# Least-Squares Mirrorsymmetric Solution for Matrix Equations (AX=B, XC

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```					NUMERICAL MATHEMATICS                                                          Vol. 15 No. 3, pp. 217-226
A Journal of Chinese Universities (English Series)                             August 2006

Least-Squares Mirrorsymmetric Solution for
Matrix Equations (AX = B, XC = D)
Fanliang Li1,2,∗, Xiyan Hu2 and Lei Zhang2
1
Department of Mathematics and Computational Science, Hunan City College, Yi
Yang 413000, China.
2
College of Mathematics and Econometrics, Hunan University, Changsha 410082,
China.
Received June 9, 2005; Accepted (in revised version) October 24, 2005

Abstract. In this paper, least-squares mirrorsymmetric solution for matrix equations (AX =
B, XC = D) and its optimal approximation is considered. With special expression of mir-
rorsymmetric matrices, a general representation of solution for the least-squares problem is
obtained. In addition, the optimal approximate solution and some algorithms to obtain the
optimal approximation are provided.

Key words: Mirrorsymmetric matrix; least-squares solution; optimal approximation.

AMS subject classiﬁcations: 15A06; 15A57

1     Introduction
Matrix equations (AX = B, XC = D) is a class of important matrix equations. The problem of
the solution of the matrix equations (AX = B, XC = D) arise in engineering and in some special
matrix inverse problems [1-3]. Many authors have been devoted to the study of this problem, and
a series of useful results have been obtained. For example, Mitra [4] gave the common solution of
minimum possible rank by using generalized inverse of matrix. Chu [5], Mitra [6] presented the
necessary and suﬃcient conditions for the solvability and general solution by using the singular
value decomposition (SVD) and generalized inverse of matrix, respectively. When the solution
matrix X is constrained and the matrix equations are not consistent, it is necessary to study the
least-squares constrained solution of the corresponding matrix equations.
The purpose of this paper is to discuss the least-squares mirrorsymmetric solution of the
matrix equations (AX = B, XC = D) by using the special structure of mirrorsymmetric ma-
trices. The background for introducing the deﬁnition of mirrorsymmetric matrices is to study
odd/even-mode decomposition of symmetric multiconductor transmission lines (MTL)[7].
We now introduce some notation. Let Rn×m be the set of all n × m real matrices; Rr       n×m
n×m                                                 n×n
be the set of all matrices in R        with rank r, R(A) denote the rank of A; OR          denote

∗ Correspondenceto: Fanliang Li, Department of Mathematics and Computational Science, Hunan City College,
Yi Yang 413000, China/College of Mathematics and Econometrics, Hunan University, Changsha 410082, China.
E-mail: lfl302@tom.com

Numer. Math. J. Chinese Univ. (English Ser.)              217               http://www.global-sci.org/nm
218                    Least-Squares Mirrorsymmetric Solution for Matrix Equations

the set of all n × n orthogonal matrices; The identity matrix of order n by In ; AT and A+
be the transpose and the Moore-Penrose generalized inverse of A, respectively. For A = (aij ),
B = (bij ) ∈ Rn×m , A ∗ B = (aij bij ) denote the Hadamard product of matrices A and B;
(A, B) = trB T A denote the inner product of matrices A and B. The induced matrix norm is
1              1
Frobenius norm, i.e ||A|| = (A, A) 2 = (tr(AT A)) 2 .

Deﬁnition 1.1. The (k, p)-mirror matrix W(k,p) is deﬁned by
                  
Jk
W(k,p) =         Ip         ,                            (1)
Jk

where Jk is the k-square backward identity matrix with ones along the secondary diagonal and
zero elsewhere.

The dimension of the (k, p)-mirror matrix is n = 2k + p, where k ≥ 1, p ≥ 0. The (k, p)-mirror
matrix W(k,p) is orthogonal and symmetric, i.e. W −1 = W T = W . When p = 0 or 1, mirror
matrix W(k,p) is backward identity matrix Jn .

