# Determinantal Representations of Solutions and Hermitian Solutions to Some System of Two-Sided Quaternion Matrix Equations

Authors: Ivan I. Kyrchei
Journal: Journal of Mathematics
 (2018)

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## Abstract

Within the framework of the theory of quaternion row-column determinants previously introduced by the author, we derive determinantal representations (analogs of Cramer’s rule) of solutions and Hermitian solutions to the system of two-sided quaternion matrix equationsA1XA1⁎=C1 and A2XA2⁎=C2. Since the Moore-Penrose inverse is a necessary tool to solve matrix equations, we use determinantal representations of the Moore-Penrose inverse previously obtained by the theory of row-column determinants.


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## Body

## 1. Introduction

The study of matrix equations and systems of matrix equations is an active research topic in matrix theory and its applications. Research on the classical system of two-sided matrix equations(1)A1XB1=C1,A2XB2=C2over the complex field, a principle domain, and the quaternion skew field has been actively ongoing for many years. For instance, Mitra [1, 2] gave necessary and sufficient conditions of system (3) over the complex field and the expression for its general solution. Navarra et al. [3] derived a new necessary and sufficient condition for the existence and a new representation of the general solution to (3) over the complex field and used the results to give a simple representation. Özgüler et al. [4] gave solutions to (3) over a principle domain. Wang [5] got solvability conditions of system (3) over the quaternion skew field and represented its general solution in terms of generalized inverses.Since quaternion matrices play an important role in quantum mechanics, signal processing, and control theory, research on quaternion matrix equations and systems of quaternion matrix equations, their general solutions, especially Hermitian solutions, has been actively developing for more recent years (see, e.g., [6–26]).Throughout the paper, we denote the real number field byR, the set of all m×n matrices over the quaternion algebra (2)H=a0+a1i+a2j+a3k∣i2=j2=k2=-1,a0,a1,a2,a3∈Rby Hm×n, and by Hrm×n its subset of matrices of a rank r. Let Mn,H be the ring of n×n quaternion matrices. For A∈Hn×m, the symbol A∗ stands for the conjugate transpose (Hermitian adjoint) matrix of A. The matrix A=aij∈Hn×n is Hermitian if A∗=A.Motivated by the wide application of quaternion matrix equations and in order to improve the theoretical development of solutions and Hermitian solutions to quaternion matrix equations, we consider a special case of (3), more specifically,(3)A1XA1∗=C1,A2XA2∗=C2.Generalized inverses are useful tools used to solve matrix equations. The definition of the Moore-Penrose inverse matrix has been extended to quaternion matrices as follows.Definition 1.
The Moore-Penrose inverse ofA∈Hm×n, denoted by A†, is the unique matrix A†∈Hn×m satisfying the following four equations: (4)1AA†A=A,2A†AA†=A†,3AA†∗=AA†,4A†A∗=A†A.The main goal of this paper is to derive determinantal representations of solutions and Hermitian solutions to system (1) over the quaternion skew field using previously obtained determinantal representations of the Moore-Penrose inverse. Evidently, determinantal representation of a solution gives a direct method of its finding analogous to the classical Cramer’s rule that has important theoretical and practical significance [27].Through the noncommutativity of the quaternion algebra when difficulties arise already in determining the quaternion determinant, the problem of the determinantal representation of generalized inverses only now can be solved due to the theory of column-row determinants introduced in [28, 29]. Within the framework of the theory of column-row determinants, determinantal representations of various kinds of generalized inverses (generalized inverses) solutions of quaternion matrix equations have been derived by the author (see, e.g., [30–39]) and by other researchers (see, e.g., [40–43]).The paper is organized as follows. In Section2, we start with preliminary introduction of row-column determinants, determinantal representations of the Moore–Penrose inverse previously obtained within the framework of the theory of row-column determinants, and Cramer’s rules for the two-sided matrix equation and of its special cases, left- and right-sided equations. We derive some simplified expressions of general and partial solutions to (3), a solvability criterion and expressions of general and partial solutions to system (1), and determinantal representations (analogs of Cramer’s rule) of its solution and Hermitian solution. A numerical example to illustrate the main results is considered in Section 4. Finally, the conclusion is drawn in Section 5.


