1. Introduction
In this paper, we consider nonhomogeneous linear system in n variables
(1)Ax=c,
where A is an m×n matrix over the field ℂ of rank a and c is an m×1 matrix over ℂ. The set of all m×n matrices over the complex field ℂ will be denoted by ℂm×n, m,n∈ℕ. The set of all m×n matrices over the complex field ℂ of rank a will be denoted by ℂam×n. For simplicity of notation, we will write Ai→ (A↓j) for the ith row (the jth column) of the matrix A∈ℂm×n.

Any matrix X satisfying the equality AXA=A is called {1}-inverse of A and is denoted by A(1). The set of all {1}-inverses of the matrix A is denoted by A{1}. It can be shown that A{1} is not empty. If the n×n matrix A is invertible, then the equation AXA=A has exactly one solution A-1, so the only {1}-inverse of the matrix A is its inverse A-1; that is, A{1}={A-1}. Otherwise, {1}-inverse of the matrix A is not uniquely determined. For more information about {1}-inverses and various generalized inverses, we recommend Ben-Israel and Greville [1] and Campbell and Meyer [2].

For each matrix A∈ℂam×n there are regular matrices P∈ℂn×n and Q∈ℂm×m such that
(2)QAP=Ea=[Ia000],
where Ia is a × a identity matrix. It can be easily seen that every {1}-inverse of the matrix A can be represented in the form
(3)A(1)=P[IaUVW]Q,
where U=[uij], V=[vij], and W=[wij] are arbitrary matrices of corresponding dimensions a×(m-a), (n-a)×a, and (n-a)×(m-a) with mutually independent entries; see Rohde [3] and Perić [4].

We will generalize the results of Urquhart [5]. Firstly, we explore the minimal numbers of free parameters in Penrose's formula
(4)x=A(1)c+(I-A(1)A)y
for obtaining the general solution of the system (1). Then, we consider relations among the elements of A(1) to obtain the general solution in the form x=A(1)c of the system (1) for c≠0. This construction has previously been used by Malešević and Radičić [6] (see also [7] and [8]). At the end of this paper, we will give an application of this results to the matrix equation AXB=C.

2. The Main Result
In this section, we indicate how a technique of an {1}-inverse may be used to obtain the necessary and sufficient condition for an existence of a general solution of a nonhomogeneous linear system.

Lemma 1.
The nonhomogeneous linear system (1) has a solution if and only if the last m-a coordinates of the vector c′=Qc are zeros, where Q∈ℂm×m is regular matrix such that (2) holds.

Proof.
The proof follows immediately from Kronecker-Capelli theorem. We provide a new proof of the lemma by using the {1}-inverse of the system matrix A. The system (1) has a solution if and only if c=AA(1)c; see Penrose [9]. Since A(1) is described by (3), it follows that
(5)AA(1)=AP[IaUVW]Q=Q-1[IaU00]Q.
Hence, we have the following equivalences:
(6)c=AA(1)c⟺(I-AA(1))c=0 ⟺(Q-1Q-Q-1[IaU00]Q)c=0⟺Q-1[0-U0In-a]Qc︸c′=0⟺[0-U0In-a]c′=0⟺c′=[ca′ cn-a′][0-U0In-a][ca′cn-a′]=0⟺[-Ucn-a′cn-a′]=0⟺cn-a′=0.
Furthermore, we conclude that c=AA(1)c⇔cn-a′=0.

Theorem 2.
The vector
(7)x=A(1)c+(I-A(1)A)y,y∈ℂn×1 is an arbitrary column, is the general solution of the system (1), if and only if the {1}-inverse A(1) of the system matrix A has the form (3) for arbitrary matrices U and W and the rows of the matrix V(ca′-ya′)+y(n-a)′ are free parameters, where Qc=c′=[ca′/0]T and P-1y=y′=[ya′/yn-a′]T.

