Solving Positive Trapezoidal Fully Fuzzy Sylvester Matrix Equation

Abstract

Many problems in systems and control theory are related to solvability of Sylvester matrix equations. In many applications, at least some of the parameters of the system should be represented by fuzzy numbers rather than crisp ones. In most of the previous literature, the solutions of fuzzy Sylvester matrix equation are only presented with triangular fuzzy numbers. In this paper, we propose two analytical methods for solving Positive Trapezoidal Fully Fuzzy Sylvester Matrix Equation (PTrFFSME). The PTrFFSME is converted to an equivalent system of crisp Sylvester Matrix Equations (SME) using the existing arithmetic fuzzy multiplication operations. The necessary and sufficient conditions for the existence and uniqueness of the positive fuzzy solutions to the PTrFFSME are investigated. In addition, the equivalency between the solution to the system of SME and the PTrFFSME are discussed. The proposed methods are illustrated by solving one example.

KEYWORDS:

• Bartels Stewart
• fully fuzzy Sylvester matrix equations
• Kronecker product
• Schur decomposition
• trapezoidal fuzzy numbers

Mathematics Subject Classifications:

• 03B52
• 94D05

1. Introduction

Sylvester matrix equation in the form $AX+XB=C$ played a very important role in many areas such as in control systems [1–3], reduction of non-linear control systems models [4,5], system design [6], theory of orbits [7] and medical imaging acquisition system [8–11]. In many scenarios, the classical linear system is not well suited to handle uncertain information ties in real-life problems, since some coefficient values may be vague and imprecise due to incomplete information [12]. Using fuzzy numbers rather than the crisp numbers is a common way of expressing such imprecision [13]. The SME can be extended to a Fuzzy Sylvester Matrix Equation (FSME) in the form $A\tilde{X}+\tilde{X}B=\tilde{C}$ if the solution matrix $\tilde{X}$ and the constant matrix $\tilde{C}$ are in fuzzy form. The FSME was studied in [14,15], where the Kronecker product was applied to convert the FSME to a fuzzy linear system. However, this method can be applied to a FSME with small size only. In order to overcome this shortcoming, authors in [16] applied Dubois and Prade’s arithmetic multiplication operations [17] to convert the FSME to a SME and then the fuzzy solution obtained by applying Bartle’s Stewart method. When all the parameters of the SME are in the form positive trapezoidal fuzzy matrices then it is called PTrFFSME.

Several studies were conducted to solve the Triangular Fully Fuzzy Sylvester Matrix Equation (TFFSME), in the form $\tilde{A}\tilde{X}+\tilde{X}\tilde{B}=\tilde{C}$. Shang et al. [18] extended the TFFSME into a system of three crisp linear matrix equations, while Malkawi et al. [19] applied the Kronecker product method to convert the TFFSME to a Fully Fuzzy Linear System (FFLS). The FFLS is then converted to a linear system and the solution obtained by direct inversion method. Similarly, Daud et al. [20–22] applied Kronecker product and Vec-operator to convert the TFFSME to a system of matrix linear equation. However, in all these methods the size of the obtained linear system equation is much larger than the TFFSME considered. Therefore, these methods are limited to TFFSME with small size. Hou et al. [23] proposed a method for solving TFFSME with parametric fuzzy numbers using $\alpha -cut$. However, the consistency of the fuzzy solution cannot be checked before applying the method. Therefore, their method needs further modifications.

In a recent study by [24], new analytical methods for solving both positive and negative Trapezoidal Fully Fuzzy Sylvester Matrix Equation (TrFFSME) were introduced. It was the first attempt to extend the concept of Kronecker product and Vec-operator for solving TrFFSME. The arbitrary fuzzy solution to the TrFFSME with arbitrary coefficients was considered by [25] where two-stage algorithm was constructed to find all possible solutions. The main limitation of this method is that it needs along computational time and, therefore, large memory storage in order to find all possible fuzzy solutions.