Deﬁnition 1.2. A matrix A ∈ R(2k+p)×(2k+p) is called the (k, p)-mirrorsymmetric matrix if and
only if
A = W(k,p) AW(k,p) .                                     (2)

We denote the set of all (k, p)-mirrorsymmetric matrices by M S(k,p) . A matrix A ∈ R(2k+p)×(2k+p)
is called the (k, p)-mirrorskewsymmetric matrix if and only if

A = −W(k,p) AW(k,p) .                                   (3)

We denote the set of all (k, p)-mirrorskewsymmetric matrices by M SS(k,p) .

From Deﬁnition 1.2, it is easy to see that the (k, 1)-mirrorsymmetric matrices and the (k, 0)-
mirrorsymmetric matrices are centrosymmetric matrices. That is to say, all centrosymmetric
matrices are the special cases of mirrorsymmetric matrices, i.e. when p = 0 or 1, mirror matrix
W(k,p) is the backward identity matrix Jn . Then (2) becomes A = Jn AJn ,which is the deﬁnition
of centrosymmetric matrices [8].
We study the following problems in this paper.
Problem I. Given A ∈ Rh×(2k+p) , B ∈ Rh×(2k+p) , C ∈ R(2k+p)×l , D ∈ R(2k+p)×l , ﬁnding
∆
X ∈ M S(k,p) such that f (X) = ||AX − B||2 + ||XC − D||2 = min.

Problem II.                                       ˜
Given X ∗ ∈ R(2k+p)×(2k+p) ﬁnding X ∈ SE such that
˜
X ∗ − X = min            X∗ − X ,
∀X∈SE

where SE is the solution set of Problem I.
The paper is organized as follows. In Section 2, we at ﬁrst discuss the structure of mirrorsym-
metric matrices. Then with the special structure of mirrorsymmetric matrices and the singular
value decomposition (SVD) of matrix, we obtain the solution set of Problem I. In Section 3,
the unique approximation solution of Problem II is presented by applying the decomposition
of space. Finally, some algorithms to obtain the optimal approximate solution and numerical
experiment to illustrate the results obtained in this paper correction are provided.
F. Li, X. Hu and L. Zhang                                      219

2    The solution of problem I
Lemma 2.1. A ∈ M S(k,p) if and only if
                                           
Ak×k               Bk×p        Ck×k Jk
A =  Dp×k                 Ep×p        Dp×k Jk  .                         (4)
Jk Ck×k            Jk Bk×p     Jk Ak×k Jk

The above result follows directly from Deﬁnition 1.2. Hence the proof is omitted. From
Deﬁnition 1.1, 1.2 and Lemma 2.1, it is easy to derive the following lemma.
Lemma 2.2. A ∈ M S(k,p) if and only if

A1    0
A=K                  KT ,                                    (5)
0     A2

where
√
Ak×k + Ck×k      2Bk×p
A1 =      √                         ∈ R(k+p)×(k+p) ,       A2 = Jk (Ak×k − Ck×k )Jk ∈ Rk×k ,
2Dp×k         Ep×p
                    
1
Ik            √0      −Jk
K = √ 0               2Ip      0 .                              (6)
2 J              0       Ik
k

Lemma 2.3. R(2k+p)×(2k+p) = M S(k,p)          M SS(k,p) .

To prove Lemma 2.3, we only prove that for any A ∈ R(2k+p)×(2k+p) , there exist a unique
A1 ∈ M S(k,p) , and a unique A2 ∈ M SS(k,p) such that A1 + A2 = A and (A1 , A2 ) = 0.
Proof    1) For any A ∈ R(2k+p)×(2k+p) , there exist A1 ∈ M S(k,p) , A2 ∈ M SS(k,p) such that