Proof.
Since {1}-inverse A(1) of the matrix A has the form (3), the solution of the system x=A(1)c+(I-A(1)A)y can be represented in the form
(8)x=P[IaUVW]Qc+(I-P[IaUVW]QA)y=P[IaUVW]c′+(I-P[IaUVW]QAPP-1)y.
According to Lemma 1 and from (2), we have
(9)x=P[IaUVW][ca′0]+(I-P[IaUVW][Ia000]P-1)y.
Furthermore, we obtain
(10)x=P[ca′Vca′]+(I-P[Ia0V0]P-1)[yayn-a]=P[ca′Vca′]+(PP-1-P[Ia0V0]P-1)[yayn-a]=P[ca′Vca′]+P(I-[Ia0V0])P-1[yayn-a]=P[ca′Vca′]+P[00-VIn-a][ya′yn-a′],
where y′=P-1y. We now conclude that
(11)x=P([ca′Vca′]+[0-Vya′+yn-a′])=P[ca′V(ca′-ya′)+yn-a′].
Therefore, since matrix P is regular, we deduce that P[ca′/(V(ca′-ya′)+yn-a′)]T is the general solution of the system (1) if and only if the rows of the matrix V(ca′-ya′)+yn-a′ are n-a free parameters.

Corollary 3.
The vector
(12)x=(I-A(1)A)y,y∈ℂn×1 is an arbitrary column, is the general solution of the homogeneous linear system Ax=0, A∈ℂm×n, if and only if the {1}-inverse A(1) of the system matrix A has the form (3) for arbitrary matrices U and W and the rows of the matrix -Vya′+y(n-a)′ are free parameters, where P-1y=y′=[ya′/yn-a′]T.

Example 4.
Consider the homogeneous linear system
(13)x1+2x2+3x3=04x1+5x2+6x3=0.
The system matrix is
(14)A=[123456].
For regular matrices
(15)Q=[10-41], P=[12310-13-2001],
equality (2) holds. Rohde's general {1}-inverse A(1) of the system matrix A is of the form
(16)A(1)=P[1001v11v12]Q.
According to Corollary 3 the general solution of the system (13) is of the form
(17)x=P[000000-v11-v121]P-1[y1y2y3],
where
(18)P-1=[1230-3-6001].
Therefore, we obtain
(19)x=P[000000-v11-v121][y1+2y2+3y3-3y2-6y3y3]=P[00-v11y1-(2v11-3v12)y2-(3v11-6v12-1)y3].
If we take α=-v11y1-(2v11-3v12)y2-(3v11-6v12-1)y3 as a parameter, we get the general solution
(20)x=[12310-13-2001][00α]=[α-2αα].

Corollary 5.
The vector
(21)x=A(1)c
is the general solution of the system (1), if and only if the {1}-inverse A(1) of the system matrix A has the form (3) for arbitrary matrices U and W and the rows of the matrix Vca′ are free parameters, where Qc=c′=[ca′/0]T.

Remark 6.
Similar result can be found in the paper by Malešević and Radičić [6].

Example 7.
Consider the nonhomogeneous linear system
(22)x1+2x2+3x3=74x1+5x2+6x3=8.
According to Corollary 5, the general solution of the system (22) is of the form
(23)x=P[1001v11v12]Q[78]=P[7-207v11-20v12].
If we take α=7v11-20v12 as a parameter, we obtain the general solution of the system
(24)x=P[7-20α]=[12310-13-2001][7-20α]=[-193+α203-2αα].

We are now concerned with the matrix equation
(25)AX=C,
where A∈ℂm×n, X∈ℂn×k, and C∈ℂm×k.

Lemma 8.
The matrix equation (25) has a solution if and only if the last m-a rows of the matrix C′=QC are zeros, where Q∈ℂm×m is regular matrix such that (2) holds.