A coupled of TrFFSME in general form was considered in [26] where the fuzzy solution is obtained analytically by the fuzzy matrix vectorization method and numerically by gradient and least square methods. However, these methods need further modification to be applied for TrFFSME in LR-form. On the other hand, there was a study on the TFFSME in the form $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}$ by [27]. The method obtained maximal and minimal symmetric positive solution for the TFFSME. However, this method is still limited for some TFFSME applications because the method used required long multiplication process and therefore long computational time.

Most of the existing literature for solving TFFSME and TrFFSME are based on Kronecker product and Vec-operator. As the size of the system increases the computational time is also increased due to the complexity of the procedure applied. Although analytical solutions, which can be computed using Vec-operator and Kronecker product, are important, the computational efforts rapidly increase with the dimensions of the matrices to be solved. For example, it required getting the inverse of $mn\times mn$ matrix for a system of size $m\times n$ which leads to computation complexity. Therefore, this method is limited to systems with small coefficients only. Consequently, the TFFSME and TrFFSME were considered with small size matrices only. Alternative ways exist which transform the matrix equations into forms for which solutions may be readily computed, such as the Jordan canonical form [28] and Hessenberg–Schur form [29]. Therefore, in dealing with this shortcoming, this paper presents two different methods for solving the PTrFFSME; the generalized Bartels Stewart’s method and matrix vectorization method. The proposed methods are able to obtain positive fuzzy solutions for large PTrFFSME with short computational timing using Matlab or Mathematica. This paper is organized as follows. In Section 2, the preliminary concepts and arithmetic operations of trapezoidal fuzzy numbers are discussed. In Section 3, two new methods for solving PTrFFSME are developed. In Section 4, numerical example is solved using the proposed methods. In Section 5, conclusion about the proposed methods and achieved results will be drawn.

2. Preliminaries

In this section some basic arithmetic operations of fuzzy numbers are introduced [17,30].

Definition 2.1:

A fuzzy number $\tilde{A}=\left(m,p,\alpha ,\text{β}\right)$ is said to be LR-TrFN, if its membership function is given by ${\mu }_{\tilde{A}}\left(x\right)=\{\begin{array}{l}1-{\displaystyle \frac{m-x}{\alpha }},m-\alpha \le x〈m,\alpha 〉0,\\ 1m<x<p,\\ 1-{\displaystyle \frac{x-n}{\text{β}}},n\le x〈n+\text{β},\text{β}〉0,\\ 0\text{otherwise}\text{.}\end{array}$

Definition 2.2:

The sign of $\tilde{A}=\left(m,p,\alpha ,\text{β}\right)$ is classified as follows:

• $\tilde{A}$ is called positive (negative) iff $m-\alpha \ge 0,\left(\text{β}+p\le 0\right).$

• $\tilde{A}$ is called zero iff $\left(m=0,p=0,\alpha =0,\text{β}=0\right).$

• $\tilde{A}$ is called near zero iff $m-\alpha \le 0\le \text{β}+n.$

Definition 2.3:

A fuzzy number $\tilde{A}=\left(m,p,\alpha ,\text{β}\right)$ is a zero fuzzy trapezoidal number iff $m=0,p=0,\alpha =0and\text{β}=0$.

Definition 2.4:

The fuzzy numbers $\tilde{A}=\left(m,p,\alpha ,\text{β}\right)$ and $\tilde{B}=\left(a,b,\gamma ,\delta \right)$ are equal iff $m=a,p=b,\alpha =\gamma and\text{β}=\delta$.

Definition 2.5:

The arithmetic operations of fuzzy numbers are presented as follows, let $\tilde{A}=\left(m,p,\alpha ,\text{β}\right)$ and $\tilde{B}=\left(a,b,\gamma ,\delta \right)$ be two trapezoidal fuzzy numbers then:

Addition: (2.1) $\left(m,p,\alpha ,\text{β}\right)+\left(a,b,\gamma ,\delta \right)=\left(m+a,p+b,\alpha +\gamma ,\text{β}+\delta \right).$(2.1) Subtraction: (2.2) $\left(m,p,\alpha ,\text{β}\right)-\left(a,b,\gamma ,\delta \right)=\left(m-b,p-a,\alpha +\delta ,\text{β}+\gamma \right).$(2.2) Multiplication [31,32]:

