EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 3, Article Number 6241
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

T-Ordering on Generalized Regular Intuitionistic Fuzzy
Matrices

P. Jenita1, M. Princy Flora2, Chiranjibe Jana3,∗, Elvis Popović4, Nikola Ivković4

1 Government Arts College, Coimbatore, India
2 Kumaraguru College of Technology, Coimbatore, India
3 Saveetha School of Engineering, Saveetha Institute of Medical and Technical Sciences
(SIMATS), Chennai 602105, Tamil Nadu, India
4 University of Zagreb Faculty of Organization and Informatics, Pavlinska 2, 42000 Varaždin,
Croatia

Abstract. In this paper, we study T -ordering on generalized regular intuitionistic fuzzy matrices
(IFM) named as k−T -ordering, as a generalization of the T -ordering on intuitionistic fuzzy matri-
ces. Some equivalent conditions for this ordering using generalized inverses are derived. Further,
we prove that k − T -ordering is not a partial ordering.

2020 Mathematics Subject Classifications: 03Gxx

Key Words and Phrases: Fuzzy Matrix, Intuitionistic Fuzzy Matrices, Partial Ordering, k-T-
ordering

1. Introduction

Atanassov first introduced the concept of intuitionistic fuzzy sets [1], building on the
foundation of fuzzy set theory. Meanwhile, Ben-Israel and Greville [2] explored the idea
of generalized inverses for complex matrices. In fuzzy algebra, defined over the interval
F = [0, 1], matrix operations are carried out using the max-min operations, where addition
is defined as a + b = max{a, b} , and multiplication as a · b = min{a, b} for all a, b ∈ F .
The set Fm×n consists of all m× n fuzzy matrices under this algebra.

A fuzzy matrix A ∈ Fm×n is said to be regular if there exists a matrix X such that
AXA = A, in which case X is termed a generalized (g-) inverse of A. Kim and Roush
[3] extended fuzzy matrix theory by drawing analogies to Boolean matrices and studying
their inverses.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6241

Email addresses: soulwinjenita@gmail.com (P. Jenita),
princyfloram@gmail.com (M. Princy Flora), jana.chiranjibe7@gmail.com (C. Jana),
elvpopovi@foi.unizg.hr (E. Popović ), nikola.ivkovic@foi.hr (N. Ivković )

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 2 of 31

Further developments include Cho’s analysis of the consistency of fuzzy matrix equa-
tions [4] and the introduction of k-regular fuzzy matrices by Meenakshi and Jenita [5],
a generalization of regular fuzzy matrices. Khan and Paul [6] introduced the concept of
generalized inverses for intuitionistic fuzzy matrices, while Pradhan and Pal [7] proposed
a method to compute such inverses using block-wise decompositions.

Pal and Khan [8] further developed the fundamental properties of intuitionistic fuzzy
matrices, and Meenakshi and Gandhimathi [9] examined their regularity. Ordering con-
cepts in fuzzy matrices have also been explored, with Sriram and Murugadas investigating
general ordering [10], Cen [11] proposing the idea of T-ordering, along with its relation to
generalized inverses. Platil and Tanaka [12] proposed a multi-criteria evaluation framework
based on set-relations for intuitionistic fuzzy sets, which may offer broader interpretive
foundations for such ordering concepts.

Expanding on this, Poongodi et al. [13] discussed ordering in k-regular interval-valued
fuzzy matrices, extending the minus ordering concept previously studied in [14]. Addi-
tionally, [15] explored special types of inverses for regular intuitionistic fuzzy matrices.
In another significant contribution, Jenita, Karuppusamy, and Thangamani introduced
k-regular intuitionistic fuzzy matrices in [16], extending the notion of regular intuitionistic
fuzzy matrices and analyzing various types of inverses for them.

Meenakshi and Inbam [17] defined minus ordering on matrices using generalized in-
verses. As a continuation of this line of research, two further studies were conducted [18],
[19], focusing on minus ordering and sharp ordering in the context of generalized regular
intuitionistic fuzzy matrices. In [20] Radio fuzzy graphs and assignment of frequency in
radio stations is discussed. Applications of edge colouring of fuzzy graphs was discussed in
[21]. In [22] and [23] Rupkumar Mahapatra , Sovan Samanta,Madhumangal Palhave dis-
cussed about the link prediction in social networks by neutrosophic graph and generalized
neutrosophic planar graphs. Detecting influential node in a network using neutrosophic
graph, Edge colouring of neutrosophic graphs, A study on linguistic z-graph and its appli-
cation in social networks, Centrality measure using linguistic Z-graph and its application
was also discussed in [24–27].

2. Preliminaries

The matrix operations on IFM as stated in [9] will be followed. For A,B ∈ (IFM)m×n,
the operations are defined as follows:

A+B = (⟨max{aijµ, bijµ},min{aijϑ, bijϑ}⟩) ,

AB =

(〈
max
k

min{aikµ, bkjµ},min
k

max{aikϑ, bkjϑ}
〉)

.

The order relation on (IFM)m×n defined as:

A ≤ B ⇔ aijµ ≤ bijµ and aijϑ ≥ bijϑ, for all i, j.

Throughout this paper we denoted right k−regular as rightk-reg, left k−regular as leftk-reg,
right k−g− inverse as rightk-g-inv, left k−g− inverse as leftk-g-inv, right k−Moore-Penrose



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 3 of 31

inverse as rightk-Moore -Penrose inv ,left k−Moore-Penrose inverse as leftk-Moore -Penrose inv
,k−regular as k− reg, k − g−inverse as k − g−inv, and k−Moore-Penrose inverse as
k−Moore-Penrose inv..

Definition 1. [16] If there exists a matrix X ∈ (IFM)n such that ukXu = uk, for some
positive integer k, then the matrix u ∈ (IFM)n is said to be rightk-reg . X is called a

rightk-g-inv of u.
Let, u{1kr} = {X | ukXu = uk}.

Definition 2. [16] If there exists a matrix Y ∈ (IFM)n such that uY uk = uk, for some
integer k, then the matrix u ∈ (IFM)n is said to be leftk-reg. Y is called a leftk-g-inv of u.
Let, u{1k} = u{1kr} ∪ u{1kl } and u{1k} = u{1kr} ∩ u{1kl }.

Theorem 1. [28] Let u ∈ (IFM)n and k be a positive integer. Then,

X ∈ u{1kr} ⇔ XT ∈ uT {1kl }.