Proof.
If we write X=[X↓1X↓2⋯X↓k] and C=[C↓1C↓2⋯C↓k], then we can observe the matrix equation (25) as the system of matrix equations
(26)AX↓1=C↓1AX↓2=C↓2⋮AX↓k=C↓k.
Each of the matrix equation AX↓i=C↓i, 1≤i≤k, by Lemma 1 has solution if and only if the last m-a coordinates of the vector C↓i′=QC↓i are zeros. Thus, the previous system has solution if and only if the last m-a rows of the matrix C′=QC are zeros, which establishes that matrix equation (25) has solution if and only if all entries of the last m-a rows of the matrix C′ are zeros.

Theorem 9.
The matrix
(27)X=A(1)C+(I-A(1)A)Y∈ℂn×k,Y∈ℂn×k is an arbitrary matrix, is the general solution of the matrix equation (25) if and only if the {1}-inverse A(1) of the system matrix A has the form (3) for arbitrary matrices U and W and the entries of the matrix
(28)V(Ca′-Ya′)+Y(n-a)′
are mutually independent free parameters, where QC=C′=[Ca′/0]T and P-1Y=Y′=[Ya′/Yn-a′]T.

Proof.
Applying Theorem 2 on each system AX↓i=C↓i, 1≤i≤k, we obtain that
(29)X↓i=P[Ca↓i′V(Ca↓i′-Ya↓i′)+Yn-a↓i′]
is the general solution of the system if and only if the rows of the matrix V(Ca↓i′-Y′a↓i)+Yn-a↓i′ are n-a free parameters. Assembling these individual solutions together we get that
(30)X=P[Ca′V(Ca′-Ya′)+Yn-a′]
is the general solution of matrix equation (25) if and only if entries of the matrix V(Ca′-Ya′)+Yn-a′ are (n-a)k mutually independent free parameters.

From now on we proceed with the study of the nonhomogeneous linear system of the form
(31)xB=d,
where B is an n×m matrix over the field ℂ of rank b and d is a 1×m matrix over ℂ. Let R∈ℂn×n and let S∈ℂm×m be regular matrices such that
(32)RBS=Eb=[Ib000].
An {1}-inverse of the matrix B can be represented in Rohde's form
(33)B(1)=S[IbMNK]R,
where M=[uij], N=[vij], and K=[wij] are arbitrary matrices of corresponding dimensions b×(n-b), (m-b)×b and (m-b)×(n-b) with mutually independent entries.

Lemma 10.
The nonhomogeneous linear system (31) has a solution if and only if the last m-b elements of the row d′=dS are zeros, where S∈ℂm×m is regular matrix such that (32) holds.

Proof.
By transposing the system (31), we obtain system BTxT=dT and by transposing matrix equation (32) we obtain that STBTRT=Eb. According to Lemma 1, the system BTxT=dT has solution if and only if the last m-b coordinates of the vector STdT are zeros, that is, if and only if the last m-b elements of the row d′=dS are zeros.

Theorem 11.
The row
(34)x=dB(1)+y(I-BB(1)),y∈ℂ1×n is an arbitrary row, is the general solution of the system (31), if and only if the {1}-inverse B(1) of the system matrix B has the form (33) for arbitrary matrices N and K and the columns of the matrix (db′-yb′)M+yn-b′ are free parameters, where dS=d′=[db′ | 0] and yR-1=y′=[yb′ | yn-b′].

Proof.
The basic idea of the proof is to transpose the system (31) and to apply Theorem 2. The {1}-inverse of the matrix BT is equal to a transpose of the {1}-inverse of the matrix B. Hence, we have
(35)(BT)(1)=(B(1))T=(S[IbMNK]R)T=RT[IbNTMTKT]ST.
We can now proceed analogously to the proof of Theorem 2 to obtain that
(36)xT=RT[d′bTMT(db′T_ -yb′T)+yn-b′T]
is the general solution of the system BTxT=dT if and only if the rows of the matrix MT(db′T-yb′T)+yn-b′T are n-b free parameters. Therefore,
(37)x=[db′∣(db′-yb′)M+yn-b′]R
is the general solution of the system (31) if and only if the columns of the matrix (db′-yb′)M+yn-b′ are n-b free parameters. Analogous corollaries hold for Theorem 11.