## 2. Preliminaries

ForA∈Hn×n, we define n row determinants and n column determinants. Suppose Sn is the symmetric group on the set In={1,…,n}.Definition 2 (see [28]).
Theith row determinant of A=(aij)∈Hn×n is defined for all i=1,…,n by putting (5)rdetiA=∑σ∈Sn-1n-raiik1aik1ik1+1⋯aik1+l1i⋯aikrikr+1⋯aikr+lrikr,σ=iik1ik1+1⋯ik1+l1ik2ik2+1⋯ik2+l2⋯ikrikr+1⋯ikr+lr,where ik2<ik3<⋯<ikr and ikt<ikt+s for all t=2,…,r and s=1,…,lt.Definition 3 (see [28]).
Thejth column determinant of A=(aij)∈Hn×n is defined for all j=1,…,n by putting (6)cdetjA=∑τ∈Sn-1n-rajkrjkr+lr⋯ajkr+1jkr⋯ajjk1+l1⋯ajk1+1jk1ajk1j,τ=jkr+lr⋯jkr+1jkr⋯jk2+l2⋯jk2+1jk2jk1+l1⋯jk1+1jk1j,where jk2<jk3<⋯<jkr and jkt<jkt+s for t=2,…,r and s=1,…,lt.Since [28] for Hermitian A we have (7)rdet1A=⋯=rdetnA=cdet1A=⋯=cdetnA∈R,the determinant of a Hermitian matrix is defined by putting detA≔rdetiA=cdetiA for all i=1,…,n.The properties of row and column determinants are completely explored in [29]. We note the following that will be required below.Lemma 4.
LetA∈Hm×n. Then(8)cdetiA∗=rdetiA¯,rdetiA∗=cdetiA¯.We shall use the following notations. Letα≔α1,…,αk⊆1,…,m and β≔β1,…,βk⊆1,…,n be subsets of the order 1≤k≤minm,n. Let Aβα be a submatrix of A whose rows are indexed by α and columns indexed by β. Similarly, let Aαα be a principal submatrix of A whose rows and columns are indexed by α. If A∈Mn,H is Hermitian, then Aαα is the corresponding principal minor of detA. For 1≤k≤n, the collection of strictly increasing sequences of k integers chosen from 1,…,n is denoted by Lk,n≔α:α=α1,…,αk,1≤α1<⋯<αk≤n. For fixed i∈α and j∈β, let Ir,mi≔α:α∈Lr,m,i∈α,Jr,nj≔β:β∈Lr,n,j∈β.Leta.j be the jth column and ai. be the ith row of A. Suppose A.jb denotes the matrix obtained from A by replacing its jth column with column b, and Ai.b denotes the matrix obtained from A by replacing its ith row with the row b. Denote by a.j∗ and ai.∗ the jth column and the ith row of A∗, respectively.Theorem 5 (see [30]).
IfA∈Hrm×n, then the Moore-Penrose inverse A†=aij†∈Hn×m has the following determinantal representations, (9)aij†=∑β∈Jr,nicdetiA∗A.ia.j∗ββ∑β∈Jr,nA∗Aββ,and (10)aij†=∑α∈Ir,mjrdetjAA∗j.ai.∗αα∑α∈Ir,mAA∗αα.Remark 6.
For an arbitrary full-rank matrixA∈Hrm×n, we put (11)cdetiA∗A.id.j=∑β∈Jn,nicdetiA∗A.id.jββ,detA∗A=∑β∈Jn,nA∗Aββwhenr=n,rdetjAA∗j.di.=∑α∈Im,mjrdetjAA∗j.di.αα,detAA∗=∑α∈Im,mAA∗ααwhenr=m,where a column vector d.j and a row vector di. have appropriate sizes.Remark 7.