A = A1 + A2 ,                                            (7)
1
where A1 = 2 (A + W(k,p) AW(k,p) ), A2 = 1 (A − W(k,p) AW(k,p) ).
2
¯               ¯
2) If there exist another A1 ∈ M S(k,p) , A2 ∈ M SS(k,p) such that
¯    ¯
A = A1 + A2 .                                            (8)

Combining (7) and (8) gives
¯           ¯
A1 − A1 = −(A2 − A2 ).                                       (9)
Multiplying (9) on the left and the right by W(k,p) , respectively, and using the deﬁnition of A1 ,
¯ ¯
A2 , A1 , A2 , we obtain
¯            ¯
A1 − A1 = (A2 − A2 ).                                    (10)
¯        ¯
It follows from (9) and (10) that A1 = A1 , A2 = A2 .
3) For any A1 ∈ M S(k,p) , A2 ∈ M SS(k,p) ,

(A1 , A2 ) =    tr(AT A1 ) = tr(AT W(k,p) W(k,p) A1 )
2            2
=    tr(W(k,p) AT W(k,p) W(k,p) A1 W(k,p) )
2
=    tr(−AT A1 ) = −tr(AT A1 ).
2             2
220                      Least-Squares Mirrorsymmetric Solution for Matrix Equations

Hence (A1 , A2 ) = tr(AT A1 ) = 0. Combining 1), 2) and 3), we have
2

R(2k+p)×(2k+p) = M S(k,p)                    M SS(k,p) .

Let A ∈ Rh×(2k+p) , B ∈ Rh×(2k+p) , C ∈ R(2k+p)×l , D ∈ R(2k+p)×l , K be given by (6).
Denote
C1                    D1
AK = (A1 , A2 ),      BK = (B1 , B2 ),             KT C =             ,        KT D =       ,    (11)
C2                    D2

A1 ∈ Rh×(k+p) ,       B1 ∈ Rh×(k+p) ,               C1 ∈ R(k+p)×l ,          D1 ∈ R(k+p)×l ,
A2 ∈ Rh×k ,      B2 ∈ Rh×k ,               C2 ∈ Rk×l ,          D2 ∈ Rk×l .
Denote the SVDs of A1 , A2 , C1 , C2 as follows, respectively.
T
Σ1    0                    Σ1                  0      V11
A1 = U1              V T = (U11 , U12 )                                                T
= U11 Σ1 V11 ,
0     0 1                  0                   0        T
V12

U1 = (U11 , U12 ) ∈ ORh×h ,                V1 = (V11 , V12 ) ∈ OR(k+p)×(k+p) ,              (12)
h×r1                    (k+p)×r1
U11 ∈ R             ,     V11 ∈ R                  ,
Σ1 = diag(σ1 , · · · , σr1 ),     σi > 0,          i = 1, · · · , r1 ,    r1 = R(A1 ).
T
Σ2    0                    Σ2                  0      V21
A2 = U2              V T = (U21 , U22 )                                                T
= U21 Σ2 V21 ,
0     0 2                  0                   0        T
V22
U2 = (U21 , U22 ) ∈ ORh×h ,                   V2 = (V21 , V22 ) ∈ ORk×k ,               (13)
U21 ∈ Rh×r2 ,                 V21 ∈ Rk×r2 ,
Σ2 = diag(δ1 , · · · , δr2 ), δi > 0, i = 1, · · · , r2 , r2 = R(A2 ).
τ1    0                   τ                    0      QT
C1 = P1             QT = (P11 , P12 ) 1
1
11
= P11 τ1 QT ,
11
0     0                   0                    0      QT
12

P1 = (P11 , P12 ) ∈ OR(k+p)×(k+p) ,                  Q1 = (Q11 , Q12 ) ∈ ORl×l ,             (14)
(k+p)×s1                       l×s1
P11 ∈ R                   ,     Q11 ∈ R            ,
τ1 = diag(α1 , · · · , αs1 ), αi > 0, i = 1, · · · , s1 , s1 = R(C1 ).
τ2    0                   τ                    0      QT
C2 = P2             QT = (P21 , P22 ) 2
2
21
= P21 τ2 QT ,
21
0     0                   0                    0      QT
22

P2 = (P21 , P22 ) ∈ ORk×k ,                   Q2 = (Q21 , Q22 ) ∈ ORl×l ,                (15)
P21 ∈ Rk×s2 ,               Q21 ∈ Rl×s2 ,
τ2 = diag(β1 , · · · , βs2 ), βi > 0, i = 1, · · · , s2 , s2 = R(C2 ).