Definition 3. [15] A matrix u ∈ (IFM)n is said to have a rightk-Moore -Penrose inv if
there exists a matrix X ∈ (IFM)n satisfying the four equations

ukXu = uk,−−−{1kr}
XuXk = Xk,−−−{2kl }
(ukX)T = ukX −−− {3k}
(Xuk)T = Xuk −−− {4k}.

This inverse is denoted as u+rk.

Definition 4. [15] A matrix u ∈ (IFM)n is said to have a leftk-Moore -Penrose inv if
there exists a matrix Y ∈ (IFM)n satisfying the four equations:

uY uk = uk −−− {1kl }
Y kuY = Y k −−− {2kr}

(ukY )T = ukY −−− {3k}
(Y uk)T = Y uk −−− {4k}.

This inverse is denoted as u+lk.

3. k - T Ordering on IFM

Theorem 2. Let u ∈ (IFM)mn. The following are equivalent:
(i) u+ exists and u+ = uT .
(ii) uT is a g-inverse of u.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 4 of 31

Proof. (i) ⇒ (ii)

u+ = uT ⇒ uT is a g-inverse of u,

(ii) ⇒ (i)
uT is a g-inverse of u ⇒ uuTu = u (1)

By taking transpose on both sides in (1), we get:

uT = uTuuT

(uuT )T = uuT and (uTu)T = uTu,

Hence u+ exists and u+ = uT .

Remark 1. In general, for a k-regular IFM, the rightk-Moore -Penrose inv u+rk is different
from leftk-Moore -Penrose inv u+lk and it is not unique.If u+rk = u+lk, let us call it as the
k-Moore–Penrose inv, and it is denoted by u+k . Thus,

u+k = u+rk = u+lk

This is shown in the following example.

Example 1. Let us consider the matrix u as follows:

u =

 ⟨0.5, 0⟩ ⟨0.2, 0.5⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 .

For the permutation matrices:

P1 =

⟨1, 0⟩ ⟨0, 0⟩ ⟨0, 1⟩
⟨0, 1⟩ ⟨1, 0⟩ ⟨0, 0⟩
⟨0, 0⟩ ⟨0, 1⟩ ⟨1, 0⟩

 ,

P2 =

⟨1, 0⟩ ⟨0, 1⟩ ⟨0, 0⟩
⟨0, 1⟩ ⟨0, 0⟩ ⟨1, 0⟩
⟨0, 0⟩ ⟨1, 0⟩ ⟨0, 1⟩

 ,

P3 =

⟨0, 0⟩ ⟨1, 0⟩ ⟨0, 1⟩
⟨1, 0⟩ ⟨0, 1⟩ ⟨0, 0⟩
⟨0, 1⟩ ⟨0, 0⟩ ⟨1, 0⟩

 ,

P4 =

⟨0, 1⟩ ⟨1, 0⟩ ⟨0, 0⟩
⟨0, 0⟩ ⟨0, 1⟩ ⟨1, 0⟩
⟨1, 0⟩ ⟨0, 0⟩ ⟨0, 1⟩

 ,



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 5 of 31

P5 =

⟨0, 0⟩ ⟨0, 1⟩ ⟨1, 0⟩
⟨1, 0⟩ ⟨0, 0⟩ ⟨0, 1⟩
⟨0, 1⟩ ⟨1, 0⟩ ⟨0, 0⟩

 ,

P6 =

⟨0, 1⟩ ⟨0, 0⟩ ⟨1, 0⟩
⟨0, 0⟩ ⟨1, 0⟩ ⟨0, 1⟩
⟨1, 0⟩ ⟨0, 1⟩ ⟨0, 0⟩

 .

uP1u =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 ̸= u,

uP2u =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩

 ̸= u.

uP3u =

⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 ̸= u,

uP4u =

⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩
⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩

 ̸= u,

uP5u =

⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩
⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩

 ̸= u

and

uP6u =

⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩

 ̸= u.

Therefore, u is not regular.
For this u,

u2 =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 .

For

X =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩


u
2
Xu =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 ⟨0.5, 0⟩ ⟨0.2, 0.5⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 ⟨0.5, 0⟩ ⟨0.2, 0.5⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩





P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 6 of 31

=

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = u
2
.

uXu
2

=

 ⟨0.5, 0⟩ ⟨0.2, 0.5⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 ⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩


=

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = u
2
.

u
2
Xu = u

2
= uXu

2
holds.

Thus u is 2-reg.
For k = 2,

X2uX =
[ ⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩

⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

] [ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

] [⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

]
=

[⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

]
= X2.

XuX2 =
[⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

] [ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩ ⟨0, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

] [⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

]
=

[⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

]
= X2.

(u2X)T =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = u2X

(Xu2)T =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = Xu2.

Hence, X is a right2-Moore -Penrose inverse as well as a left2-Moore -Penrose inverse.
Therefore, X = u+2 = u+r2 = u+l2 exists.

uT =

 ⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0, 0.5⟩
⟨0.2, 0.5⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0, 0⟩ ⟨0.2, 0.5⟩ ⟨0.5, 0⟩

 .

u2uTu =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = u2.

uuTu2 =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = u2.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 7 of 31

(uT )2uuT =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = (uT )2,

uTu(uT )2 =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = (uT )2,

(u2uT )T =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = u2uT

and

(uTu2)T =

⟨0.5, 0⟩ ⟨0.2, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.5, 0⟩ ⟨0.2, 0⟩
⟨0.2, 0⟩ ⟨0.2, 0⟩ ⟨0.5, 0⟩

 = uTu2.

Therefore, X and uT are 2-Moore-Penrose inv of u.
Hence u+2 exists, but it is not unique.

Definition 5. Let u ∈ (IFM)−mn and v ∈ (IFM)mn, the minus ordering denoted as u ≤ v
is defined as:

u ≤ v ⇔ uX = vX and Xu = Xv, for some X ∈ u{1}.

u{1} − set of generalized inverses

Remark 2. Let u ∈ (IFM)−mn and v ∈ (IFM)mn, if u
+ exists, then u+ is unique and

u+=uT .Then we have the following definition.

Definition 6. The T-ordering u <T v in (IFM)mn is defined as:

u <T v ⇐⇒ uuT = vuT and uTu = uT v

Here, uT is a g-inverse of u.

Lemma 1. Let u = ⟨uµ, uν⟩ ∈ (IFM)mn and v ∈ (IFM)mn. Then:

u <T v ⇔ uµ <T vµ and uν <T vν .