First note thatA∗†=A†∗. Because of symbol equivalence, we shall use the denotation A†,∗≔(A∗)† as well. So, by Lemma 4, for the Hermitian adjoint matrix A∗∈Hrn×m determinantal representations of its Moore-Penrose inverse (A∗)†=aij∗†∈Hm×n are (12)aij∗†=aji†¯=∑α∈Ir,njrdetjA∗Aj.ai.αα∑β∈Ir,nA∗Aαα,and (13)aij∗†=∑β∈Jr,micdetiAA∗.ia.jββ∑β∈Jr,mAA∗ββ.Corollary 8.
IfA∈Hrm×n, then the projection matrix A†A≕PA=pijn×n has the determinantal representation (14)pij=∑β∈Jr,nicdetiA∗A.ia˙.jββ∑β∈Jr,nA∗Aββ,where a˙.j is the jth column of A∗A∈Hn×n.Corollary 9.
IfA∈Hrm×n, then the projection matrix AA†≕QA=qijm×m has the determinantal representation(15)qij=∑α∈Ir,mjrdetjAA∗j.a¨i.αα∑α∈Ir,mAA∗αα,where a¨i. is the ith row of AA∗∈Hm×m.The orthogonal projectorsLA≔I-A†A and RA≔I-AA† induced by A will be used below.Theorem 10 (see [44]).
LetA∈Hm×n, B∈Hr×s, C∈Hm×s be known and X∈Hn×r be unknown. Then the matrix equation (16)AXB=Cis consistent if and only if AA†CB†B=C. In this case, its general solution can be expressed as (17)X=A†CB†+LAV+WRB,where V, W are arbitrary matrices over H with appropriate dimensions.Theorem 11 (see [31]).
LetA∈Hr1m×n, B∈Hr2r×s. Then the partial solution X0=A†CB†=(xij0)∈Hn×r to (16) has determinantal representations, (18)xij0=∑β∈Jr1,nicdetiA∗A.id.jBββ∑β∈Jr1,nA∗Aββ∑α∈Ir2,rBB∗αα,or (19)xij0=∑α∈Ir2,rjrdetjBB∗j.di.Aαα∑β∈Jr1,nA∗Aββ∑α∈Ir2,rBB∗αα,where (20)d.jB=∑α∈Ir2,rjrdetjBB∗j.c~k.αα∈Hn×1,k=1,…,n,di.A=∑β∈Jr1,nicdetiA∗A.ic~.lββ∈H1×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∗CB∗.Corollary 12.
LetA∈Hkm×n, C∈Hm×s be known and X∈Hn×s be unknown. Then the matrix equation AX=C is consistent if and only if AA†C=C. In this case, its general solution can be expressed as X=A†C+LAV, where V is an arbitrary matrix over H with appropriate dimensions. The partial solution X0=A†C has the following determinantal representation, (21)xij0=∑β∈Jk,nicdetiA∗A.ic^.jββ∑β∈Jk,nA∗Aββ.where c^.j is the jth column of C^=A∗C.Corollary 13.
LetB∈Hkr×s, C∈Hn×s be given and X∈Hn×r be unknown. Then the equation XB=C is solvable if and only if C=CB†B and its general solution is X=CB†+WRB, where W is any matrix with conformable dimension. Moreover, its partial solution X=CB† has the determinantal representation, (22)xij=∑α∈Ik,rjrdetjBB∗j.c^i.αα∑α∈Ik,rBB∗αα.where c^i. is the ith row of C^=CB∗.


