IMPACT: International Journal of Research in Engineering & Technology (IMPACT: IJRET) ISSN(E): 2321-8843; ISSN(P): 2347-4599 Vol. 2, Issue 8, Aug 2014, 77-90 © Impact Journals
ON INVERSES AND GENERALIZED INVERSES OF BIMATRICES RAMESH G1 & ANBARASI N2 1
Associate Professor of Mathematics, Government Arts College (Autonomous), Kumbakonam, Tamilnadu, India 2
Assistant Professor in Mathematics, Arasu Engineering College, Kumbakonam, Tamilnadu, India
ABSTRACT The concept of singular, Semi-singular and non-singular bimatrices are introduced. The concept of inverse bimatrices, reverse order law and some properties of inverses bimatrices are studied. Also the notion of generalized inverses and some properties of generalized inverse of bimatrices are discussed. The solution of homogeneous and non-homogeneous system of equations are studied.
KEYWORDS: Bimatrix, Inverse Bimatrix, Generalized Inverse of Bimatrix, Singular, Semi- Singular, Non-Singular Bimatrices
AMS Classification: 15A09, 15A15, 15A57. INTRODUCTION Let
be the space of nxn complex matrices of order n. For
,
,
let
,
∗
,
,
,ℛ
denote the
inverse if | | ≠ 0 that is, A is non singular. Generalized inverse is a great tool in solving linearly dependent and inverse, transpose, conjugate transpose, Moore-penrose inverse, rank and range space of
=A is denoted by
respectively. A matrix has its
unbalanced system of linear equations. It has the ability to find the solution of square matrix when it is singular and non-square. A solution ,
=
of the equation
the Moore-penrose inverse ∗
and is called generalized inverse of A. For = ,
of A is the unique solution of the four equation
= ,
∗
,
=
. The concept of a generalized inverse was first introduced by Fredholm (1903), he called a
particular generalized inverse as pseudo inverse which serve as integral operator. However, the concept of an inverse of a =
∪
singular matrix seems to have been first introduced by Moore [4,5] in 1920. If matrix
is said to be bimatrix [7]. A bimatrix
is said to be EP if
=
and
[8].
are any two matrices then the ∗
In this paper the concept of singular, semi-singular, non-singular bimatrices are introduced. The concept of inverse bimatrices, reverse order law and some properties of inverse bimatrices are studied. Also, the notion of generalized inverses and some properties of generalized inverses of bimatrices are discussed. The solution of homogeneous and non-homogeneous system of equations are analysed.
INVERSES OF BIMATRICES Definition 2.1 Let
be a square bimatrix of order n. Then,
is said to be invertible if there exists a square bimarix
order n such that
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of
78
Ramesh G & Anbarasi N
=
=
,
is called the inverse of
and Example 2.2
2 = !2 1
Let
2 1 3
2 = !−9 5
Now,
2 = !2 1
1 = !0 0
0 1 0
.
=
3 2 1& ∪ !0 5 5
1 2 2
3 2 1& !−9 5 5
−1 7 −4
−1 7 −4
It is verified that,
2 1 3
and is denoted by
0 1 0& ∪ !0 1 0
−1 1& −3
−1 8 4 & ∪ !−5 −2 10
0 1 0
0 0& 1
−1 1 −1
.
−3 2& −4
−1 2 4 & ∪ !0 −2 5
1 2 2
−1 8 1 & !−5 −3 10
−1 1 −1
−3 2& −4
Definition 2.3 (That is, |
| = 0 and |
| =0).
A square bimatrix
Definition 2.4
is said to be singular if the determinant value of both the components are zero.
=
A square bimatrix non zero. Definition 2.5
=
A square bimatrix component is zero.
∪
is said to be non-singular if the determinant value of both the components are
∪
is said to be semi-singular if the determinant value of either one of the
Properties of the Conjugate Transpose of Bimatrices 2.6 • • •
∗∗
,
=
+
∗
• •
∗
• •
∗
∗
=
∗
∗
=
∗
∗
∗
=
=,̅
∗
= 0 implies =
∗
+
∗
∗
=0
implies implies
∗
= =
∗
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79
On Inverses and Generalized Inverses of Bimatrices
Properties of Inverse of Bimatrices 2.7 Let
and
• •
.
• •
=
=
be the two bimatrices, then the following holds:
=.
=
/
Proof of (i) Given
=
∪
=
/
∪ ∪
=
∪
= =
∪
∪
=
Proof of (ii)
∪
=
= =
Proof of (iii) .
= .
= .
∪
∪
=
)
∪ =
∪
∪ =
(since
= = [.
∪.
∪
]
∪ .
=.
∪.
.
=.
=.
