∗
C
1
A
η
=
1
3
16
i
16
-
16
j
-
16
16
i
-
16
k
16
j
16
k
16
i
.and (determinantal) r
(
A
1
∗
A
1
)
=
r
(
A
)
=
1, then, by (83), (141)
v
1
.
1
=
1
3
16
i
16
-
16
j
,
v
2
.
1
=
1
3
-
16
16
i
-
16
k
,
v
3
.
1
=
1
3
16
j
16
k
16
i
.Further, by (81), we obtain (142)
y
11
1
=
-
η
∑
α
∈
I
1,3
j
r
d
e
t
1
A
∗
A
j
.
v
1
.
η
α
α
η
∑
β
∈
J
1,3
A
∗
A
β
β
2
=
i
12
.By continuing of similar computations, we get (143)
Y
1
=
1
12
i
1
-
j
-
1
i
-
k
j
k
i
.Now, we findY
2 by (85). Since (144)
B
^
=
A
1
∗
B
1
M
∗
=
1
12
-
4
-
10
i
-
8
j
+
16
k
-
12
+
4
i
+
7
j
+
3
k
-
12
+
4
i
+
7
j
+
3
k
10
-
4
i
-
16
j
-
8
k
-
4
-
12
i
-
3
j
+
7
k
-
4
-
12
i
-
3
j
+
7
k
-
16
+
8
i
-
10
j
-
4
k
-
3
-
7
i
+
4
j
-
12
k
-
3
-
7
i
+
4
j
-
12
k
,and r
(
A
1
)
=
1, then (145)
φ
1
.
A
=
1
12
-
4
-
10
i
-
8
j
+
16
k
-
12
+
4
i
+
7
j
+
3
k
-
12
+
4
i
+
7
j
+
3
k
,
φ
2
.
A
=
1
12
10
-
4
i
-
16
j
-
8
k
-
4
-
12
i
-
3
j
+
7
k
-
4
-
12
i
-
3
j
+
7
k
,
φ
3
.
A
=
1
12
-
16
+
8
i
-
10
j
-
4
k
-
3
-
7
i
+
4
j
-
12
k
-
3
-
7
i
+
4
j
-
12
k
.Further, we construct the matrix Φ by (86). Since r
(
M
)
=
1, then (146)
ϕ
1
.
=
φ
1
.
A
,
ϕ
2
.
=
φ
2
.
A
,
ϕ
3
.
=
φ
3
.
A
.From this by direct multiplication, it follows that(147)
Φ
~
=
Φ
η
C
1
η
A
1
=
1
9
4
+
10
i
-
8
j
+
16
k
10
-
4
i
-
16
j
-
8
k
16
+
8
i
+
10
j
-
4
k
-
10
+
4
i
-
16
j
-
8
k
4
+
10
+
8
j
-
16
k
-
8
+
16
i
+
4
j
+
10
k
16
-
8
i
-
10
j
-
4
k
-
8
-
16
i
+
4
j
-
10
k
-
4
+
10
i
-
8
j
-
16
k
.Finally, by (85), we have(148)
y
11
2
=
-
η
∑
α
∈
I
1,3
j
r
d
e
t
1
A
∗
A
1
.
ϕ
~
1
.
α
α
η
∑
β
∈
J
1,3
A
∗
A
β
β
2
∑
α
∈
I
r
3
,
m
M
M
∗
α
α
=
1
504
2
+
5
i
+
4
j
-
8
k
,
Y
2
=
1
504
2
+
5
i
+
4
j
-
8
k
5
-
2
i
+
8
j
+
4
k
2
+
4
i
-
5
j
+
2
k
-
5
+
2
i
+
8
j
+
4
k
2
+
5
i
-
4
j
+
8
k
-
4
+
8
i
-
2
j
-
5
k
8
-
4
i
+
5
j
+
2
k
-
4
-
8
i
-
2
j
+
5
k
-
2
+
5
i
+
4
j
+
8
k
.Now, we findY
3 by (89). The following algorithm of finding Y
3 is given by Theorem 24.(1)
Find the matrixS
1
=
A
1
∗
S
B
1
∗. (149)
S
1
=
16
21
-
2
i
3
j
-
k
j
-
3
k
2
-
j
-
3
k
3
j
+
k
-
2
j
-
1
+
3
i
3
-
i
.