Proof.
u <T v ⇐⇒ uuT = vuT and uTu = uT v

⇐⇒ ⟨uµ, uν⟩⟨uTµ , uTν ⟩ = ⟨vµ, vν⟩⟨uTµ , uTν ⟩
⇐⇒ ⟨uµuTµ , uνuTν ⟩ = ⟨vµuTµ , vνuTν ⟩
⇐⇒ uµu

T
µ = vµu

T
µ and uνu

T
ν = vνu

T
ν

Similarly,



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 8 of 31

uTu = uT v ⇐⇒ uTµuµ = uTµvµ and

uTν uν = uTν vν

Remark 3. u <T v ⇐⇒ u ≤ v with respect to u+

⇐⇒
uTu = uT v and uuT = vuT

where uT is a g-inverse of u ,which is the definition of T-ordering

u <T v =⇒ u ≤ v

but the converse

u ≤ v =⇒ u <T v

need not be true. This is illustrated in the following example.

Example 2. Let u =

[
⟨1, 0⟩ ⟨1, 0⟩
⟨0, 1⟩ ⟨0, 1⟩

]
and v =

[
⟨1, 0⟩ ⟨0, 1⟩
⟨0, 1⟩ ⟨1, 0⟩

]
.

uµ =

[
1 1
0 0

]
, uν =

[
0 0
1 1

]
, vµ =

[
1 0
0 1

]
, vν =

[
0 1
1 0

]

uT =

[
⟨1, 0⟩ ⟨0, 1⟩
⟨1, 0⟩ ⟨0, 1⟩

]

uTµ =

[
1 0
1 0

]
, uTν =

[
0 1
0 1

]
Here,

uµu
T
µuµ =

[
1 1
0 0

] [
1 0
1 0

] [
1 1
0 0

]
= uµ

uνu
T
ν uν =

[
0 0
1 1

] [
0 1
0 1

] [
0 0
1 1

]
= uν

Hence,

uT = ⟨uTµ , uTν ⟩

is a g-inverse of u = ⟨uµ, uν⟩, u+ exists and u+ = uT , also u is idempotent.

Since u2 = u, u itself is a g-inverse of u,



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 9 of 31

uv =

[
⟨1, 0⟩ ⟨1, 0⟩
⟨0, 1⟩ ⟨0, 1⟩

]
= u

vu =

[
⟨1, 0⟩ ⟨1, 0⟩
⟨0, 1⟩ ⟨0, 1⟩

]
= u

Hence u ≤ v.

But,

uTµuµ =

[
1 0
1 0

] [
1 0
0 0

]
=

[
1 1
1 1

]

uTµvµ =

[
1 0
1 0

] [
1 0
0 1

]
=

[
1 0
1 1

]
⇒ uTµuµ ̸= uTµvµ

uTν uν =

[
0 1
0 1

] [
0 0
1 1

]
=

[
0 0
0 0

]

uTν vν =

[
0 1
0 1

] [
0 1
1 0

]
=

[
0 1
0 1

]
⇒ uTν uν ̸= uTν vν

Also,

uµu
T
µ =

[
1 1
0 0

] [
1 0
1 0

]
=

[
1 0
0 0

]

vµu
T
µ =

[
1 0
0 1

] [
1 0
1 0

]
=

[
1 0
1 0

]
⇒ uµu

T
µ ̸= vµu

T
µ

uνu
T
ν =

[
0 0
1 1

] [
0 1
0 1

]
=

[
0 1
1 1

]
vνu

T
ν =

[
0 1
1 0

] [
0 1
0 1

]
=

[
0 1
0 1

]
⇒ uvu

T
v ̸= vvu

T
v

⇒ uTu ̸= uT v and uuT ̸= vuT

Hence u ≤ v need not imply u <T v.

Definition 7. The k-T ordering u <T
k v in (IFM)n is defined as:

u <T
k v ⇐⇒ ukuT = vkuT and uTuk = uT vk

where, uT ∈ u{1k} and uT ∈ (u{3k} or u{4k})
(i.e.) uT ∈ u{1kr} ∩ u{1kl } and uT ∈

(
u{3k}oru{4k}

)
.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 10 of 31

Remark 4. For k = 1 Definition 7, reduces to the definition of T-ordering for IFM: Also,
from Definition 6 and Definition 7, it is to be noted that uk <T vk ⇔ u <T

k v

Definition 8. [18] For u ∈ (IFM)−n and v ∈ (IFM)n, the k-minus ordering, denoted as
u <−

k v, and is defined by

u <−
k v ⇔ ukX = vkX for some X ∈ u{1kr} and Y uk = Y vk for some Y ∈ {1kl }.

Remark 5.
u <T

k v ⇔ u <−
k v with respect to A+

k .

Thus,
u <T

k v ⇒ u <−
k v.

But the converse,
u <−

k v ⇒ u <T
k v

need not be true. This is given in the following example.

Example 3.

u =

[
< 0.5, 0.1 > < 0.2, 0.3 >
< 0.3, 0.2 > < 0.1, 0.5 >

]
, v =

[
< 0.6, 0.1 > < 0.2, 0.3 >
< 0.3, 0.2 > < 0.5, 0.3 >

]

uµ =

[
0.5 0.2
0.3 0.1

]
, uν =

[
0.1 0.3
0.2 0.5

]

u2µ =

[
0.5 0.2
0.3 0.1

] [
0.5 0.2
0.3 0.1

]
=

[
0.5 0.2
0.3 0.2

]
̸= uµ

u2ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
̸= uν

P1 =

[
< 1, 0 > < 0, 1 >
< 0, 1 > < 1, 0 >

]
P2 =

[
< 0, 1 > < 1, 0 >
< 1, 0 > < 0, 1 >

]
uµP1µuµ ̸= uµ

uµP2µuµ ̸= uµ

uνP1νuν ̸= uν

uνP2νuν ̸= uν

Therefore, u is not regular.
For,

X =

[
< 0.5, 0.1 > < 0.1, 0.5 >
< 0.1, 0.2 > < 0.2, 0.3 >

]



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 11 of 31

Xµ =

[
0.5 0.1
0.1 0.2

]
, Xν =

[
0.1 0.5
0.2 0.3

]

u2µXµuµ =

[
0.5 0.2
0.3 0.2

]
= u2µ

u2νXνuν =

[
0.1 0.3
0.2 0.3

]
= u2ν

∴ u2Xu = u2

Hence, u is 2-reg and X is a 2-g-inv of u.
For,

vµ =

[
0.6 0.2
0.3 0.5

]
, vν =

[
0.1 0.3
0.2 0.3

]

v2µ =

[
0.6 0.2
0.3 0.5

] [
0.6 0.2
0.3 0.5

]
=

[
0.6 0.2
0.3 0.5

]
= vµ

v2ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.2 0.3

]
= vν

Therefore, v = v2.

u2µXµ =

[
0.5 0.2
0.3 0.2

] [
0.5 0.1
0.1 0.2

]
=

[
0.5 0.2
0.3 0.2

]

v2µXµ =

[
0.6 0.2
0.3 0.5

] [
0.5 0.1
0.1 0.2

]
=

[
0.5 0.2
0.3 0.2

]

u2νXν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.5
0.2 0.3

]
=

[
0.1 0.3
0.2 0.3

]

v2νXν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.5
0.2 0.3

]
=

[
0.1 0.3
0.2 0.3

]
Therefore, u2X = v2X.