• Case I: If $\tilde{A}>0$ and, $\tilde{B}>0$ then: (2.3) $\tilde{A}\cdot \tilde{B}=\left(m,n,\alpha ,\text{β}\right)\cdot \left(p,q,\gamma ,\delta \right)=\left(mp,nq,m\gamma +p\alpha ,n\delta +q\text{β}\right).$(2.3)

• Case II: If $\tilde{A}<0$ and $\tilde{B}<0$ then: (2.4) $\tilde{A}\cdot \tilde{B}=\left(m,n,\alpha ,\text{β}\right)\cdot \left(p,q,\gamma ,\delta \right)=\left(mp,nq,-\left(m\gamma +p\alpha \right),-\left(n\delta +q\text{β}\right)\right).$(2.4)

Equality: Two fuzzy numbers ${\tilde{A}}_1=\left(a_1,n_1,{\alpha }_1,{\text{β}}_1\right)$ and ${\tilde{A}}_2=\left(a_2,n_2,{\alpha }_2,{\text{β}}_2\right)$ are equal, if and only (2.5) $\text{if }{a}_1=a_2,n_1=n_2,{\alpha }_1={\alpha }_{2}and{\text{β}}_1={\text{β}}_2.$(2.5) Scalar multiplication: Let $\lambda \in \mathbb{R}$ then: $\lambda \otimes \left(a,b,\alpha ,\text{β}\right)=\{\begin{array}{cc}\left(\lambda a,\lambda b,\lambda \alpha ,\lambda \text{β}\right)& \lambda \ge 0,\\ \left(\lambda b,\lambda a,-\lambda \text{β},-\lambda \alpha \right)& \lambda <0.\end{array}$

Definition 2.6:

Let the two matrices $A$ and $X$ be $n\times n$ and $m\times m$ respectively. Then the matrix, (2.6) $A\otimes X=\left(a_{ij}X\right)=\left(\begin{array}{ccc}{a}_{11}X& \cdots & a_{1m}X\\ ⋮& \ddots & ⋮\\ a_{m1}X& \cdots & a_{nn}X\end{array}\right).$(2.6) is called the Kronecker product of $A$ and $X$, it is also called the direct product or tensor product.

Definition 2.7:

The Vec-operator generates a column vector from a matrix $A$ by stacking the column vectors of $A=\left(\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right)$ as, (2.7) $Vec\left(A\right)=\left(\begin{array}{c}{a}_{11}\\ a_{12}\\ \begin{array}{c}⋮\\ a_{nn}\end{array}\end{array}\right).$(2.7) In addition to that, If $A=Ve{c}^{-1}\left(\begin{array}{c}{a}_{11}\\ a_{12}\\ \begin{array}{c}⋮\\ a_{nn}\end{array}\end{array}\right)$, then $A=\left(\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right)$.

Definition 2.8

[33,34]: Let $A=(a_{ij}{)}_{q\times q},B=(b_{ij}{)}_{p\times p},$ $X=(x_{ij}{)}_{q\times p}$ and $I_p,I_q$ are identity matrices of order $p$ and $q$ respectively, then:

I)	$Vec\left[AX\right]=\left[I_p\otimes A\right]Vec\left(X\right)$.
II)	$Vec\left[XB\right]=\left[B^T\otimes I_q\right]Vec\left(X\right)$.

Definition 2.9:

The Kronecker difference of two matrices $\ominus$ can be considered as a matrix difference defined by (2.8) $A\ominus B=A\otimes I_q-I_p\otimes B.$(2.8) where $A$ is a square matrix of order $p$ and $B$ is a square matrix of order $q$ and $I_p,I_q$ are identity matrices of order $p$ and $q$ respectively and $\otimes$ represents the Kronecker product.