Lemma 2.4 ([9]). If B ∈ Rn×n , C ∈ Rn×n , Σ = diag(α1 , · · · , αn ) ∈ Rn×n , τ = diag(β1 , · · · , βn ) ∈
Rn×n , then the solution for problem XΣ − B 2 + τ X − C 2 = minX∈Rn×n can be expressed as
1
X = Φ ∗ (BΣ + τ C), where Φ = (ϕij ), ϕij =                                   2.
α2 + βi
j
F. Li, X. Hu and L. Zhang                                                              221

Lemma 2.5. If A1 , B1 , C1 , D1 are given by (11), and the SV Ds of A1 , C1 be (12) and (14)
respectively. Then the general representation of solution for problem A1 X − B1 2 + XC1 −
D1 2 = minX∈R(k+p)×(k+p) can be expressed as

Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12
T              T
1
T
T
X = V1                      T         −1                            P1 ,                                           (16)
V12 D1 Q11 τ1                   X22

1
∀ X22 ∈ R(k+p−r1 )×(k+p−s1 ) ,               Φ = (ϕij ),          ϕij =              ,            1 ≤ i ≤ r1 ,     1 ≤ j ≤ s1 .
2
σi + α2
j

Proof     From (12) and (14), we have
2                          2
A1 X − B1           + XC1 − D1
2                                             2
Σ1     0                                          τ1   0
=       U1              V T X − B1                + XP1                 QT − D1
0      0 1                                        0    0  1

2                                                 2
Σ1      0                                                            τ1     0
=                              T
V T XP1 − U1 B1 P1                     + V1T XP1                    − V1T D1 Q1          .
0       0 1                                                          0      0

Denote

X11        X12                                B11      B12                               D11   D12
V1T XP1 =                    ,         T
U1 B1 P1 =                         ,          V1T D1 Q1 =                  .
X21        X22                                B21      B22                               D21   D22

Hence
2                        2
A1 X − B1         + XC1 − D1
2                                                                2
Σ1    0     X11       X12           B11        B12                  X11         X12      τ1        0   D11         D12
=                                      −                              +                                       −
0     0     X21       X22           B21        B22                  X21         X22      0         0   D21         D22
=      Σ1 X11 − B11 2 + X11 τ1 − D11 2 + Σ1 X12 − B12                                2
+ X21 τ1 − D21        2

+ B21 2 + B22 2 + D12 2 + D22 2 .
2                        2
So A1 X − B1         + XC1 − D1               = minX∈R(k+p)×(k+p) if and only if
                         2                             2
     Σ1 X11 − B11            + X11 τ1 − D11                =         min      ,
X11 ∈Rr1 ×s1

2
Σ1 X12 − B12            = min,
2

X21 τ1 − D21            = min .


From Lemma 2.4, we have
T             T
X11 = Φ ∗ (Σ1 B11 + D11 τ1 ) = Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ),

1
Φ = (ϕij ),        ϕij =              ,               1 ≤ i ≤ r1 ,         1 ≤ j ≤ s1 .
2
σi + α2
j

X12 = Σ−1 B12 = Σ−1 U11 B1 P12 , X21 = D21 τ1 = V12 D1 Q11 τ1 .
1         1
T                            T

This completes the proof of Lemma 2.5.
222                      Least-Squares Mirrorsymmetric Solution for Matrix Equations