(2)
Find the matrixB
1
B
1
∗. (150)
B
1
B
1
∗
=
1
9
17
3
-
6
i
-
7
j
-
6
k
-
6
-
3
i
+
6
j
+
10
k
3
+
6
i
+
7
j
+
6
k
14
-
6
-
7
i
-
6
j
+
3
k
-
6
+
3
i
-
6
j
-
10
k
-
6
+
7
i
+
6
j
-
3
k
17
.
(3)
By (90), construct the matrix the matrix Ω. Since r
(
B
1
B
1
∗
)
=
2, then (151)
ω
11
=
∑
α
∈
I
2,3
1
r
d
e
t
1
B
1
B
1
∗
1
.
s
1
.
1
α
α
=
16
189
r
d
e
t
1
-
2
i
3
j
-
k
3
+
6
i
+
7
j
+
6
k
14
+
r
d
e
t
1
-
2
i
j
-
3
k
-
6
+
3
i
-
6
j
-
10
k
17
=
16
189
39
-
59
i
+
12
j
+
6
k
.
By similar computing, we get(152)
Ω
=
16
189
39
-
59
i
+
12
j
+
6
k
27
-
9
i
+
141
j
-
59
k
-
15
-
15
i
+
22
j
-
108
k
9
+
9
i
+
36
j
+
18
k
9
+
27
i
-
9
j
-
63
k
9
+
27
i
+
54
j
36
-
18
i
-
9
j
+
9
k
-
9
+
63
i
-
6
j
+
27
k
54
-
9
j
+
27
k
.
(4)
Obtain the matrixC
2
=
Ω
C
1. (153)
C
2
=
16
567
20
+
i
+
78
j
-
84
k
33
+
99
i
+
52
j
+
37
k
15
-
111
i
-
36
j
+
32
k
-
9
+
26
i
+
18
j
-
36
k
45
+
45
i
-
207
j
+
117
k
9
-
63
i
+
144
j
-
198
k
12
+
33
i
+
9
j
+
261
k
-
207
-
117
i
-
45
j
+
39
k
144
+
198
i
-
6
j
-
57
k
.
(5)
By (93), construct the matrix Ω
2. Since r
(
A
1
∗
A
1
)
=
1, then Ω
2
=
C
2.
(6)
By (92), construct the matrix Ψ. Since r
a
n
k
M
M
∗
=
1, then Ψ
=
B
^
∗, where B
^ is obtained in (144).
​
(7)
Get the matrixΨ
2
=
Ω
2
η
Ψ.
​(154)
Ψ
2
=
4
1701
-
289
+
2683
i
+
668
j
-
3544
k
2683
+
289
i
+
3544
j
+
668
k
-
3544
-
668
i
+
2683
j
+
289
k
-
252
+
630
i
+
504
j
-
1008
k
630
+
252
i
+
1008
j
+
504
k
-
1008
-
504
i
+
630
j
+
252
k
-
525
-
999
i
-
216
j
-
642
k
-
999
+
525
i
+
642
j
-
216
k
-
642
+
216
i
-
999
j
+
525
k
.
(8)
Finally, by (89) and due to r
(
A
1
∗
A
1
)
=
1, we have Y
3
=
(
1
/
448
)
Ψ
2
η. So, (155)
Y
3
=
1
190512
-
289
+
2683
i
-
668
j
+
3544
k
2683
+
289
i
-
3544
j
-
668
k
-
3544
-
668
i
-
2683
j
-
289
k
-
252
+
630
i
-
504
j
+
1008
k
630
+
252
i
-
1008
j
-
504
k
-
1008
-
504
i
-
630
j
-
252
k
-
525
-
999
i
+
216
j
+
642
k
-
999
+
525
i
-
642
j
+
216
k
-
642
+
216
i
+
999
j
-
525
k
.Hence,(156)
Y
=
Y
1
-
1
2
Y
2
-
Y
2
η
∗
-
1
2
Y
3
-
Y
3
η
∗
=
1
12
i
1
-
j
-
1
i
-
k
j
k
i
-
1
504
5
i
5
-
3
-
5
j
-
5
5
i
-
5
k
3
+
5
j
5
k
5
i
-
1
381024
2683
i
2935
+
919
i
-
3040
j
-
1676
k
-
3292
-
1067
i
-
2899
j
-
931
k
-
2935
+