For example, the Kronecker difference of two $2\times 2$ matrices $a_{ij}$ and $b_{ij}$ is given by: (2.9) $\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)\ominus \left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right)=\left(\begin{array}{cccc}{a}_{11}-b_{11}& -b_{12}& a_{12}& 0\\ -b_{21}& a_{11}-b_{22}& 0& a_{12}\\ a_{21}& 0& a_{22}-b_{11}& -b_{12}\\ 0& a_{21}& -b_{21}& a_{22}-b_{22}\end{array}\right).$(2.9)

Definition 2.10:

The trapezoidal fully fuzzy matrix equation that can be written as (2.10) $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}.$(2.10) where $\tilde{A}=({\tilde{a}}_{ij}{)}_{n\times n}$, $\tilde{B}=({\tilde{b}}_{ij}{)}_{m\times m}$, $\tilde{C}=({\tilde{c}}_{ij}{)}_{n\times m}$ and $\tilde{X}=({\tilde{x}}_{ij}{)}_{n\times m}$ is called a TrFFSME and it can also be written in the form: (2.11) $\sum _{i,j=1}^n{\tilde{a}}_{ij}{\tilde{x}}_{ij}-\sum _{i,j=1}^m{\tilde{x}}_{ij}{\tilde{b}}_{ij}={\tilde{c}}_{ij}.$(2.11)

Definition 2.11

[35]: The Schur factorization of a matrix $A$ is the factorization: $A=QR{Q}^T.$ where,

• $R$ is upper triangular matrix which is called a Schur form of A.

• $Q$ is a unitary matrix ($Q{Q}^T=I$).

3. Proposed Method

In this section, the solution to the PTrFFSME is discussed. The PTrFFSME in Equation (2.10) is extended to a system of four SME using arithmetic fuzzy multiplication operations. The positive fuzzy solution afterwards is obtained by developing two different methods. In the following Theorem 3.1, the PTrFFSME is converted to an equivalent system of matrix equations.

Theorem 3.1:

If $\tilde{A}=({\tilde{a}}_{ij}{)}_{n\times n}=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right)>0$, $\tilde{B}=({\tilde{b}}_{ij}{)}_{m\times m}=\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right)>0$,$\tilde{X}=({\tilde{x}}_{ij}{)}_{n\times m}=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right)>0$ and $\tilde{C}=({\tilde{c}}_{ij}{)}_{n\times m}=\left(c_{ij},g_{ij},h_{ij},f_{ij}\right)$, then the PTrFFSME in $\tilde{A}\tilde{X}-\tilde{X}\tilde{D}=\tilde{E}$ is equivalent to the following SME: (3.1) $\{\begin{array}{l}{m}_{ij}{x}_{ij}-y_{ij}{b}_{ij}=c_{ij},\\ n_{ij}{y}_{ij}-x_{ij}{a}_{ij}=g_{ij},\\ m_{ij}{z}_{ij}+{\alpha }_{ij}{x}_{ij}+y_{ij}{\delta }_{ij}+q_{ij}{b}_{ij}=h_{ij},\\ n_{ij}{q}_{ij}+{\text{β}}_{ij}{y}_{ij}+x_{ij}{\gamma }_{ij}+z_{ij}{a}_{ij}=f_{ij}.\end{array}$(3.1)

Proof:

Let $\tilde{A}=({\tilde{a}}_{ij}{)}_{n\times n}=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right)>0$, $\tilde{B}=({\tilde{b}}_{ij}{)}_{m\times m}=\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right)>0$,$\tilde{X}=({\tilde{x}}_{ij}{)}_{n\times m}=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right)>0$ and $\tilde{C}=({\tilde{c}}_{ij}{)}_{n\times m}=\left(c_{ij},g_{ij},h_{ij},f_{ij}\right).$

We have from Definition 2.5 and by Equation (2.3), $\begin{array}{rl}\tilde{A}\tilde{X}& =\left({\tilde{a}}_{ij}\right)\left({\tilde{x}}_{ij}\right)=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right)\left(x_{ij},y_{ij},z_{ij},q_{ij}\right).\\ & =\left(m_{ij}{x}_{ij},n_{ij}{y}_{ij},m_{ij}{z}_{ij}+{\alpha }_{ij}{x}_{ij},n_{ij}{q}_{ij}+{\text{β}}_{ij}{y}_{ij}\right).\end{array}$

and $\begin{array}{rl}\tilde{X}\tilde{B}& =\left({\tilde{x}}_{ij}\right)\left({\tilde{b}}_{ij}\right)=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right)\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right).\\ & =(x_{ij}{a}_{ij},y_{ij}{b}_{ij},x_{ij}{\gamma }_{ij}+z_{ij}{a}_{ij},y_{ij}{\delta }_{ij}+q_{ij}{b}_{ij}.)\end{array}$