Theorem 2.1. If A, B, C, D are given by (11). Let the SVDs of A1 , A2 , C1 , C2 be (12)-(15),
respectively. Then Problem I has a solution in M S(k,p) . Moreover, the general solution can be
expressed as
X1 0
X=K                 KT ,                                (17)
0 X2
where
Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12
T              T
1
T
T
X1 = V1                T         −1                            P1 ,                                           (18)
V12 D1 Q11 τ1                   X22
1
∀ X22 ∈ R(k+p−r1 )×(k+p−s1 ) ,         Φ = (ϕij ),         ϕij =      2        ,           1 ≤ i ≤ r1 ,       1 ≤ j ≤ s1 ,
σi   + α2
j

Φ′ ∗ (Σ2 U21 B2 P21 + V21 D2 Q21 τ2 ) Σ−1 U21 B2 P22
T             T
2
T
T
X2 = V2                T        −1                      ′      P2 ,                                           (19)
V22 D2 Q21 τ2                    X22
1
∀ X22 ∈ R(k−r2 )×(k−s2 ) ,
′
Φ′ = (ϕ′ ),
ij          ϕ′ =
ij         2      2,           1 ≤ i ≤ r2 ,          1 ≤ j ≤ s2 .
δi   + βj

Proof For any X ∈ M S(k,p) , from Lemma 2.2, X can be expressed as

X1    0
X=K                    KT .                                                        (20)
0     X2

From (11), (20), we have
2                  2
f (X) =      AX − B           + XC − D
2                                                 2
X1        0                                  X1      0
=    AK                  KT − B              + K                    KT C − D
0         X2                                 0       X2
2                                                     2
X1        0                            X1    0
=    AK                  − BK            +                  KT C − KT D
0         X2                           0     X2
2                                                2
X1       0                                     X1   0           C1           D1
=    (A1 , A2 )                   − (B1 , B2 )          +                                  −
0        X2                                    0    X2          C2           D2
2                        2                          2
=    A1 X1 − B1           + A2 X2 − B2             + X1 C1 − D1                 + X2 C2 − D2 2 .

Hence, there exists X ∈ M S(k,p) such that f (X) = minX∈MS(k,p) is equivalent to the fact that
there exist X1 ∈ R(k+p)×(k+p) and X2 ∈ Rk×k satisfying
2                           2
A1 X1 − B1            + X1 C1 − D1                =        min              ,
X1 ∈R(k+p)×(k+p)
2                           2
A2 X2 − B2            + X2 C2 − D2                =     min      .
X2 ∈Rk×k

According to Lemma 2.5, the general solution of X1 can be expressed by (18), and the general
solution of X2 can be expressed by (19). Substituting (18) and (19) into (20) gives (17).

3      The solution of Problem II
Denote SE the solution set of Problem I. From (16), it is easy to see that the SE is a nonempty
closed convex set. We can claim that for any given X ∗ ∈ R(2k+p)×(2k+p) , there exists a unique
optimal approximation for Problem II.
F. Li, X. Hu and L. Zhang                                    223

Theorem 3.1. Let X ∗ ∈ R(2k+p)×(2k+p) , A, B, C and D are given by (11). Problem II has a
˜                 ˜
unique solution X ∈ SE . Moreover X can be expressed as

˜
X1           0
˜
X=K                 T
(21)
0            ˜ K ,
X2

where
˜       Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12
T              T
1
T
T
X1 = V1             T         −1                T    ∗      P1 ,                     (22)
V12 D1 Q11 τ1               V12 X11 P12

˜       Φ′ ∗ (Σ2 U21 B2 P21 + V21 D2 Q21 τ2 ) Σ−1 U21 B2 P22
T             T
2
T
T
X2 = V2              T         −1                  T   ∗        P2 ,                 (23)
V22 D2 Q21 τ2                 V22 X22 P22
            
Ik √ 0                  X ∗ + W(k,p) X ∗ W(k,p)
∗      1 Ik √ 0      Jk      ∗                      ∗
X11   =                     X1  0       2Ip  , X1 =                             ,
2 0     2Ip 0                                               2
Jk     0
      
−Jk
∗     1               ∗
X22 = (−Jk , 0, Ik )X1  0  ,                                        (24)
2
Ik
where Φ and Φ′ are given by Theorem 2.1.