## 3. Cramer’s Rules for the Solution and Hermitian Solution to System (3)

First, consider the general system (1).Lemma 14 (see [5]).
LetA1∈Hm×n, B1∈Hr×s, C1∈Hm×s, A2∈Hk×n, B2∈Hr×p, C2∈Hk×p be given and X∈Hn×r is to be determined. Put H=A2LA1, N=RB1B2, T=RHA2, F=B2LN. Then system (1) is consistent if and only if (23)AiAi†CiBi†Bi=Ci,i=1,2;(24)TA2†C2B2†-A1†C1B1†F=0.In that case, the general solution to (1) can be expressed as the following: (25)X=A1†C1B1†+LA1H†A2LTA2†C2B2†-A1†C1B1†B2B2†+T†TA2†C2B2†-A1†C1B1†B2N†RB1+LA1Z-H†HZB2B2†-LA1H†A2LTWNB2†+W-T†TWNN†RB1,where Z and W are arbitrary matrices over H with compatible dimensions.Some simplification of (25) can be derived due to the quaternionic analogue of the following proposition.Lemma 15 (see [45]).
IfA∈Hn×n is Hermitian and idempotent, then the following equation holds for any matrix B∈Hm×n, (26)ABA†=BA†.It is evident that ifA∈Hn×n is Hermitian and idempotent, then the following equation is true as well: (27)AB†A=AB†.Since LA1, RB1, and RH are projectors, then by (26) and (27), we have, respectively, (28)LA1H†=LA1A2LA1†=A2LA1†=H†,N†RB1=RB1B2†RB1=RB1B2†=N†,T†T=RHA2†RHA2=RHA2†A2=T†A2,LT=I-T†T=I-T†A2.Using (28) and (23), we obtain the following expressions of (25): (29)X=A1†C1B1†+H†A2I-T†A2A2†C2B2†-A1†C1B1†B2B2†+T†A2A2†C2B2†-A1†C1B1†B2N†+LA1Z-H†HZB2B2†-H†A2LTWNB2†+W-T†TWNN†RB1=A1†C1B1†+H†C2B2†+H†A2T†-IA2A1†C1B1†QB2-H†A2T†C2B2†+T†C2N†-T†A2A1†C1B1†B2N†+LA1Z-H†HZB2B2†-H†A2LTWNB2†+W-T†TWNN†RB1.By putting Z, W as zero-matrices in (29), we obtain the following partial solution of (25): (30)X=A1†C1B1†+H†C2B2†+T†C2N†+H†A2T†A2A1†C1B1†QB2-H†A2A1†C1B1†QB2-H†A2T†C2B2†-T†A2A1†C1B1†B2N†.Now consider system (1). Since (31)PAi∗=Ai∗†Ai∗=AiAi†∗=QAi,so QAi∗=PAi, LAi∗=I-PAi∗=I-QAi=RAi, and RAi∗=LAi for i=1,2. Moreover, substituting Bi=Ai∗, we have N=RA1∗A2∗=A2LA1∗=H∗; similarly, F=A2∗LH∗=RHA2∗=T∗. Due to the above, we obtain the following analog of Lemma 14.Lemma 16.
LetA1∈Hm×n, A2∈Hk×n, C1∈Hm×m, C2∈Hk×k be given and X∈Hn×n is to be determined. Then system (3) is consistent if and only if (32)QAiCiQAi=Ci,i=1,2;TA2†C2A2†,∗-A1†C1A1†,∗T∗=0.In that case, the general solution to (3) can be expressed as follows: (33)X=A1†C1A1∗†+H†C2A2∗†+H†A2T†-IA2A1†C1A1∗†PA2-H†A2T†C2A2∗†+T†C2H∗†-T†A2A1†C1A1∗†A2∗H∗†+LA1Z-H†HZA2∗A2∗†-H†A2LTWH∗A2∗†+W-T†TWH∗H∗†LA1.Z and W are arbitrary matrices over H with compatible dimensions.By puttingZ, W as zero-matrices, the partial solution of (3) is (34)X=A1†C1A1∗†+H†C2A2∗†+T†C2H∗†+H†A2T†A2A1†C1A1∗†PA2-H†A2A1†C1A1∗†PA2-H†A2T†C2A2∗†-T†A2A1†C1A1∗†A2∗H∗†.Further we give determinantal representations of (34).LetA1∈Hr1m×n, A2∈Hr2k×n, C1=cij(1)∈Hm×m, C2=cij(2)∈Hk×k, rankH=r3, and rankT=r4. There exist A1†=aij(1),†∈Hn×m, (A1∗)†=aij(1),∗,†∈Hm×n, A2†=aij(2),†∈Hk×n, (A2∗)†=aij(2),∗,†∈Hn×k, H†=hij†∈Hn×k, and T†=tij†∈Hn×k.Consider each term of (34) separately.