Therefore, $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\left(m_{ij}{x}_{ij},n_{ij}{y}_{ij},m_{ij}{z}_{ij}+{\alpha }_{ij}{x}_{ij},n_{ij}{q}_{ij}+{\text{β}}_{ij}{y}_{ij}\right)-\left(x_{ij}{a}_{ij},y_{ij}{b}_{ij},x_{ij}{\gamma }_{ij}+z_{ij}{a}_{ij},y_{ij}{\delta }_{ij}+q_{ij}{b}_{ij}\right).$

Which can be written as $\begin{array}{rl}\tilde{A}\tilde{X}-\tilde{X}\tilde{B}& =(m_{ij}{x}_{ij}-y_{ij}{b}_{ij},n_{ij}{y}_{ij}-x_{ij}{a}_{ij},m_{ij}{z}_{ij}+{\alpha }_{ij}{x}_{ij}+y_{ij}{\delta }_{ij}+q_{ij}{b}_{ij},n_{ij}{q}_{ij}\\ & +{\text{β}}_{ij}{y}_{ij}+x_{ij}{\gamma }_{ij}+z_{ij}{a}_{ij}).\end{array}$

Since $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}$ can be written as $\sum _{i,j=1}^n{\tilde{a}}_{ij}{\tilde{x}}_{ij}-\sum _{i,j=1}^m{\tilde{x}}_{ij}{\tilde{b}}_{ij}={\tilde{c}}_{ij}.$ Therefore, the PTrFFSME $\tilde{A}\tilde{X}-\tilde{X}\tilde{D}=\tilde{E}$ is equivalent to the following system of SME: $\{\begin{array}{l}{m}_{ij}{x}_{ij}-y_{ij}{b}_{ij}=c_{ij},\\ n_{ij}{y}_{ij}-x_{ij}{a}_{ij}=g_{ij},\\ m_{ij}{z}_{ij}+{\alpha }_{ij}{x}_{ij}+y_{ij}{\delta }_{ij}+q_{ij}{b}_{ij}=h_{ij},\\ n_{ij}{q}_{ij}+{\text{β}}_{ij}{y}_{ij}+x_{ij}{\gamma }_{ij}+z_{ij}{a}_{ij}=f_{ij}.\end{array}$ To solve the PTrFFSME in Equation (2.10), we consider the corresponding SME in Equation (3.1). The analytical solution of the system of SME in Equation (3.1) can be obtained by many classical methods. In the following, the Bartels Stewart Method (BSM) [36] is generalized in the fuzzy environment.

Method 1: Generalized BSM for solving PTrFFSME:

Step 1: Suppose $m_{ij}$, $a_{ij}$, $n_{ij}$ and $b_{ij}$ are real and have real Schur decompositions $m_{ij}=U_1{R}_1{U}_1^T$, $a_{ij}=V_1{S}_1{V}_1^T$, $n_{ij}=U_2{R}_2{U}_2^T$, $b_{ij}=V_2{S}_2{V}_2^T$, where $U$ and $V$ are orthogonal and $R$ and $S$ are upper quasi-triangular. Then the first two equations in Equation (3.1) can be transformed to: $U_1^T{m}_{ij}{U}_1\cdot U_2^T{x}_{ij}{V}_1-U_1^T{y}_{ij}{V}_2.V_2^T{b}_{ij}{V}_2=U_1^T{c}_{ij}{V}_2,$ $U_2^T{n}_{ij}{U}_2\cdot U_1^T{y}_{ij}{V}_2-U_2^T{x}_{ij}{V}_1.V_1^T{a}_{ij}{V}_1=U_2^T{g}_{ij}{V}_1.$

Consequently, they can be written as $\{\begin{array}{c}{R}_1{W}_1-W_2{S}_2=D_1,\\ R_2{W}_2-W_1{S}_1=D_2.\end{array}$

where $R_1=U_1^T{m}_{ij}{U}_1$, $R_2=U_2^T{n}_{ij}{U}_2$, $W_1=U_2^T{x}_{ij}{V}_1$, $W_2=U_1^T{y}_{ij}{V}_2$, $S_1=V_1^T{a}_{ij}{V}_1$, $S_2=V_2^T{b}_{ij}{V}_2$, $D_1=U_1^T{c}_{ij}{V}_2$ and $D_2=U_2^T{g}_{ij}{V}_1$.