We now deal with the matrix equation
(38)XB=D,
where X∈ℂk×n, B∈ℂn×m, and D∈ℂk×m.

Lemma 12.
Matrix equation (38) has a solution if and only if the last m-b columns of the matrix D′=DS are zeros, where S∈ℂm×m is regular matrix such that (32) holds.

Theorem 13.
The matrix
(39)X=DB(1)+Y(I-BB(1))∈ℂk×n,Y∈ℂk×n is an arbitrary matrix, is the general solution of the matrix equation (38) if and only if the {1}-inverse B(1) of the system matrix B has the form (33) for arbitrary matrices N and K and the entries of the matrix
(40)(Db′-Yb′)M+Y(n-b)′
are mutually independent free parameters, where DS=D′=[Db′ | 0] and YR-1=Y′=[Yb′ | Yn-b′].

3. An Application
In this section we will briefly sketch properties of the general solution of matrix equation
(41)AXB=C,
where A∈ℂm×n, X∈ℂn×k, B∈ℂk×l, and C∈ℂm×l. If we denote by Y matrix product XB, then the matrix equation (41) becomes
(42)AY=C.
According to Theorem 9, the general solution of the system (42) can be presented as a product of the matrix P and the matrix which has the first a=rank(A) rows same as the matrix QC and the elements of the last m-a rows are (m-a)n mutually independent free parameters; P and Q are regular matrices such that QAP=Ea. Thus, we are now turning on to the system of the form
(43)XB=D.
By Theorem 13, we conclude that the general solution of the system (43) can be presented as a product of the matrix which has the first b=rank(B) columns equal to the first b columns of the matrix DS and the rest of the columns have mutually independent free parameters as entries, and the matrix R, for regular matrices R and S such that RBS=Eb. Therefore, the general solution of the system (41) is of the form
(44)X=P[GabFHL]R,
where Gab is a submatrix of the matrix QCS corresponding to the first a rows and the first b columns and the entries of the matrices F, H, and L are nk-ab free parameters. We will illustrate this on the following example.

Example 14.
We consider the matrix equation
(45)AXB=C,
where A=[1-2-24], B=[121121121] and C=[121-2-4-2]. If we take Y=XB, we obtain the system
(46)AY=C.
It is easy to check that the matrix A is of the rank a=1 and for matrices Q=[1021] and P=[1201] the equality QAP=Ea holds. Based on Theorem 9, the equation AY=C can be rewritten in the system form
(47)AY↓1=[1-2],AY↓2=[2-4],AY↓3=[1-2].
Combining Theorem 2 with the equality
(48)[c1′c2′c3′000]=[1021][121-2-4-2]=[121000]
yields
(49)Y↓1=P[1v-vz11+z21︸α],Y↓2=P[22v-2vz12+z22︸β],Y↓3=P[1v-vz13+z23︸γ],
for an arbitrary matrix Z=[z11z12z13z21-z22z23]. Therefore, the general solution of the system AY=C is
(50)Y=P[121αβγ].
From now on, we consider the system
(51)XB=D
for
(52)D=P[121αβγ]=[1+2α2+2β1+2γαβγ].
There are regular matrices R=[100-110-101] and S=[1-2-1010001] such that RBS=Eb holds. Since the rank of the matrix B is b=1, according to Lemma 12 all entries of the last two columns of the matrix D′=DS are zeros; that is, we have γ=α, β=2α. Hence, we get that the matrix D′ is of the form D′=[1+2α00α00]. Applying Theorem 13, we obtain
(53)X=[1+2α(1+2α-t11)m11+t12︸β1(1+2α-t11)m12+t13︸β2α(α-t21)m11+t22︸γ1(α-t12)m12+t23︸γ2]R,
for an arbitrary matrix T=[t11t12t13t21t22t23]. Finally, the solution of the system AXB=C is
(54)X=[1+2α-β1-β2β1β2α-γ1-γ2γ1γ2].