919
i
+
3040
j
+
1676
k
252
i
-
9
+
19
i
+
12
j
-
468
k
3292
-
1067
i
+
2899
j
+
931
k
9
+
19
i
-
12
j
+
468
k
216
i
=
1
381024
25289
i
28437
-
919
i
+
3040
j
+
1676
k
5560
+
1067
i
-
25073
j
+
931
k
-
28437
-
919
i
-
3040
j
-
1676
k
27720
i
9
-
19
i
-
12
j
-
27504
k
-
5560
+
1067
i
+
25073
j
-
931
k
-
9
-
19
i
+
12
j
+
27504
k
27756
i
. After rounding coefficients, we obtain (157)
Y
=
0.0664
i
0.0746
-
0.0024
i
+
0.0080
j
+
0.0044
k
0.0146
+
0.0028
i
-
0.0658
j
+
0.0024
k
-
0.0746
-
0.0024
i
-
0.0080
j
-
0.0044
k
0.0728
i
-
0.0722
k
-
0.0146
+
0.0028
i
+
0.0658
j
-
0.0024
k
0.0722
k
0.0728
i
.Now, we findZ
1 by (94). First, we obtain the matrix C
^
=
M
∗
C
1
B
1
∗, (158)
C
^
=
1
36
12
+
7
i
-
3
k
3
-
3
i
-
9
j
-
11
k
-
3
i
+
4
j
+
3
k
-
3
-
7
i
-
12
k
-
11
-
9
i
+
3
j
-
3
k
3
+
4
i
+
3
j
3
+
2
i
-
j
-
17
k
-
5
-
16
i
-
7
k
1
+
4
i
+
5
j
+
3
k
.Since r
(
M
)
=
1, then, by (96), (159)
v
1
.
2
=
c
^
1
.
,
v
2
.
2
=
c
^
2
.
,
v
3
.
2
=
c
^
3
.
.So, by (94), we have (160)
z
11
1
=
-
η
∑
α
∈
I
2,3
1
r
d
e
t
1
B
1
∗
B
1
1
.
v
1
.
2
,
η
α
α
η
∑
β
∈
J
1,3
M
∗
M
β
β
∑
α
∈
I
2,3
B
1
∗
B
1
α
α
=
1
108
67
+
123
i
-
15
j
-
53
k
.Continuing in like manner yields (161)
Z
1
=
1
108
67
+
123
i
-
15
j
-
53
k
43
-
33
i
-
133
j
-
51
k
-
7
-
79
i
+
73
j
+
63
k
-
53
-
15
i
-
123
j
-
53
k
-
51
-
133
i
+
33
j
-
43
k
63
+
73
i
+
79
j
+
7
k
-
12
+
20
i
-
97
j
-
155
k
37
-
175
i
-
12
j
-
52
k
23
+
65
i
+
112
j
+
78
k
.Further, we findZ
2 by (98). The algorithm of finding Z
2 is given by Theorem 24 as well. By (101), we have(162)
μ
11
=
∑
α
∈
J
2,3
t
c
d
e
t
1
B
1
∗
B
1
.
1
c
^
.
1
∗
β
β
=
185
-
17
i
-
27
j
+
23
k
,where c
^
.
1
∗ is the first column of C
^
∗ that is the conjugate transpose of C
^ obtained in (158). By similar computing, we get (163)
μ
1
.
=
185
-
17
i
-
27
j
+
23
k
-
23
-
27
i
+
17
j
+
185
k
56
-
24
i
-
89
j
+
205
k
,
μ
2
.
=
23
+
21
i
+
49
j
+
183
k
-
183
+
49
i
-
21
j
+
23
k
-
147
+
151
i
-
8
j
+
104
k
,
μ
3
.
=
23
+
41
i
-
75
j
-
13
k
13
-
75
i
-
41
j
+
23
k
-
13
-
61
i
-
90
j
-
4
k
.By (99), we construct the matrix Υ
=
(
υ
t
g
). Since r
(
M
∗
M
)
=
1, then (164)
υ
t
g
=
-
η
∑
α
∈
I
r
6
,
p
g
r
d
e
t
g
M
∗
M
g
.
μ
t
.