Then, this system can be written as $P_1{w}_1=d_1$

where $P_1=\left(\begin{array}{c}\begin{array}{cc}{I}_n\otimes R_1& -S_2^T\otimes I_m\end{array}\\ \begin{array}{cc}-S_1^T\otimes I_m& I_n\otimes R_2\end{array}\end{array}\right)$, $w_1=\left(\begin{array}{c}vec\left(W_1\right)\\ vec\left(W_2\right)\end{array}\right)$ and $d_1=\left(\begin{array}{c}vec\left(D_1\right)\\ vec\left(D_2\right)\end{array}\right).$

Gaussian elimination and back substitution are applied to obtain $w_1$.

Step 2: The values of $x_{ij}$ and $y_{ij}$ can be computed as follows: $x_{ij}=U_2{W}_1{V}_1^T$ $y_{ij}=U_1{W}_2{V}_2^T$

Step 3: The third and fourth equations in Equation (3.1) can be written as follows: (3.2) $\{\begin{array}{l}{m}_{ij}{z}_{ij}+q_{ij}{b}_{ij}=h_{ij}-{\alpha }_{ij}{x}_{ij}-y_{ij}{\delta }_{ij},\\ n_{ij}{q}_{ij}+z_{ij}{a}_{ij}=f_{ij}-{\text{β}}_{ij}{y}_{ij}-x_{ij}{\gamma }_{ij}.\end{array}$(3.2) If we let, $h_1^{\alpha }=h_{ij}-{\alpha }_{ij}{x}_{ij}-y_{ij}{\delta }_{ij}$ and $f_1^{\alpha }=f_{ij}-{\text{β}}_{ij}{y}_{ij}-x_{ij}{\gamma }_{ij}.$ Then Equation (3.2) can be written as (3.3) $\{\begin{array}{l}{m}_{ij}{z}_{ij}+q_{ij}{b}_{ij}=h_1^{\alpha },\\ n_{ij}{q}_{ij}+z_{ij}{a}_{ij}=f_1^{\alpha }.\end{array}$(3.3)

Since Equation (3.3) has the same structure as the first two equations in Equation (3.1), it can be transformed to: $\begin{array}{r}{U}_1^T{m}_{ij}{U}_1\cdot U_2^T{z}_{ij}{V}_1+U_1^T{q}_{ij}{V}_2.V_2^T{b}_{ij}{V}_2=U_1^T{h_1}_{ij}^{\alpha }{V}_2.\\ U_2^T{n}_{ij}{U}_2\cdot U_1^T{q}_{ij}{V}_2+U_2^T{z}_{ij}{V}_1.V_1^T{a}_{ij}{V}_1=U_2^T{f_1}_{ij}^{\alpha }{V}_1.\end{array}$

that is, $\{\begin{array}{c}{R}_1{W}_3+W_4{S}_2=D_3,\\ R_2{W}_4+W_3{S}_1=D_4.\end{array}$

or equivalently $P_2{w}_2=d_2$

where $P_2=\left(\begin{array}{c}\begin{array}{cc}{I}_n\otimes R_1& S_2^T\otimes I_m\end{array}\\ \begin{array}{cc}{S}_1^T\otimes I_m& I_n\otimes R_2\end{array}\end{array}\right)$, $w_2=\left(\begin{array}{c}vec\left(W_3\right)\\ vec\left(W_4\right)\end{array}\right)$ and $d_2=\left(\begin{array}{c}vec\left(D_3\right)\\ vec\left(D_4\right)\end{array}\right).$

Gaussian elimination and back substitution are applied to obtain $w_2$.