(i) DenoteC11≔A1∗C1A1. For the first term of (34) X1=A1†C1A1∗†=(xij(1)), we have (35)xij1=∑l=1m∑t=1mail1,†clt1atj1,∗,†.By using determinantal representations (9) and (13) of the Moore-Penrose inverses A1† and A1∗†, respectively, we obtain(36)xij1=∑l=1m∑t=1m∑β∈Jr1,nicdetiA1∗A1.ia.l1,∗ββclt1∑α∈Ir1,njrdetjA1∗A1j.at.1αα∑α∈Ir1,nA1∗A1αα∑β∈Jr1,nA1∗A1ββSuppose el. and e.l are the unit row vector and the unit column vector, respectively, such that all their components are 0, except the lth components, which are 1. Since ∑l=1m∑t=1mafl(1),∗clt(1)ats(1)=cfs(11), then (37)xij1=∑f=1n∑s=1n∑β∈Jr1,nicdetiA1∗A1.ie.fββcfs11∑α∈Ir1,njrdetjA1∗A1j.es.αα∑α∈Ir1,nA1∗A1αα∑β∈Jr1,nA1∗A1ββIf we denote by (38)vis1≔∑f=1n∑β∈Jr1,nicdetiA1∗A1.ie.fββcfs11=∑β∈Jr1,nicdetiA1∗A1.ic.s11ββthe sth component of a row vector vi.(1)=vi1(1),…,vin(1), then (39)∑s=1mvis1∑α∈Ir1,njrdetjA1∗A1j.es.αα=∑α∈Ir1,njrdetjA1∗A1j.vi.1αα.Further, it is evident that ∑β∈Jr1,nA1∗A1ββ=∑α∈Ir1,nA1∗A1αα, so the first term of (34) has the determinantal representation (40)xij1=∑α∈Ir1,njrdetjA1∗A1j.vi.1αα∑α∈Ir1,nA1∗A1αα2,where (41)vi.1=∑β∈Jr1,nicdetiA1∗A1.ic.s11ββ∈H1×n,s=1,…,n.If we denote by (42)vfj2≔∑s=1ncfs11∑α∈Ir1,njrdetjA1∗A1j.es.αα=∑α∈Ir1,njrdetjA1∗A1j.cf.11ααthe fth component of a column vector v.j(2)=v1j(2),…,vnj(2), then (43)∑f=1n∑β∈Jr1,nicdetiA1∗A1.ie.fββvfj2=∑β∈Jr1,nicdetiA1∗A1.iv.j2ββ.So, another determinantal representation of the first term of (34) is (44)xij1=∑β∈Jr1,nicdetiA1∗A1.iv.j2ββ∑β∈Jr1,nA1∗A1ββ2,where(45)v.j2=∑α∈Ir1,njrdetjA1∗A1j.cf.11αα∈Hn×1,f=1,…,n,are the column vector.(ii) Similarly above, for the second termX2H†C2A2∗†=xij(2) of (34), we have (46)xij2=∑β∈Jr3,nicdetiH∗H.id.jA2ββ∑β∈Jr3,nH∗Hββ∑α∈Ir2,nA2∗A2αα,or (47)xij2=∑α∈Ir2,njrdetjA2∗A2j.di.Hαα∑β∈Jr3,nH∗Hββ∑α∈Ir2,nA2∗A2αα,where (48)d.jA2=∑α∈Ir2,njrdetjA2∗A2j.cq.21αα∈Hn×1,q=1,…,n,di.H=∑β∈Jr3,nicdetiH∗H.ic.l21ββ∈H1×n,l=1,…,n,are the column vector and the row vector, respectively. cq.(21) and c.l(21) are the qth row and the lth column of C21=H∗C2A2. Note that H∗H=A2LA1∗A2LA1=LA1A2∗A2LA1.(iii) The third term of (34) can be obtained similarly as well. So, (49)xij3=∑β∈Jr4,nicdetiT∗T.id.jHββ∑β∈Jr4,nT∗Tββ∑α∈Ir3,nH∗Hαα,or (50)xij3=∑α∈Ir3,rjrdetjH∗Hj.di.Tαα∑β∈Jr4,nT∗Tββ∑α∈Ir3,nH∗Hαα,where (51)d.jH=∑α∈Ir3,nfrdetjH∗Hj.cq.22αα∈Hn×1,q=1,…,n,di.T=∑β∈Jr4,nicdetiT∗T.ic.l22ββ∈H1×n,l=1,…,n,are the column vector and the row vector, respectively. cq.