Y =

[
⟨0.5, 0.1⟩ ⟨0.2, 0.5⟩
⟨0.1, 0.2⟩ ⟨0.2, 0.3⟩

]

Yµ =

[
0.5 0.2
0.1 0.2

]
, Yν =

[
0.1 0.5
0.2 0.3

]

uµYµu
2
µ =

[
0.5 0.2
0.3 0.1

] [
0.5 0.2
0.1 0.2

] [
0.5 0.2
0.3 0.2

]
=

[
0.5 0.2
0.3 0.2

]
= u2µ

uνYνu
2
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.5
0.2 0.3

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.1
0.2 0.3

]
= u2ν



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 12 of 31

Therefore uY u2 = u2, Y is a left 2-g-inv. of u.

Yµu
2
µ =

[
0.5 0.2
0.1 0.2

] [
0.5 0.2
0.3 0.2

]
=

[
0.5 0.2
0.2 0.2

]

Yνu
2
ν =

[
0.1 0.5
0.2 0.3

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.5
0.2 0.3

]

Yµv
2
µ =

[
0.5 0.2
0.1 0.2

] [
0.6 0.2
0.3 0.5

]
=

[
0.5 0.2
0.2 0.2

]

Yνv
2
ν =

[
0.1 0.5
0.2 0.3

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.2 0.3

]
Therefore Y u2 = Y v2. Hence u <−

k v.

Here uTµ =

[
0.5 0.3
0.2 0.1

]
, uTν =

[
0.1 0.2
0.3 0.5

]

u2µu
T
µuµ =

[
0.5 0.2
0.3 0.2

] [
0.5 0.3
0.2 0.1

] [
0.5 0.2
0.3 0.1

]
=

[
0.5 0.2
0.3 0.2

]
= u2µ

uνu
T
ν u

2
ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.2 0.3

]
= u2ν

uµu
T
µu

2
µ =

[
0.5 0.2
0.3 0.1

] [
0.5 0.3
0.2 0.1

] [
0.5 0.2
0.3 0.2

]
=

[
0.5 0.2
0.3 0.2

]
= u2µ

u2νu
T
ν uν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
= u2ν

Therefore u2uTu = u2 and uuTu2 = u2.

u2µu
T
µ =

[
0.5 0.2
0.3 0.2

] [
0.5 0.3
0.2 0.1

]
=

[
0.5 0.3
0.3 0.3

]

v2µu
T
µ =

[
0.6 0.2
0.3 0.5

] [
0.5 0.3
0.2 0.1

]
=

[
0.5 0.3
0.3 0.3

]

u2νu
T
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]

v2νu
T
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]
Therefore u2uT = v2uT .

uTµu
2
µ =

[
0.5 0.3
0.2 0.1

] [
0.5 0.2
0.3 0.2

]
=

[
0.5 0.2
0.2 0.2

]



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 13 of 31

uTµv
2
µ =

[
0.5 0.3
0.2 0.1

] [
0.6 0.2
0.3 0.5

]
=

[
0.5 0.3
0.2 0.2

]

uTν u
2
ν =

[
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.3 0.3

]

uTν v
2
ν =

[
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.3 0.3

]
Therefore uTu2 ̸= uT v2.

u2µu
T
µ =

[
0.5 0.2
0.3 0.2

] [
0.5 0.3
0.2 0.1

]
=

[
0.5 0.3
0.3 0.3

]

(u2µu
T
µ )

T =

[
0.5 0.3
0.3 0.3

]

u2νu
T
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]

(u2νu
T
ν )

T =

[
0.1 0.2
0.2 0.2

]
Therefore (u2uT )T = u2uT .
Therefore uT is a 2-Moore-Penrose inv of u.

But u2uT = v2uT and uTu2 ̸= uT v2

Hence u̸<T
k v.

Lemma 2. Let < uµ, uν >∈ (IFM)n and v =< vµ, vν >∈ (IFM)n. u <T
k v ⇐⇒ uµ <T

k

vµ and uν <T
k vν .

Proof.

u <T
k v ⇔ ukuT = vkuT and uTuk = uT vk

uTuk = uT vk ⇒ ⟨uTµ , uTν ⟩⟨ukµ, ukν⟩
= ⟨uTµ , uTν ⟩⟨vkµ, vkν ⟩
⇔ ⟨uTµukµ, uTν ukν⟩ = ⟨uTµukµ, uTν vkν ⟩
⇐⇒ uTµu

k
µ = uTµv

k
µ and uTν u

k
ν = uTν v

k
ν

Similarly,
ukuT = vkuT ⇔ ukµu

T
µ = vkµu

T
µ and ukνu

T
ν = vkνu

T
ν

Hence, u <T
k v ⇐⇒ uµ <T

k vµ and uν <T
k vν .



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 14 of 31

Theorem 3. For u ∈ (IFM)n, we have the following:

u+k = uT ⇔ uT ∈ u{1rk} ∩ u{1lk} and [uT ∈ u{3k} or uT ∈ u{4k}].

Proof. Since u+k = uT ,
the result directly follows from the Definition 3 and 4.
Conversely, let uT ∈ u

{
1kr
}
∩ u

{
1kl
}
.

Then, by Definition 1,

ukuTu = uk (2)

and by Definition 2,
uTuuk = uk (3)

By taking the transpose on both sides in Equation (2) and (3), we have

uTu(uT )k = (uT )k (4)

and
(uT )kuuT = (uT )k (5)

⇒ uT ∈ u
{
2kr

}
and uT ∈ u

{
2kl

}
.

Let uT ∈ u
{
3k
}
.

We claim that uT is a solution of equation {4k}.
Since uT ∈ u{3k},

(ukuT )T = ukuT

⇒ u(uT )k = ukuT .

Pre-multiplying by uT and post-multiplying by u, we get

uTu(uT )ku = uTukuTu.