(since
=
=
∪
∪
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80
Ramesh G & Anbarasi N
Proof of (iv) /
=
∪
/
=
/
=
/
= = /
/
∪
=
/
∪
/ /
∪
/
∪
∪ /
/
=
/
(since
/
=
/
)
/
GENERALIZED INVERSES OF BIMATRICES Definition 3.1
AB is the unique solution of the following equations:
Moore – Penrose inverse of a bimatrix =
• • ∗
•
∗
•
= = =
Definition 3.2 Group inverse of =
• • • •
If
#
=
AB , denoted as AB # satisfying the equations,
.
=
. .
exists, then it is unique.
Example 3.3 Let
4 = !1 3
1 1 1
Generalized inverse of 13 3 = 2−13 3 0
such that
−13 3 43 3 0
=
2 1 5& ∪ !5 3 2 is,
3 2 6
2 6& 4
−23 13 4 ∪ 2 5 0 313 0 0
0
33 13 −13 13 0
0
04 0
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On Inverses and Generalized Inverses of Bimatrices
Lemma 3.4 be an n × n complex bimatrix. Then the following holds :
Let =
•
• •
=
∗
•
If
,
•
∗
=,
=
∗
Proof of (i)
=
5
6 =5
=
∪
=5
∪
6 ∪5
5
6
6
(since
=A)
6 =
Proof of (ii) ∗
∗
∪
∪
=
=
=
is non singular, then
=
∗
∗
=
=5
∪ ∗
=5
=
=5
∗
Proof of (iii) Given
∗
=
∪
∪
∪
6
∪
∗
=5
∪
∗
6 ∪5 ∗
∗
6
∗
∗
6 (Since ∗
∗
∗
=
∗
)
∗
6
∗
AB is non singular bimatrix ⇒ both A1 and A2 are non singular matrices.
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82
Ramesh G & Anbarasi N
=
∪
∪
=
By lemma (1.3) of [8], =
⇒
=
=
=
∪
∪
Proof of (iv)
= [,
,
= ,
∪,
=,
∪,
,
=,
=, 5
Proof of (v) ∗
= ∗
=
∗
∪
=
∪
∗
=
∗
=
∗
∪
=5 =5 ∗
∪
∪
Lemma 3.5 [10]
( since ,
6
∗
∪ 65 65
=
]
∪
∪ ,
= ,
=
and
∪
∗
∗
∪
∗
∗
∗
)
∪ ∗
( since
∪
∪
=,
∗
∗
6
6
∗
=
∗
)
∗
Let A be an n × n complex matrix. Then the following statements are equivalent: • •
A is EP. ∗
is EP.
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On Inverses and Generalized Inverses of Bimatrices
[ ,
•
∗]
= [ ].
is EP.
•
=
•
.
=
•
#
.
Theorem 3.6
Let AB be an n × n complex bimatrix. Then the following are equivalent: is EP bimatrix
• • • Proof
5
6 =
Let
=
5
6 =5
=5
= = =
=
5
To Prove (ii)
=
6 ∪5 65
5
(By lemma 3.5)
6 =5
5
6
∪
∪
6 =
6
6∪5
∪
6
5
∪ 5
∪
6
65
6
6
5
6
∪
6
⇒ (i)
Suppose
satisfies (ii)
Note that 5 That is, ⇒
be an n × n complex bimatrix.
is EP then
Hence 5 =5
6
6
∪
To Prove (i) ⇒(ii) If
5
∪
6 = =
5
=
6
∪
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83
84
Ramesh G & Anbarasi N
∪
5
6
∪
∗
∪
∗
=
∗
Now use the fact that ∪
∗
∪
∪
∗
=
∗
=
∗
=
∗ ∗
∗
∗
∪
∗
∪
=
and
∪
5
∗
∗
∪
=5
∪
∗
=
∗
Post-multiply both sides by
∗
∪
=
6
∗
we get 6
∪
∗
we get
∪
∗
∗
∗
∗
Then it follows by equation (1.4) of lemma (1.2) of [10] that 5
= 85
= 5
= :
−
∗
∗
∗
−
∗
−
∗
6= 8
∗
∗
∗
∗
∗
:
∗
and <
∗
∗
;5
∗;
=
Hence, ℛ :
6 =< ∗
=:
∗ ∗
∗
∗
6∪ 5
6 ∪ 5
Observe that ∗
∗
∗
− ∗
∗
∗
;=ℛ<
∗
∪
=;: ∗
∪
∗
∗ ∗
;;<
∗
−
−
∗
∗
∗
∗
:
∗
;5
=
= ;ℛ:
− 5
∗
6
∗
69
∗
∗
6 =< ∗
∗ ∗
=: ∗
∗
∗
∗
∗; ∗
∗
∗
;=ℛ<
∗
∪
−
∗
69
(3.1)
= ;
∗
=
According to equation (1.12) of lemma (1.2) in [10], then the equation (3.1) is reduced to ∗
r5 = >: = =
[
[
=@ [ =
[
∗
−
∗
;[
∗ ]∗ [ ∗]
∗]
∗]
∗ ]?