Proof For any X ∗ ∈ R(2k+p)×(2k+p) , from Lemma 2.3, there exist only X1 ∈ M S(k,p) , only
∗
1
∗                          ∗    ∗
X2 ∈ M SS(k,p) such that X = X1 + X2 , (X1 , X2 ) = 0, where X1 = 2 (X + W(k,p) X ∗ W(k,p) ).
∗    ∗   ∗              ∗       ∗

Therefore, for any X ∈ SE , from Theorem 2.1, we have

||X ∗ − X||2    =    ||X1 + X2 − X||2 = ||X1 − X||2 + ||X2 ||2
∗    ∗             ∗             ∗

2
X1
=        ∗
X1 − K                  KT        + ||X2 ||2
∗
0     X2
2
X1       0
=       K T X1 K −
∗
+ ||X2 ||2 .
∗
0        X2

∗     ∗
X11   X12
Denote K T X1 K =
∗
∗     ∗  , where
X21   X22
                                                   
1
Ik    √0               1 Ik √ 0
−Jk
∗
X11   =
Ik    √0      Jk      ∗
X1  0                ∗
2Ip  , X12 =
Jk     ∗
X1  0  ,
2     0     2Ip    0                                 2 0       2Ip 0
Jk       0                                         Ik
                                                      
∗      1                ∗
Ik        √0                ∗    1              ∗
−Jk
X21 =     (−Jk , 0, Ik )X1  0         2Ip  ,         X22 = (−Jk , 0, Ik )X1  0  .
2                                                    2
Jk         0                                         Ik

Then, we have

∗               ∗          2
X11 − X1       X12
||X ∗ − X||2     =             ∗         ∗
∗
+ X2       2
X21        X22 − X2
=   ||X11 − X1 ||2 + ||X22 − X2 ||2 + ||X12 ||2 + ||X21 ||2 + ||X2 ||2 .
∗                ∗                ∗           ∗           ∗
224                         Least-Squares Mirrorsymmetric Solution for Matrix Equations

From this, it is obvious that
∗
X11 − X1 = min,
X ∗ − X = min            is equivalent to         ∗                                  (25)
X∈SE                            X22 − X2 = min .

From (18),we have
∗           2
X11 − X1
2
∗
T              T
Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 )       Σ−1 U11 B1 P12
1
T
T
=    X11 − V1                 T         −1                                  P1
V11 D1 Q12 τ1                         X22
2
Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) Σ−1 U11 B1 P12
T              T                  T
=    V1T X11 P1
∗
−             T         −1
1
V11 D1 Q12 τ1                   X22
2
V11 X11 P11 − Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 ) V11 X11 P12 − Σ−1 U11 B1 P12
T  ∗                 T              T              T  ∗
1
T
=                 T  ∗          T         −1                    T   ∗
V12 X11 P11 − V11 D1 Q12 τ1                   V12 X11 P12 − X22
2                                    2
=      T  ∗                 T             T
V11 X11 P11 − Φ ∗ (Σ1 U11 B1 P11 + V11 D1 Q11 τ1 )          + V11 X11 P12 − Σ−1 U11 B1 P12
T  ∗
1
T

2
T  ∗          T         −1
+ V12 X11 P11 − V11 D1 Q12 τ1            + V12 X11 P12 − X22 2 .
T  ∗

∗                               T  ∗
Hence, X11 − X1 = min is equivalent to V12 X11 P12 − X22 = min∀X22 ∈R(k+p−r1 )×(k+p−s1 ) .
Then, we have
T  ∗
X22 = V12 X11 P12 .                               (26)
∗
Similarly, from X22 − X2 = min we can also obtain
′      T  ∗
X22 = V22 X11 P22 .                                        (27)

Substituting (26) and (27) into (17) yields the conclusion.
Algorithm
1. Input A, B, C, D, X ∗ .
2. Compute A1 , A2 , B1 , B2 , C1 , C2 , D1 , D2 according to (11).
∗    ∗     ∗
3. Compute X1 , X11 , X22 according to (24).
˜ ˜
4. Compute X1 , X2 according to (22), (23), respectively.
˜
5. According to (21) calculate X.