By using Equations (2) and (4), we have

(uT )ku = uTuk

⇒ (uTuk)T = uTuk

⇒ uT ∈ u{4k}.

Hence u+r and u+ℓ exist and they are equal.
By Remark 1 u+k = u+rk = u+ℓk = uT .
Hence the Theorem.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 15 of 31

Theorem 4. Let u, v ∈ (IFM)n and u+k exists. Then the following conditions are equiv-
alent:

(i) u <T
k v.

(ii) u+k u
k = u+k v

k and uku+k = vku+k .

Proof. The proof of this theorem directly follows from Definition 7 and Theorem 3

Theorem 5. For u ∈ (IFM)+n and v ∈ (IFM)n, we have:

u <T
k v ⇒ uk = uuT vk = vkuTu.

Proof.
u <T

k v ⇔ ukuT = vkuT and

uTuk = uT vk,

ukuT = vkuT ⇒ uk = vkuTu,

uTuk = uT vk ⇒ uk = uuT vk.

Hence, the theorem.

The converse of the above theorem is need not be true. This illustrated in the following
example.

Example 4. let,

uµ =

[
0.5 0.5
0.5 0.1

]
, uν =

[
0.1 0.3
0.2 0.5

]

u2µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.5

]
̸= uµ

uµP1µuµ ̸= uµ

uµP2µuµ ̸= uµ

uµP3µuµ ̸= uµ

uµP4µuµ ̸= uµ

uµP5µuµ ̸= uµ

uµP6µuµ ̸= uµ

u2µu
T
µuµ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]
= u2µ

Thus, uµ is 2-reg and uTµ is the 2-g-inv of uµ.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 16 of 31

u2ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
̸= uν

uνP1νuν ̸= uν

uνP2νuν ̸= uν

uνP3νuν ̸= uν

uνP4νuν ̸= uν

uνP5νuν ̸= uν

uνP6νuν ̸= uν

u2νu
T
ν uν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
= u2ν

Therefore, uν is 2-reg and uTν is the 2-g-inv of uν . Let,

vµ =

[
0.7 0.6
0.5 0.5

]
and Bν =

[
0 0.3
0.2 0.3

]

v2µ =

[
0.7 0.6
0.5 0.5

] [
0.7 0.6
0.5 0.5

]
=

[
0.7 0.6
0.5 0.5

]
= vµ

vν2 =

[
0 0.3
0.2 0.3

] [
0 0.3
0.2 0.3

]
=

[
0 0.3
0.2 0.3

]
= vν

Therefore, v is regular.

u2µu
T
µuµ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]

=

[
0.5 0.5
0.5 0.5

]
= u2µ.

u2νu
T
ν uν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.5

]

=

[
0.1 0.3
0.2 0.3

]
= u2ν

uµu
T
µu

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.5

]



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 17 of 31

=

[
0.5 0.5
0.5 0.5

]
= u2µ

uνu
T
ν u

2
ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]

=

[
0.1 0.3
0.2 0.3

]
= u2ν

(u2µu
T
µ )

T =

[
0.5 0.5
0.5 0.5

]

u2µu
T
µ =

[
0.5 0.5
0.5 0.5

]

(u2νu
T
ν )

T =

[
0.1 0.2
0.2 0.2

]

u2νu
T
ν =

[
0.1 0.2
0.2 0.2

]
Therefore, u2uTu = u2 and uuTu2 = u2.

(u2uT )T = u2uT .

So, uT is a 2-Moore-Penrose inv of u.

uµu
T
µv

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

] [
0.7 0.6
0.5 0.5

]
=

[
0.5 0.5
0.5 0.5

]
= u2µ

uνu
T
ν v

2
ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]
=

[
0.1 0.3
0.2 0.3

]
= u2ν

v2µu
T
µuµ =

[
0.7 0.6
0.5 0.5

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.5

]
= u2µ

v2νu
T
ν uν =

[
0 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
= u2ν

u2µu
T
µ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.5

]
v2µu

T
µ =

[
0.7 0.6
0.5 0.5

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.1

]
u2νu

T
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 18 of 31

v2νu
T
ν =

[
0 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]
uTµu

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.5

]
=

[
0.5 0.5
0.5 0.5

]
uTµv

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.7 0.6
0.5 0.5

]
=

[
0.5 0.5
0.5 0.5

]
uTν u

2
ν =

[
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.3 0.3

]
uTν v

2
ν =

[
0.1 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]
=

[
0.1 0.3
0.3 0.3

]
Therefore, u̸<T

k v

Remark 6. For k = 1, Theorem 5 and Theorem 4 reduces to the following.

Theorem 6. Let A,B ∈ (IFM)mn and A+ exists. Then the following conditions are
equivalent:

(i) A <T B

(ii) A+A = A+B and AA+ = BA+

(iii) AA+B = A = BA+A

Theorem 7. For u, v ∈ (IFM)n, u
+
k and v+k both exist:

u <T
k v ⇒

(
uT

)k
= (vT )kvvT = vT v(uT )k.

Proof.

ukuT = vkuT and uTuk = uT vk,

u+k and v+k exist.
Take u+k = uT and v+k = vT .



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 19 of 31

uTuk =
(
uTuk

)T
,

=
(
uT vk

)T
,

=
(
uT

(
vkvT v

))T
,

=
(
vT v

)T (
uT vk

)T
,

= vT v
(
uTuk

)T
,(

uTuk
)T

= vT v
(
uTuk

)T
,(

uT
)k

u = vT v
(
uT

)k
u,(

uT
)k

uuT = vT v
(
uT

)k
uuT .

Thus,
(uτ )k = vT v (uτ )k .

Similarly, (
uT

)k
=

(
uT

)k
vvT .

Theorem 8. In (IFM)+n , the set of all matrices u ∈ (IFM)n for which u+k exists, <T
k is

not a partial ordering.

Proof. u <T
k u is obvious.

Hence <T
k is reflexive.

By Theorem 5,
u <T

k v ⇒ uk = vkuTu = uuT vk

and v <T
k u ⇒ vk = ukvT v = vvTuk

Now,
uk = vkuTu

=
(
vvTuk

)
uTu

= vvT
(
ukuTu

)
= vvTuk

= vk

Hence, <T
k is anti-symmetric.

u <T
k v and v <T

k w ⇒ u̸<T
kw

Hence <T
k is not transitive.