−
−
∪
∗
6= :
−
∗]
[
−
∪
∗]
∪ [
∗
∪
A−@
>:
∗]
∗
∗
[
−
∗; −
;[
∪
∗ ]∗ [
∗ ]?
−
∪
∗]
:
∗
∗
∗
∗;
−
−
A
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85
On Inverses and Generalized Inverses of Bimatrices
⇒
5
∗
Hence
−
∗
=
∗
6= [ ∗
∗]
−
is equivalent to
By (i) and (iii) of lemma (3.5),
[
∗]
=
is EP bimatrix.
To Prove (i) ⇒(iii) If
=
is EP then
Hence 5 =5
=5 =
5
=
5
6 ∪5
∪
=
=5
6 =5
65
5
6 6
Suppose
6
65
6
5
6
5
6
6
∪
satisfies (iii) 6 =
We have 5 ⇒
∪
6∪5
∪
∪
To Prove (iii) ⇒(i)
6
∪
6 =
That is,
(By lemma 3.5)
=
∪
5
=
6
Pre-multiply both sides by
∗
∗
∪
∗
∗
∪
∗
∗
Now apply
∪
5
∪
∪
∗
∗
∗
∪
∗
=
∪
6=
=
∗
∗
5
=
= and
=
∗
=
∗
∗
∗
5
∗
=
∗
∗
∪
∪
6 ∪
∪
∗
we get ∗
∪
∗
∗
5
6 ∪
∗
5
∗
5
∪
6
6
6
we get
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86
Ramesh G & Anbarasi N
⇒ ∗
∪
∗
∪
∗
∗
∗
=
=
∪
∗
∗
=
∗
∪
∗
∪
5
∗
∪
6
∪
Then it follows by equation (1.4) of lemma (1.2) of [10] that 5
∗
= 85
= 5
∗
∗
∗
= :
∗
:
−
−
−
∗
∗
∗
∗ ∗
∗
∗
∗
∗
6[ ∗
and
∗[
∗
∗
∗
5
6[ ∗
and
∗[
∗
Hence, ℛ[ and ℛ[
∗
∗
−
∗
∗
∗
∗
∗
∪
;− ∗
]=[
∗
]=[
]=[ ∗
∗
∗
∗
∗
∗
∗
∗
]
∗
∗
6
69
∪
∗
69
∗
] = ℛ[
∗
∗
]
]
∗
] = ℛ[
∗
− 5
∗
(3.2)
]=[
similarly,
−
∗
;−
Observe that 5
−
∗
6 ∪ 5
∗
∪
∗
6∪5
∗
∗
∗
6= 8
] ]
]
According to equation (1.12) of lemma (1.2) in [10], then the equation(3.2) is reduced to r5
= >: = =
[
[
=@ [
= [ ⇒
5
∗; [
∗ ∗ ∗
∗
∗
6= :
∗ ∗
]∗ [
∗
]? −
] −
] −
] ∪
]− −
[ ∗
∗
∪
∗
∗
∪
[
∗
∗
∪
>:
] A−@ 6= [
∗
∗
[
;−
∗; [ ∗
] −
∪
∗
]∗ [
∪
∗
]? −
:
∗
∗ ∗
∗
;−
] −
}
]−
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87
On Inverses and Generalized Inverses of Bimatrices
=
∗
Hence
∗
⇒
is EP bimatrix.
Let
=
Theorem 3.7
∗ ∗
∪
=
[
is equivalent to
]=
∗
be a bimatrix, then the four equations
3.3
=
3.4
=
3.6
=
3.5
have a unique solution for any
.
Proof First to show that equations (3.4) and (3.5) are equivalent to the single equation, ∗
∗
=
On Substituting (3.5) in (3.4) we get,
∗
∗
∗
=
=
=
(Since by (3.5)) 3.7
Conversely, 3.7 ⇒
∗
∗
∗
=
∗
∗
=
= =
=
(Since by (3.5))
which gives (3.4) Thus, (3.4) and (3.5) are equivalent to Similarly from (3.3) and (3.6) =
=
∗
∗
∗
= ∗
∗
∗
=
∗ ∗ ∗
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88
Ramesh G & Anbarasi N
=
∗
∗
Thus, it is sufficient to find an ∗
∗
For then ∗
∗
=
∗
=
=
∗
∗
∗
,
=
∗
+,
∗
,
∗
∗
+,…,+,D
where , ,, ,…,,D are not all zero.