Example. Let k = 2, p = 3, n = 2k + p = 7, h = 5, l = 6. Giving
                                             
12.1 8.3 −5.6 6.3 9.4 10.7            7.6
 11.8 2.9 8.6        6.9 9.5 −7.8 3.5 
                                             
A=    10.6 2.7 11.5 7.8 6.7 8.9              4.8  ,

 3.6 7.8 4.9 11.9 9.3 5.9 −2.7 
4.5 6.3 7.8        3.1 5.6 11.6 10.8
                                           
8.5 9.3 3.6 7.8 6.3 4.7 −4.3
 2.5 3.6 7.9 9.4 5.6 7.8 1.9 
                                           
B =  3.7 6.7 8.6 9.8 3.1 2.9 8.7  ,
                                           
 −4.3 6.1 5.7 7.4 5.4 9.5 6.3 
2.9 3.9 −5.1 6.3 7.8 4.6 6.4
F. Li, X. Hu and L. Zhang                                     225

                                      
5.4    6.4    3.7    5.6 9.5   7.5

      3.5    4.2    7.8    7.8 −6.3 6.1 

      6.7    3.5   −4.6    2.9 2.7 −3.8 
C=
      −2.7   7.1   10.8    3.7 3.8   8.5  ,


      1.9    3.9    8.1    5.6 11.2  7.8 

      8.9    7.8    9.3    7.9 5.6   2.8 
9.5    5.1    4.7    8.5 1.7   5.6
                                      
7.9    9.5 5.3        2.8 8.6 1.9

      8.6    2.6 6.7        8.4 8.1 2.8 

      5.7    3.9 −2.9       5.1 1.9 3.6 
D=
      4.8    5.8 1.8       −7.1 5.8 −4.3  .


      2.7    7.9 4.5        6.7 9.6 5.8 
      9.5    4.1 3.4        9.7 3.9 9.7 
−6.3   6.7 3.8        7.6 5.7 5.8
                                         
2.9  1.9 5.6        1.6 3.8 1.5 3.3

      3.5  3.6 1.9        1.4 5.4 2.5 4.1 
      1.8 −2.7 4.7        4.6 1.7 3.6 5.2 
X∗ = 
                                           
      4.3  1.7 5.1        5.3 4.6 −2.3 1.6  .


      1.9  1.4 2.1        2.9 3.8 1.7 4.5 
      −1.8 5.3 3.7        3.5 2.5 5.4 6.7 
3.6  2.8 1.7        4.6 1.8 2.9 1.8
˜
Using MATLAB, we can obtain the unique solution X of Problem II given by
                                                                                 
0.1929 −0.1608 −0.0367 0.3664 −0.0732 −0.2489 0.0367
 0.3204 −0.1139 −0.1980 −0.2026 0.3445            0.2523     0.2614              
                                                                                 
 0.1639   0.1307 −0.4161 −0.0133 0.0820           0.1307     0.1639              
˜
                                                                                 
X =  −0.5199 −0.6649 0.0779
                               0.7340    0.0054 −0.6649 −0.5199                  .

 0.2299 −0.0157 0.1387 −0.1238 0.5227 −0.0157 0.2299                             
                                                                                 
 0.2614   0.2523 −0.1980 −0.2026 0.3445 −0.1139 0.3204                           
0.0367 −0.2489 −0.0367 0.3664           0.0732 −0.1608 0.1929

Acknowledgments
This research was supported by National Natural Science Foundation of China (1057,1047).

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