Step 4: The values of $z_{ij}$ and $q_{ij}$ can be computed as follows: $\begin{array}{rl}{z}_{ij}& =U_2{W}_3{V}_1^T,\\ q_{ij}& =U_1{W}_4{V}_2^T.\end{array}$

Step 5: Combining the values of $x_{ij},y_{ij},z_{ij}$ and $q_{ij}$ which obtained in step 2 and step 4. The solution of TrFFSME is represented by: $\tilde{X}=({\tilde{x}}_{ij}{)}_{n\times m}=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right),\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$ The following Matrix Vectorization Method (MVM) is based on the concept of Kronecker product, Kronecker difference and Vec-operator. It is an extension of the method proposed for solving PTrFFSME in the form $\tilde{A}\tilde{X}+\tilde{X}\tilde{B}=\tilde{C}$ by Elsayed et al. [24].

Method 2: MVM for solving PTrFFSME.

In this method we apply the concept of Kronecker product and Vec-operator to Equation (3.1) as follows:

Step1: Applying subtraction property of equality on the third and fourth equations in Equation (3.1) we get: (3.4) $\{\begin{array}{l}{m}_{ij}{x}_{ij}-y_{ij}{b}_{ij}=c_{ij},\\ n_{ij}{y}_{ij}-x_{ij}{a}_{ij}=g_{ij},\\ m_{ij}{z}_{ij}+q_{ij}{b}_{ij}=h_{ij}-{\alpha }_{ij}{x}_{ij}-y_{ij}{\delta }_{ij},\\ n_{ij}{q}_{ij}+z_{ij}{a}_{ij}=f_{ij}-{\text{β}}_{ij}{y}_{ij}-x_{ij}{\gamma }_{ij}.\end{array}$(3.4) By taking Vec-operator for both sides of Equation (3.4), we have (3.5) $\{\begin{array}{l}Vec\left(m_{ij}{x}_{ij}-y_{ij}{b}_{ij}\right)=Vec\left(c_{ij}\right),\\ Vec\left(n_{ij}{y}_{ij}-x_{ij}{a}_{ij}\right)=Vec\left(g_{ij}\right),\\ Vec\left(m_{ij}{z}_{ij}+q_{ij}{b}_{ij}\right)=Vec\left(h_{ij}-{\alpha }_{ij}{x}_{ij}-y_{ij}{\delta }_{ij}\right),\\ Vec\left(n_{ij}{q}_{ij}+z_{ij}{a}_{ij}\right)=Vec\left(f_{ij}-{\text{β}}_{ij}{y}_{ij}-x_{ij}{\gamma }_{ij}\right).\end{array}$(3.5)