Thus, <T
k is not a partial ordering. This is explained in the following example.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 20 of 31

Example 5. Let,

uµ =

[
0.5 0.5
0.5 0.1

]
, uν =

[
0.1 0.3
0.2 0.5

]

u2µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.5

]
̸= uµ

uµP1µuµ ̸= uµ

uµP2µuµ ̸= uµ

uµP3µuµ ̸= uµ

uµP4µuµ ̸= uµ

uµP5µuµ ̸= uµ

uµP6µuµ ̸= uµ

u2µu
T
µuµ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]
= u2µ

Thus, uµ is 2-reg and uTµ is the 2-g-inv of uµ.

u2ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
̸= uν

uνP1νuν ̸= uν

uνP2νuν ̸= uν

uνP3νuν ̸= uν

uνP4νuν ̸= uν

uνP5νuν ̸= uν

uνP6νuν ̸= uν

u2νu
T
ν uν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.5

]
=

[
0.1 0.3
0.2 0.3

]
= u2ν

Therefore, uν is 2-reg and uTν is the 2-g-inv of uν .
Let,

vµ =

[
0.7 0.6
0.5 0

]
, vν =

[
0 0.3
0.2 0.5

]

v2µ =

[
0.7 0.6
0.5 0

] [
0.7 0.6
0.5 0

]
=

[
0.7 0.6
0.5 0.5

]
̸= vµ



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 21 of 31

vµP1µvµ ̸= vµ

vµP2µvµ ̸= vµ

vµP3µvµ ̸= vµ

vµP4µvµ ̸= vµ

vµP5µvµ ̸= vµ

vµP6µvµ ̸= vµ

v2µv
T
µ vµ =

[
0.7 0.6
0.5 0.5

] [
0.7 0.5
0.6 0

] [
0.7 0.6
0.5 0

]

=

[
0.7 0.6
0.5 0.5

]
= v2µ

Thus, vµ is 2-reg and vTµ is the 2-g-inv of vµ.

v2ν =

[
0 0.3
0.2 0.5

] [
0 0.3
0.2 0.5

]
=

[
0 0.3
0.2 0.3

]
̸= vν

vνP1νvν ̸= vν

vνP2νvν ̸= vν

vνP3νvν ̸= vν

vνP4νvν ̸= vν

vνP5νvν ̸= vν

vνP6νvν ̸= vν

v2νv
T
ν vν =

[
0 0.3
0.2 0.3

] [
0 0.2
0.3 0.5

] [
0 0.3
0.2 0.5

]
=

[
0 0.3
0.2 0.3

]
= v2ν

Therefore vν is 2-reg and vTν is the 2-g inv of vν .
Therefore, v is 2-reg and vT is the 2-g inv of v.

uTµu
2
µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.5

]

=

[
0.5 0.5
0.5 0.5

]

uTµv
2
µ =

[
0.5 0.5
0.5 0.1

] [
0.7 0.6
0.5 0.5

]



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 22 of 31

=

[
0.5 0.5
0.5 0.5

]

uTν u
2
ν =

[
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]

=

[
0.1 0.3
0.3 0.3

]

uTν v
2
ν =

[
0.1 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]

=

[
0.1 0.3
0.3 0.3

]

u2µu
T
µ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

]

=

[
0.5 0.5
0.5 0.5

]

v2µu
T
µ =

[
0.7 0.6
0.5 0.5

] [
0.5 0.5
0.5 0.1

]

=

[
0.5 0.5
0.5 0.5

]

u2νu
T
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]

=

[
0.1 0.2
0.2 0.2

]

v2νu
T
ν =

[
0 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]

=

[
0.1 0.2
0.2 0.2

]
Therefore, u2uT = v2uT and uTu2 = uT v2.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 23 of 31

So, u <T
k v.

u2µu
T
µuµ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

]

=

[
0.5 0.5
0.5 0.5

]
= u2µ.

u2νu
T
ν uν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.5

]

=

[
0.1 0.3
0.2 0.3

]
= u2ν

uµu
T
µu

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.5

]

=

[
0.5 0.5
0.5 0.5

]
= u2µ

uνu
T
ν u

2
ν =

[
0.1 0.3
0.2 0.5

] [
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]

=

[
0.1 0.3
0.2 0.3

]
= u2ν

(u2µu
T
µ )

T =

[
0.5 0.5
0.5 0.5

]

u2µu
T
µ =

[
0.5 0.5
0.5 0.5

]

(u2νu
T
ν )

T =

[
0.1 0.2
0.2 0.2

]

uν2uTν =

[
0.1 0.2
0.2 0.2

]
Therefore, u2uTu = u2 and uuTu2 = u2.

(u2uT )T = u2uT .

So, uT is a 2-Moore-Penrose inv of u.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 24 of 31

Let,

wµ =

[
0.7 0.6
0.5 0.5

]
and wν =

[
0 0.3
0.2 0.3

]

w2
µ =

[
0.7 0.6
0.5 0.5

] [
0.7 0.6
0.5 0.5

]
=

[
0.7 0.6
0.5 0.5

]
= wµ

w2
ν =

[
0 0.3
0.2 0.3

] [
0 0.3
0.2 0.3

]
=

[
0 0.3
0.2 0.3

]
= wν

Therefore, w is regular.

w2
µv

T
µ =

[
0.7 0.6
0.5 0.5

] [
0.7 0.5
0.6 0.5

]
=

[
0.7 0.5
0.5 0.5

]

v2νv
T
ν =

[
0 0.3
0.2 0.3

] [
0 0.2
0.3 0.5

]
=

[
0 0.2
0.2 0.2

]

w2
νv

T
ν =

[
0 0.3
0.2 0.3

] [
0 0.2
0.3 0.5

]
=

[
0 0.2
0.2 0.2

]

vTµ v
2
µ =

[
0.7 0.5
0.6 0.5

] [
0.7 0.6
0.5 0.5

]
=

[
0.7 0.6
0.6 0.6

]

vTµw
2
µ =

[
0.7 0.5
0.6 0.5

] [
0.7 0.6
0.5 0.5

]
=

[
0.7 0.6
0.6 0.6

]

vTν v
2
ν =

[
0 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]
=

[
0 0.3
0.3 0.3

]

vTν w
2
ν =

[
0 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]
=

[
0 0.3
0.3 0.3

]
Therefore, v <T

k w.

v2µv
T
µ vµ =

[
0.7 0.5
0.5 0.5

] [
0.7 0.6
0.5 0.5

]
=

[
0.7 0.6
0.5 0.5

]
= v2µ

v2νv
T
ν vν =

[
0 0.2
0.2 0.2

] [
0 0.3
0.2 0.5

]
=

[
0 0.3
0.2 0.3

]
= v2ν

vµv
T
µ v

2
µ =

[
0.7 0.5
0.5 0.5

] [
0.7 0.6
0.5 0.5

]
=

[
0.7 0.6
0.5 0.5

]
= v2µ

vνv
T
ν v

2
ν =

[
0 0.3
0.2 0.5

] [
0 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]
=

[
0 0.3
0.2 0.3

]
= v2ν



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 25 of 31

(v2µv
T
µ )