(22) are the qth row, and c.l(22) are the lth column of C22=T∗C2H. The following expressions give some simplifications in computing. Since T∗T=RHA2∗RHA2=A2∗RH∗RHA2=A2∗RHA2 and RH=I-HH†=I-A2LA1A2LA1†=I-A2A2LA1†, then T∗T=A2∗I-A2A2LA1†A2.(iv) Using (9) for determinantal representations of H† and T† in the fourth term of (34), we obtain (52)xij4=∑q=1n∑z=1n∑f=1n∑β∈Jr3,nicdetiH∗H.ia.q2,Hββ∑β∈Jr4,nqcdetqT∗T.qa.z2,Tββxzf1pfj∑β∈Jr3,nH∗Hββ∑β∈Jr4,nT∗Tββ,where a.i(2,H), a.i(2,T) are the ith columns of the matrices H∗A2 and T∗A2, respectively; xzf(1) is the first term; pfj is the (fj)th element of PA2 with determinantal representation by (14) as (53)pfj=∑β∈Jr,nfcdetfA2∗A2.fa˙.j2ββ∑β∈Jr,nA2∗A2ββ,where a˙.j(2) is the jth column of A2∗A2. Note that H∗A2=LA1A2∗A2 and T∗A2=A2∗RHA2=A2∗I-A2A2LA1†A2.(v) Similarly to the previous case,(54)xij5=∑q=1n∑f=1n∑β∈Jr3,nicdetiH∗H.ia.q2,Hββxqf1pfj∑β∈Jr3,nH∗Hββ,(vi) Consider the sixth term by analogy to the fourth term. So,(55)xij6=∑q=1n∑β∈Jr3,nicdetiH∗H.ia.q2,Hββφqj∑β∈Jr3,nH∗Hββ∑β∈Jr4,nT∗Tββ∑β∈Jr2,nA2∗A2ββ,where (56)φqj=∑β∈Jr4,nqcdetqT∗T.qψ.jA2ββ,φqj=∑α∈Ir2,njrdetjA2∗A2j.ψq.Tαα,and (57)ψ.jA2=∑α∈Ir2,nfrdetjA2∗A2j.cq.23αα∈H1×n,q=1,…,n,ψq.T=∑β∈Jr4,nqcdetqT∗T.qc.l23ββ∈Hn×1,l=1,…,n,are the column vector and the row vector, respectively. cq.(23) and c.l(23) are the qth row and the lth column of C23=T∗C2A2.(vii) Using (9) for determinantal representations of T† and (12) for H∗,† in the seventh term of (34), we obtain (58)xij7=∑q=1n∑f=1n∑β∈Jr4,nicdetiT∗T.ia.q2,Tββxqf1∑α∈Ir3,njrdetjH∗Hj.af.2,H,∗αα∑β∈Jr4,nT∗Tββ∑α∈Ir3,rH∗Hαα,where a.q(2,T), af.(2,H,∗) are the qth column of T∗A2 and the fth row of A2∗H=A2∗A2LA1, respectively.Hence, we prove the following theorem.Theorem 17.
LetA1∈Hr1m×n, A2∈Hr2k×n, and rankH=rankA2LA1=r3, rankT=rankRHA2=r4. Then for the partial solution (34) to system (3), we have (59)xij=∑δxijδ,where the term xij(1) has the determinantal representations (40) and (44); xij(2), (46) and (47); xij(3), (49) and (50); xij(4), (52); xij(5), (54); xij(6), (55); and xij(7), (58).Due to Khatri and Mitra [46], the next lemma can be generalized to H.Lemma 18.
LetA∈Hm×n and B∈Hm×m and B=B∗ be known and X∈Hn×n be unknown. Then the matrix equation (60)AXA∗=Bhas a Hermitian solution if and only if QAB=B. In that case, the general Hermitian solution of (60) is (61)X=A†BA†,∗+LAV+V∗LA,where V∈Hn×n is any matrix.As it follows from the above, ifrankA=r, then the determinantal representation of the partial Hermitian solution X=A†BA†,∗=(xij) is (62)xij=∑α∈Ir,njrdetjA∗Aj.vi.1αα∑α∈Ir,nA∗Aαα2=∑β∈Jr,nicdetiA∗A1.iv.j2ββ∑β∈Jr,nA∗Aββ2,where (63)vi.1=∑β∈Jr,nicdetiA∗A.ib.s1ββ∈H1×n,s=1,…,n,v.j2=∑α∈Ir,njrdetjA∗Aj.bf.1αα∈Hn×1,f=1,…,n,are the row vector and the column vector and b.s(1) and bf.