∗
,
∗
D
∗
+ ,EF
and
∗
∗
=
To show that I =
∗
EF
∗
Thus, (3.9) gives Apply
∗
=
C
,…cannot all be linearly independent that is, there exists a
= 0, 3.9
Let ,E be the first non zero , and put = −,E @,EF
can be found satisfying,
∗
also satisfies (3.7).
Now, the expressions relation
exists if a
satisfies (3.8). Also we have seen that (3.8) implies,
∗
and therefore Thus,
∗
satisfying (3.7) and (3.8). Such an
3.8
+ ⋯ + ,D ∗
E
∗
H E
A
of (2.6) repeatedly we obtain, ∗
is unique, we suppose that
I ∗ I and
∗
=
∗
I
satisfies (3.7) and (3.8) and that I satisfies,
These relations are obtained by substituting (3.6) in (3.4) and (3.5) in (3.3) respectively. Now, ∗
= = = =
∗
∗
I
I∗I
=
∗
I
∗
∗
I∗ I
=I
The unique solution of (3.3),(3.4),(3.5)and (3.6) is called the generalized inverse of written as Note 3.8
=
(abbreviated as g.i) and
.
In the above lemma,
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89
On Inverses and Generalized Inverses of Bimatrices
• •
need not be a square bimatrix and may even be zero.
Use the notation , for scalars, where , means ,
Theorem 3.9
=
Let equation
∪
=
,
=
∪
=
and
to have a solution is,
=
∪
if , ≠ 0 and 0 if , = 0. are bimatrices. A necessary and sufficient condition for the
in which case the general solution is, =
Proof
+I -
I
Where I is arbitrary. Suppose Then =
=
=
satisfies
=
,
=
Conversely, if
, then
For the general solution, we have to solve Now, any expression of the form =I −
I
Satisfies
And conversely if =
−
Theorem 3.10 =
Let ∪
= 0.
=
=0
=
= 0 then, .
and
=
be two system of equations and can be written as
be a co-efficient bimatrix,
bimatrix. And let
is a particular solution of
JK = [
] ∪ [
=
=
, where
∪ I be a unknown bimatrix and BB = B1 ∪ B2 be a column
] be the augmented bimatrix of the two systems. If the components of
augmented bimatrix are equivalent then both the systems has the same solution.
In particular, for the homogeneous system of equations, if the components of co-efficient bimatrix are equivalent then the system must have the unique solution.
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Note 3.11 For the non homogeneous system of equations, if the components of augmented bimatrix are not equivalent then both the systems need not have same solution. Example 3.12 Let the two system of equations be, 2L + 3L − LC = 5 ;
4L + 4L − 3LC = 3 ;
2L − 3L + 2LC = 2 ; This can be written as =
Where
2 = !2 1
2 1 3
L + 2L + LC = 8
2L + 3L + 4LC = 20 4L + L + 2LC = 12
3 2 1& ∪ !0 5 5
The solution is x1=1, x2 = 2, x3=3
1 2 2
L −1 8 5 1 &, X = !L &, BB = !3& ∪ !20& LC −3 12 2
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Adi Ben-Isreal and Thomos N.E Greville, ‘Generalized inverses: Theory and Application’ 2nd Ed. Network, NY: springer2003.
2.
Awni M.Abu-Saman,’Solution of linearly independent equations by generalized inverse of a matrix’,int J.sci Emerging Tech,vol-4 No.2 August 2012 pp: 138-142.
3.
Hartwig. R.E, Katz.I.J, ‘On products of EP matrices’ Linear Algebra Appl. 252(1997)338-345.
4.
Moore.E.H,‘Generalized analysis’,Philadelphia, American Philosophical society,1935.
5.
Moore. E.H, ‘On the Reciprocal of the general algebraic Matrix,’ Bull amer Math. soc., vol. 26(1920), PP(394395)
6.
Penrose. R, ‘A Generalized inverse of matrices’ proc cambridge philos.soc. vol 51(1955), PP(406-413).
7.
Ramesh. G, Anbarasi. N, ‘On Bihermitian matrices’ IOSR journal of Mathematics, vol 9 Issue 6 (Jan 2014).
8.
Ramesh. G, Anbarasi.N, ‘On EPr Bimtarices’ Int. jour. of Mathematics and Statistics Invention (IJMSI) vol. 2 Issue 5 May 2014, PP(44-52).
9.
Rao. C.R. and Maitra. S.K, ‘Generalized inverse of matrices and its applications’ wiley New york,(1971).
10. Shizhen cheng, yongge Tian, ‘Two sets of new characterizations for normal and EP matrices’ Elsevier, Linear Algebra and its applications 375 (2003) (181-195).
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