α
α
η
=
-
i
μ
t
g
ifor all t
,
g
=
1,2
,
3
. So, (165)
Υ
=
185
-
17
i
+
27
j
-
23
k
-
23
-
27
i
-
17
j
-
185
k
56
-
24
i
+
89
j
-
205
k
23
+
21
i
-
49
j
-
183
k
-
183
+
49
i
+
21
j
-
23
k
-
147
+
151
i
+
8
j
-
104
k
23
+
41
i
+
75
j
+
13
k
13
-
75
i
+
41
j
-
23
k
-
13
-
61
i
+
90
j
+
4
k
.We obtain the following matrices: (166)
S
∗
S
=
8
49
3
3
k
-
2
+
2
j
-
4
k
-
3
k
3
-
4
+
2
i
+
2
k
-
2
-
2
j
+
4
k
-
4
-
2
i
-
2
k
8
,
Υ
~
=
S
∗
S
Υ
=
16
49
29
-
303
i
+
7
j
-
5
k
23
-
147
i
+
292
j
+
178
k
23
-
147
i
+
292
j
+
178
k
-
5
+
7
i
+
303
j
-
29
k
178
+
292
i
+
147
j
-
23
k
178
+
292
i
+
147
j
-
23
k
-
8
+
196
i
-
428
j
-
160
k
-
58
-
410
i
-
58
j
-
186
k
-
58
-
410
i
-
58
j
-
186
k
.Finally, since(167)
∑
β
∈
J
1,3
S
∗
S
β
β
=
112
49
,
∑
β
∈
J
2,3
B
1
∗
B
1
β
β
=
4
,
∑
α
∈
I
r
6
,
p
M
∗
M
α
α
=
7
4
,then by (98),(168)
Z
2
=
1
49
29
-
303
i
+
7
j
-
5
k
23
-
147
i
+
292
j
+
178
k
23
-
147
i
+
292
j
+
178
k
-
5
+
7
i
+
303
j
-
29
k
178
+
292
i
+
147
j
-
23
k
178
+
292
i
+
147
j
-
23
k
-
8
+
196
i
-
428
j
-
160
k
-
58
-
410
i
-
58
j
-
186
k
-
58
-
410
i
-
58
j
-
186
k
.Hence, (169)
Z
=
1
2
Z
1
-
Z
1
η
∗
+
1
2
Z
2
-
Z
2
η
∗
=
1
216
246
i
96
-
48
i
-
10
j
+
2
k
5
-
59
i
+
170
j
+
218
k
-
96
-
48
i
+
10
j
-
2
k
-
266
i
26
-
102
i
+
91
j
+
59
k
-
5
-
59
i
-
170
j
-
218
k
-
26
-
102
i
-
91
j
-
59
k
130
i
+
1
98
-
606
i
28
-
140
i
-
11
j
+
207
k
31
+
49
i
+
720
j
+
338
k
-
28
-
140
i
+
11
j
-
207
k
584
i
236
-
118
i
+
205
j
+
163
k
-
31
+
49
i
-
720
j
-
338
k
-
236
-
118
i
-
205
j
-
163
k
-
820
i
=
1
10584
-
53394
i
7728
-
17472
i
-
1678
j
+
22454
k
3593
+
2401
i
+
86090
j
+
45226
k
-
7728
-
17472
i
+
1678
j
-
22454
k
50038
i
26762
-
17742
i
+
26599
j
+
20495
k
-
3593
+
2401
i
-
86090
j
-
45226
k
-
26762
-
17742
i
-
26599
j
-
20495
k
-
82190
i
.By rounding coefficients, we have(170)
Z
=
-
5.0448
i
0.7302
-
1.6508
i
-
0.1585
j
+
2.1215
k
0.3395
+
0.2269
i
+
8.1340
j
+
4.2731
k
-
0.7302
-
1.6508
i
+
0.1585
j
-
2.1215
k
4.7277
i
2.5285
-
1.6763
i
+
2.5131
j
+
1.9364
k
-
0.3395
+
0.2269
i
-
8.1340
j
-
4.2731
k
-
2.5285
-
1.6763
i
-
2.5131
j
-
1.9364
k
-
7.7655
i
.Further, we findX
1 by (102). So,(171)
D
1
=
0.78
+
18.58
i
+
8.36
j
+
4.81
k
12.02
+
1.501
i
-
12.49
j
-
1.56
k
-
7.73
-
8.98
i
-
7.96
j
-
4.88
k
-
8.18
-
1.96
i
+
5.36
j
+
10.1
k
-
4.25
+
10.56
i
-
3.31
j
-
1.49
k
-
1.46
-
1.7
i
-
2.98
j
-
10.7
k
20.14
+
2.26
i
+
11.77
j
+
10.28
k
-
3.99
-
5.59
i
-
11.50
j
+
14.05
k
-
20.54
+
3.14
i
+
3.98
j
-
1.91
k
,
D
^
=
A
∗
D
1
=
40.45
-
5.70
i
-
13.37
j
+
2.08
k
-
11.48
-
29.38
i
-
8.13
j
+
11.15
k
-
15.7
+
6.66
i
+
17.36
j
+
12.56
k
-
5.7
-
40.45
i
+
2.08
j
+
13.37
k
-
29.38
+
11.48
i
+
11.15
j
+
8.13
k
6.66
+
15.7
i
+
12.56
j
-
17.36
k
.Sincer
a
n
k
A
∗
A
=
1, then by (102) (172)
x
11
1
=
∑
β
∈
J
1,2
1
c
d
e
t
1
A
∗
A
.