(1) are the sth column and the fth row of B1≔A∗BA, respectively.The general Hermitian solution to system (3) can be expressed as Y=1/2X+X∗, where X is an arbitrary solution of (3). Since by Lemma 18, the existence of Hermitian solutions (1) needs C1 and C2 to be Hermitian, then (64)X∗=xij∗=xji¯=A1†C1A1∗†+A2†C2H∗†+H†C2T∗†+PA2A1∗†C1A1†A2T∗†A2∗H∗†-PA2A1∗†C1A1†A2∗H∗†-A2†C2T∗†A2∗H∗†-H†A2A1†C1A1∗†A2∗T∗†.So, the determinantal representation of the partial Hermitian solution Y=(yij) can be obtained as yij=1/2xij+xji¯ for all i,j=1,…,n, where xij is determined by Theorem 17 and xji¯=∑δxji(δ)¯ so that(65)ixji1¯=xij1;iixji2¯=∑α∈Ir3,njrdetjH∗Hj.di.A2αα∑α∈Ir3,nH∗Hαα∑β∈Jr2,nA2∗A2ββ=∑β∈Jr2,nicdetiA2∗A2.id.jHββ∑α∈Ir3,nH∗Hαα∑β∈Jr2,nA2∗A2ββ,where (66)di.A2=∑β∈Jr2,nicdetiA2∗A2.ic.q21,∗βδ∈H1×n,q=1,…,n,d.jH=∑α∈Ir3,njrdetjH∗Hj.cl.21,∗αα∈Hn×1,l=1,…,n,are the row vector and the column vector, respectively. c.q(21,∗) and cl.(21,∗) are the qth column and the lth row of C21∗=A2∗C2H; (67)iiixji3¯=∑α∈Ir4,njrdetjT∗Tj.di.Hαα∑α∈Ir4,nT∗Tαα∑β∈Jr3,nH∗Hββ=∑β∈Jr3,ricdetiH∗H.id.jTββ∑α∈Ir4,nT∗Tαα∑β∈Jr3,nH∗Hββ,where (68)di.H=∑β∈Jr3,nicdetiH∗H.ic.q22,∗ββ∈H1×n,q=1,…,n,d.jT=∑α∈Ir4,njrdetjT∗Tj.cl.22,∗αα∈Hn×1,l=1,…,n,are the row vector and the column vector, respectively. c.q(22,∗) are the qth column and cl.(22,∗) are the lth row of C22∗=H∗C2T;(69)ivxji4¯=∑q=1n∑z=1n∑f=1npif¯xfz1∑α∈Ir4,nqrdetqT∗Tq.az.2,T,∗αα∑α∈Ir3,nirdetiH∗Hi.aq.2,H,∗αα∑α∈Ir3,nH∗Hαα∑α∈Ir4,nT∗Tαα,where az.(2,T,∗) is the z row of A2∗T and aq.(2,H,∗) is the qth row of A2∗H, respectively; xzf(1) can be obtained by (40) or (44); evidently, pif¯=pif and pif is the (if)th element of PA2 with determinantal representation by (14) as (70)pif=∑β∈Jr,nicdetiA2∗A2.ia˙.f2ββ∑β∈Jr,nA2∗A2ββ,where a˙.f(2) is the fth column of A2∗A2;(71)vxji5¯=∑q=1n∑f=1npifxfq1∑β∈Ir3,njrdetjH∗Hj.aq.2,H,∗αα∑α∈Ir3,nH∗Hαα;vixji06¯=∑q=1nηiq∑α∈Ir3,nirdetiH∗Hi.aq.2,H,∗αα∑α∈Ir3,nH∗Hαα∑α∈Ir4,nT∗Tαα∑β∈Ir2,nA2∗A2ββ,where (72)ηiq=∑α∈Ir4,nqrdetqT∗Tq.ζi.A2αα=∑β∈Jr2,nicdetiA2∗A2.iζ.qTββ,and (73)ζi.A2=∑β∈Jr2,nicdetiA2∗A2.ic.q23,∗ββ∈Hn×1,q=1,…,n,ζ.qT=∑α∈Ir4,nqrdetqT∗Tq.cl.23,∗αα∈H1×n,l=1,…,n,are the row vector and the column vector, respectively. c.q(23),∗ and cl.(23),∗ are the qth column and the lth row of C23∗=A2C2T∗;(74)viixji7¯=∑q=1n∑f=1n∑β∈Jr3,nicdetiH∗H.ia.f2,Hββxfg1∑α∈Ir4,njrdetjT∗Tj.aq.2,Tαα∑α∈Ir4,nT∗Tαα∑β∈Jr3,nH∗Hββ,where aq.(2,T,∗), a.f(2,H) are the qth row of A2∗T and the fth column of H∗A2, respectively.