T =

[
0.7 0.5
0.5 0.5

]
= v2µv

T
µ

(v2νv
T
ν )

T =

[
0 0.2
0.2 0.2

]
= v2νv

T
ν

Therefore vT is a 2-Moore Penrose inv of v.

u2µu
T
µ =

[
0.5 0.5
0.5 0.5

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.5

]
w2
µu

T
µ =

[
0.7 0.6
0.5 0.5

] [
0.5 0.5
0.5 0.1

]
=

[
0.5 0.5
0.5 0.1

]
u2νu

T
ν =

[
0.1 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]
w2
νu

T
ν =

[
0 0.3
0.2 0.3

] [
0.1 0.2
0.3 0.5

]
=

[
0.1 0.2
0.2 0.2

]
uTµu

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.5 0.5
0.5 0.5

]
=

[
0.5 0.5
0.5 0.5

]
uTµw

2
µ =

[
0.5 0.5
0.5 0.1

] [
0.7 0.6
0.5 0.5

]
=

[
0.5 0.5
0.5 0.5

]
uTν u

2
ν =

[
0.1 0.2
0.3 0.5

] [
0.1 0.3
0.2 0.3

]
=

[
0.1 0.3
0.3 0.3

]
uTν w

2
ν =

[
0.1 0.2
0.3 0.5

] [
0 0.3
0.2 0.3

]
=

[
0.1 0.3
0.3 0.3

]
Therefore, u̸<T

kw
So, T-ordering is not a partial ordering.

Theorem 9. Let u ∈ (IFM)+n and v ∈ (IFM)n. Then the following are equivalent:
(i) u <T

k v ⇐⇒ uT <T
k vT

(ii) u <T
k v ⇐⇒ QuQT <T

k QvQT for some permutation matrix Q

Proof. (i)
u <T

k v ⇐ ukuT = vkuT and uTuk = uT vk

Since u+k exists, u+k = uT . By Theorem 1,

uT ∈ u{1kr} ⇐⇒ u ∈ uT {1kl }

uT ∈ u{1kl } ⇐⇒ u ∈ uT {1kr}



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 26 of 31

uT ∈ u{3k} ⇐⇒ (ukuT )T = ukuT

⇐⇒ u(uT )k = ukuT

⇐⇒ (u(uT )k)T = (ukuT )T = u(uT )k

⇐⇒ u ∈ uT {4k}

Similarly,
uT ∈ u{4k} ⇐⇒ u ∈ uT {3k}

ukuT = vkuT ⇐⇒ (ukuT )T = (vkuT )T

⇐⇒ u(uT )k = u(vT )k

and uTuk = uT vk ⇐⇒ (uTuK)T = (uT vk)T

⇐⇒ (uT )ku = (vT )ku

Here, u <T
k v ⇐⇒ uT <T

k vT

(ii) Claim:
If u+k = uT , then (QuQT )+k = (QuQT )T = QuTQT

(QuQT )k(QuTQT )(QuQT )

= (QukQT )(QuTQT )(QuQT )

= Quk(QTQ)uT (QTQ)uQT

= QukuTuQT

= QukQT

= (QuQT )k

Similarly,



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 27 of 31

(QuQT )(QuTQT )(QuQT )k = (QuQT )k

((QuQT )k(QuTQT ))T = (QuTQT )T ((QuQT )T )k

= (QuQT )(QuTQT )k

= (QuQT )
(
Q(uT )kQT

)
= Qu(QTQ)(uT )kQT

= Qu(uT )kQT

= Q(ukuT )TQT

= QukuTQT

(QuQT )k(QuQT ) = (QukQT )QuTQT

= Quk(QTQ)uTQT

= QukuTQT

Therefore,
((QuQT )k(QuTQT ))T = (QuQT )k(QuTQT )

Similarly,

((QuTQT )(QuQT )k)T = (QuTQT )(QuQT )k

Now,
(QuTQT )(QuQT )k = (QuTQT )(QukQT )

= QuT (QTQ)ukQT

= Q(uTuk)QT

= Q(uT vk)QT



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 28 of 31

= (QuTQT )(QvkQT )

= (QuTQT )(QvQT )k

Similarly,

(QuQT )k(QuTQT ) = (QvQT )k(QuTQT )

Hence,
u <T

k u ⇒ QuQT <T
k QvQT

Conversely,

QuQT <T
k QvQT ⇒ QT (QuQT )Q <T

k QT (QvQT )Q ⇒ u <T
k v

Theorem 10. Let u, v ∈ (IFM)+n .

v ∈ uT {1k, 3k, 4k} ⇔ vT ∈ u{1k, 3k, 4k}

Proof. By Theorem 1,
v ∈ uT {1k} ⇔ vT ∈ u{1k}

v ∈ uT {3k} ⇔ ((uT )kv)T = (uT )kv

⇔ vTuk = (uT )kv

⇔ (vTuk)T = ((uT )kv)T = vTuk

⇔ vT ∈ u{4k}

Similarly,

v ∈ uT {4k} ⇔ vT ∈ u{3k}

Hence the proof.

Theorem 11. Let u ∈ (IFM)+n and v ∈ (IFM)n, u <T
k v ⇒ u ∈ vT {3k, 4k}

Proof.
u <T

k v ⇔ ukuT = vkuT and uTuk = uT vk



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 29 of 31

(uT vk)T = (uTuk)T

= uTuk

= uT vk

⇒ uT ∈ v{4k}
(vkuT )T = (ukuT )T

= ukuT

= vkuT

⇒ uT ∈ v{3k}

By Theorem 10, uT ∈ v{3k, 4k} ⇔ u ∈ vT {3k, 4k}.
Hence the proof.

Theorem 12. If u <T
k v, then we have the following:

(i) If
(
vk
)2

= 0, then
(
uk

)2
= 0.

(ii) If vk =
(
vk
)2
, then uk =

(
uk

)2
.