1
d
^
.
1
β
β
∑
β
∈
J
1,2
A
∗
A
β
β
=
1
6
40.45
-
5.70
i
-
13.37
j
+
2.08
k
.By similar computing, we get(173)
X
1
=
6.74
-
0.95
i
-
2.23
j
+
0.35
k
-
1.91
-
4.90
i
-
1.36
j
+
1.86
k
-
2.62
+
1.11
i
+
2.90
j
+
2.10
k
-
0.95
-
6.74
i
+
0.35
j
+
2.23
k
-
4.90
+
1.91
i
+
1.86
j
+
1.36
k
1.11
+
2.62
i
+
2.10
j
-
2.90
k
.Now, we findX
2 by (103). So,(174)
D
~
=
A
∗
D
1
A
A
∗
=
39.52
+
45.62
i
-
17.10
j
+
31.52
k
17.10
-
31.52
i
+
39.52
j
+
45.62
k
31.52
+
17.10
i
+
45.62
j
-
39.52
k
45.62
-
39.52
i
+
31.52
j
+
17.10
k
-
31.52
-
17.10
i
+
45.62
j
-
39.52
k
17.10
-
31.52
i
-
39.52
j
-
45.62
k
,
v
1
.
3
=
d
~
1
.
,
v
2
.
3
=
d
~
1
.
,
v
3
.
3
=
d
~
3
.
. Then by (103), we have (175)
x
11
2
=
-
η
∑
α
∈
I
1,3
1
r
d
e
t
1
A
A
∗
1
.
v
1
.
3
,
η
α
α
η
∑
β
∈
I
1,3
A
A
∗
β
β
2
=
1
36
39.52
+
45.62
i
-
17.10
j
+
31.52
k
.Continuing in like manner yields (176)
X
2
=
1.10
+
1.27
i
-
0.48
j
+
0.88
k
0.48
-
0.88
i
+
1.10
j
+
1.27
k
0.88
+
0.48
i
+
1.27
j
-
1.10
k
1.27
-
1.10
i
+
0.88
j
+
0.48
k
-
0.88
-
0.48
i
+
1.27
j
-
1.10
k
0.48
-
0.88
i
-
1.10
j
-
1.27
k
.Finally, (177)
X
=
X
1
-
0.5
X
2
=
6.19
-
1.59
i
-
1.99
j
-
0.09
k
-
2.15
-
3.46
i
-
1.91
j
+
1.22
k
-
3.06
+
0.87
i
+
2.26
j
+
2.65
k
-
1.59
-
6.19
i
-
0.09
j
+
1.99
k
-
3.46
+
2.15
i
+
1.22
j
+
1.91
k
0.87
+
3.06
i
+
2.65
j
-
2.26
k
.Note that we used Maple with the package CLIFFORD in the calculations.


## 6. Conclusion

Some necessary and sufficient conditions for the existence of the general solution of the skew-η-Hermitian solution to (5) are constructed in this paper. We not only obtain the explicit form of the η-skew-Hermitian solution in terms of generalized inverses but also give its explicit determinantal representation formulas that are an analog of Cramer’s rule. Our proposed Cramer’s rule is based on the theory of row-column noncommutative determinants. A numerical example demonstrates applying the obtained Cramer’s rule to finding the skew-η-Hermitian solution to the equation of (5).



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*Source: 7939238-2019-07-30.xml*