## 4. An Example

In this section, we give an example to illustrate our results. Let us consider the system of matrix equations(75)A1XA1∗=C1,A2XA2∗=C2where (76)A1=0.52-ki2j-2i-2k,C1=0.5kii-k,A2=j-k1i1k,C2=-k1ij.Since (77)detA1A1∗=det1.5-2j2j3=0.5,detA2A2∗=det3-kk3=8,then rankA1=2, rankA2=2. By Theorem 5, one can find(78)A1†=2j-kiik,A1†A1=0.510001j0-j1,LA1=0.500001-j0j1,A2†=0.25-2j-2ik11-k,H=A2LA1=0.50j-k1-i01-i-j+k,H†=0.2500-j+k1+i1+ij-k,RH=0.51k-k1,T=RHA2=j00i00.Since QA1=I, QA2=I, TA2†C2A2∗,†T∗=TA1†C1A1∗,†T∗, then, by Lemma 16, system (75) is consistent.First, we can find the solution of (75) by direct calculation. Since X=∑δXδ, where (79)X1=A1†C1A1∗†=0.5k00000000,X2=H†C2A2∗†=0.1250002j+2kj+k-1+i-2+2i1+i-j-k,X3=T†C2H∗†=0.250j-k1-j000000,X4=H†A2T†A2A1†C1A1∗†PA2=0,X5=-H†A2A1†C1A1∗†PA2=0,X6=-H†A2T†C2A2∗†=0,X7=-T†A2A1†C1A1∗†A2∗H∗†=0,then (80)X=0.1254k2j-2k2-2j2j+2kj+k-1+i-2+2i1+i-j-k.Now, we find the solution of (75) by its determinantal representation by Theorem 17. Since (81)C11=A1∗C1A1=0.1254k1212j-129k-9i12i-9i-9k,A1∗A1=0.12516-12k12i12k1010j-12i-10j10,then by (41) (82)v111=∑β∈J2,31cdet1A1∗A1.1c.111ββ=cdet10.5k-1.5k-1.51.25+cdet10.5k1.5i1.5j1.25=0.5k.Similarly, v12(1)=0.375, v13(1)=0.375j. So, v1.(1)=0.5k0.3750.375j. Further, by (40), (83)x111=∑α∈I2,31rdet1A1∗A11.v1.1αα∑α∈I2,3A1∗A1αα2=4rdet10.5k0.3751.5k1.25+rdet10.5k0.375j-1.5j1.25=0.5k.Since (84)C21=H∗C2A2=000j+kj+k-1+i-1+i-1+i-j-k,C22=T∗C2H=0j-k1-i000000,H∗H=00001-j0j1,and rankH=rankT=1, then (85)di.H=000,d.jH=000T.So, by (47) and by (49), x11(2)=0 and x11(3)=0, respectively. Moreover, x11(δ)=0 for all δ=4,…,7. So, x11=x11(1)=0.5k.So,x11 obtained by Cramer’s rule and the matrix method (80) are equal.Similarly, we can obtain for all the remainder solutions.Note that we used Maple with the package CLIFFORD in the calculations.


## 5. Conclusions

Within the framework of the theory of quaternion row-column determinants previously introduced by the author, we have derived determinantal representations (analogs of Cramer’s rule) of the general and Hermitian solutions to the system of two-sided quaternion matrix equationsA1XA1∗=C1 and A2XA2∗=C2. Since the Hermitian solution is Y=1/2X+X∗, where X is an arbitrary solution, the determinantal representation of X∗ is derived as well. To accomplish that goal, we have used the determinantal representations of the Moore-Penrose matrix inverse which were previously introduced by the author.



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