Proof. By Theorem 5,

u <T
k v ⇒ uk = uuT vk = vkuTu

(i) (
uk

)2
= ukuk

= (uuT vk)(vkuTu)

= uuT
(
vk
)2

uTu

= 0

(ii) (
uk

)2
= ukuk

= (uuT vk)(vkuTu)

= uuT
(
vk
)2

uTu

= uuT vkuTu

= ukuTu = uk



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 30 of 31

4. Conclusion

The concept of intuitionistic fuzzy matrix was defined as a generalization of fuzzy
matrix utilizing the notion of intuitionistic fuzzy sets. We proved that T -ordering is
identical for certain class of intuitionistic fuzzy matrices. And also, we learned about
the k − T ordering on intuitionistic fuzzy matrices as a generalization of the T-ordering
on intuitionistic fuzzy matrices. In many applications, the parameters of the system
should be represented by intuitionistic fuzzy rather than crisp or fuzzy numbers. Hence,
it is important to develop the mathematical procedures that would appropriately treat
intuitionistic fuzzy linear systems to solve them. Further, we can introduce the concept
of k-regularity for fuzzy and intuitionistic fuzzy soft matrices and neutrosophic matrices.

References

[1] Krassimir T. Atanassov. Intuitionistic Fuzzy Sets. Springer, 1999.
[2] Adi Ben-Israel and Thomas N. E. Greville. Generalized Inverses: Theory and Appli-

cations. Springer Science & Business Media, 2006.
[3] Ki Hang Kim and Fred W. Roush. Inverses of boolean matrices. Linear Algebra and

Its Applications, 22:247–262, 1978.
[4] Han Hyuk Cho. Regular fuzzy matrices and fuzzy equations. Fuzzy Sets and Systems,

105(3):445–451, 1999.
[5] A. R. Meenakshi and P. Jenita. Generalized regular fuzzy matrices. Iranian Journal

of Fuzzy Systems, 8(2):133–141, 2011.
[6] Susanta K. Khan and Anita Pal. The generalised inverse of intuitionistic fuzzy ma-

trices. Journal of Physical Sciences, 2007.
[7] Rajkumar Pradhan and Madhumangal Pal. Some results on generalized inverse of

intuitionistic fuzzy matrices. Fuzzy Information and Engineering, 6:133–145, 2014.
[8] Madhumangal Pal, Susanta Khan, and Amiya K. Shyamal. Intuitionistic fuzzy ma-

trices. Notes on Intuitionistic Fuzzy Sets, 8(2):51–62, 2002.
[9] A. R. Meenakshi and T. Gandhimathi. On regular intuitionistic fuzzy matrices.

International Journal of Fuzzy Mathematics, 19(2):599–605, 2011.
[10] S. Sriram and P. Murugadas. The moore-penrose inverse of intuitionistic fuzzy ma-

trices. International Journal of Mathematical Analysis, 4(36):1779–1786, 2010.
[11] Jianmiao Cen. Fuzzy matrix partial orderings and generalized inverses. Fuzzy Sets

and Systems, 105(3):453–458, 1999.
[12] Longrio Platil and Tamaki Tanaka. Multi-criteria evaluation for intuitionistic fuzzy

sets based on set-relations. Journal of the Operations Research Society of Japan, 2023.
[13] P. Poongodi, C. Padmavathi, R. Vinitha, and G. Hema. Orderings on generalized

regular interval valued fuzzy matrices. International Journal of Engineering and
Advanced Technology, 10:194–198, 2020.

[14] A. R. Meenakshi. Fuzzy Matrix: Theory and Applications. MJP Publisher, 2019.
[15] P. Jenita and E. Karuppusamy. Special type of inverses of k-regular intuitionistic

fuzzy matrices. TEST Engineering and Management, 83:20035–20049, 2022.



P. Jenita et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6241 31 of 31

[16] P. Jenita, E. Karuppusamy, and D. Thangamani. k-pseudo similar intuitionistic fuzzy
matrices. Annals of Fuzzy Mathematics and Informatics, 14(5):433–443, 2017.

[17] A. R. Meenakshi and C. Inbam. The minus partial order in fuzzy matrices. Journal
of Fuzzy Mathematics, 12(3):695–700, 2004.

[18] P. Jenita, E. Karuppusamy, and M. Princy Flora. A study on ordering on generalized
regular intuitionistic fuzzy matrices. Journal of Algebraic Statistics, 13(3):1415–1422,
2022.

[19] P. Jenita, M. Princy Flora, and E. Karuppusamy. K - sharp ordering on intuitionistic
fuzzy matrices. Advances in Nonlinear Variational Inequalities, 28:350–364, 2024.

[20] Rupkumar Mahapatra, Sovan Samanta, Tofigh Allahviranloo, and Madhumangal Pal.
Radio fuzzy graphs and assignment of frequency in radio stations. Computational and
Applied Mathematics, 38:1–20, 2019.

[21] Rupkumar Mahapatra, Sovan Samanta, and Madhumangal Pal. Applications of edge
colouring of fuzzy graphs. Informatica, 31(2):313–330, 2020.

[22] Rupkumar Mahapatra, Sovan Samanta, Madhumangal Pal, and Qin Xin. Link predic-
tion in social networks by neutrosophic graph. International Journal of Computational
Intelligence Systems, 13(1):1699–1713, 2020.

[23] Rupkumar Mahapatra, Sovan Samanta, and Madhumangal Pal. Generalized neu-
trosophic planar graphs and its application. Journal of Applied Mathematics and
Computing, 65(1):693–712, 2021.

[24] Rupkumar Mahapatra, Sovan Samanta, and Madhumangal Pal. Edge colouring of
neutrosophic graphs and its application in detection of phishing website. Discrete
Dynamics in Nature and Society, 2022(1):1149724, 2022.

[25] Rupkumar Mahapatra, Sovan Samanta, and Madhumangal Pal. Detecting influential
node in a network using neutrosophic graph and its application. Soft Computing,
27(14):9247–9260, 2023.

[26] Rupkumar Mahapatra, Sovan Samanta, Madhumangal Pal, Tofigh Allahviranloo, and
Antonios Kalampakas. A study on linguistic z-graph and its application in social
networks. Mathematics, 12(18):2898, 2024.

[27] Rupkumar Mahapatra, Prasenjit Mandal, Sovan Samanta, Vivek Kumar Dubey,
Madhumangal Pal, and Tofigh Allahviranloo. Centrality measure using linguistic
z-graph and its application. In Management of Uncertainty Using Linguistic Z-
Numbers: Applications for Decision-Making, Granular Computing and Social Net-
works, pages 219–240. Springer, 2024.

[28] P. Jenita and E. Karuppusamy. Inverses of k-regular intuitionistic fuzzy matrices.
International Journal of Pure and Applied Mathematics, 119(12):2341–2359, 2018.