Using Definition 2.8 on Equation (3.5) we get (3.6) $\{\begin{array}{l}\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right)=\left(\begin{array}{c}Vec\left(c_{ij}\right)\\ Vec\left(g_{ij}\right)\end{array}\right)\\ \left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(z_{ij}\right)\\ Vec\left(q_{ij}\right)\end{array}\right)=\left(\begin{array}{c}Vec\left(h_{ij}\right)\\ Vec\left(f_{ij}\right)\end{array}\right)-\left(\begin{array}{cc}{I}_n\otimes {\alpha }_{ij}& {{\delta }_{ij}}_1^T\otimes I_m\\ {{\gamma }_{ij}}_1^T\otimes I_m& I_n\otimes {\text{β}}_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right)\end{array}$(3.6) Step 2: Let, $\begin{array}{rl}{R}_1& =\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)=\left(\begin{array}{cccccc}{\left(m_{ij}\right)}_{11}& 0& 0& b_{11}{I}_n& \dots & b_{m1}{I}_n\\ 0& {\left(m_{ij}\right)}_{22}& 0& ⋮& \ddots & ⋮\\ 0& 0& {\left(m_{ij}\right)}_{mm}& b_{1m}{I}_n& \dots & b_{mm}{I}_n\\ -a_{11}{I}_n& \dots & -a_{m1}{I}_n& {\left(n_{ij}\right)}_{11}& 0& 0\\ ⋮& \ddots & ⋮& 0& {\left(n_{ij}\right)}_{22}& -b_{21}\\ -a_{1m}{I}_n& \dots & -a_{mm}{I}_n& 0& -b_{12}& {\left(n_{ij}\right)}_{mm}\end{array}\right),\\ R_2& =\left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)=\left(\begin{array}{cccccc}{\left(m_{ij}\right)}_{11}& 0& 0& b_{11}{I}_n& \dots & b_{m1}{I}_n\\ 0& {\left(m_{ij}\right)}_{22}& 0& ⋮& \ddots & ⋮\\ 0& 0& {\left(m_{ij}\right)}_{mm}& b_{1m}{I}_n& \dots & b_{mm}{I}_n\\ a_{11}{I}_n& \dots & a_{m1}{I}_n& {\left(n_{ij}\right)}_{11}& 0& 0\\ ⋮& \ddots & ⋮& 0& {\left(n_{ij}\right)}_{22}& -b_{21}\\ a_{1m}{I}_n& \dots & a_{mm}{I}_n& 0& -b_{12}& {\left(n_{ij}\right)}_{mm}\end{array}\right),\\ S_1& =\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right)=\left(\begin{array}{c}{x}_{11}\\ ⋮\\ x_{nm}\\ y_{11}\\ ⋮\\ y_{nm}\end{array}\right),T_1=\left(\begin{array}{c}Vec\left(c_{ij}\right)\\ Vec\left(g_{ij}\right)\end{array}\right)=\left(\begin{array}{c}{c}_{11}\\ ⋮\\ c_{nm}\\ g_{11}\\ ⋮\\ g_{nm}\end{array}\right),S_2=\left(\begin{array}{c}Vec\left(z_{ij}\right)\\ Vec\left(q_{ij}\right)\end{array}\right)=\left(\begin{array}{c}{z}_{11}\\ ⋮\\ z_{nm}\\ q_{11}\\ ⋮\\ q_{nm}\end{array}\right)\\ \text{and}\\ T_2& =\left(\begin{array}{c}Vec\left(h_{ij}\right)\\ Vec\left(f_{ij}\right)\end{array}\right)-\left(\begin{array}{cc}{I}_n\otimes {\alpha }_{ij}& {{\delta }_{ij}}_1^T\otimes I_m\\ {{\gamma }_{ij}}_1^T\otimes I_m& I_n\otimes {\text{β}}_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right)=\left(\begin{array}{c}T{{}_1^{\alpha }}_{11}\\ ⋮\\ T{{}_1^{\alpha }}_{nm}\\ T{{}_2^{\alpha }}_{11}\\ ⋮\\ T{{}_2^{\alpha }}_{nm}\end{array}\right)\end{array}$

Step 3: With the assumption that $R_1$ and $R_2$ are non- singular, the system of equations in Equation (3.6) can be written as follows: (3.7) $\{\begin{array}{c}{R}_1{S}_1=T_1\\ R_2{S}_2=T_2\end{array}$(3.7) By the multiplicative inverse of $R_1$ and $R_2$ we obtain the following: (3.8) $\{\begin{array}{c}{S}_1={R_1}^{-1}\cdot T_1\\ S_2={R_2}^{-1}\cdot T_2\end{array}$(3.8)

Step 4: The solution of the TrFFSME is represented by: $\tilde{X}=({\tilde{x}}_{ij}{)}_{n\times m}=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right),\mathrm{\forall }\{1\le i,j\le n,m.$

In the following Definition 3.1, the positive fuzzy solution to the PTrFFSME in LR form is defined.

Definition 3.1:

Positive Fuzzy Solution to PTrFFSME in LR Form

A trapezoidal fuzzy solution matrix $\tilde{X}=({\tilde{x}}_{ij}{)}_{n\times m}=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right)$ where $x_{ij}>0,y_{ij}>0,z_{ij}>0,q_{ij}>0,x_{ij}\le y_{ij}$ and $x_{ij}-z_{ij}>0$ is called a positive fuzzy solution of the PTrFFSME in LR form.

The existence and uniqueness of the positive fuzzy solution to the PTrFFSME are discussed in the following Corollary 3.1.

Corollary 3.1:

Suppose that $\tilde{A}=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right)$, $\tilde{B}=\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right)$ and $\tilde{C}=\left(c_{ij},g_{ij},h_{ij},f_{ij}\right)$ are three positive trapezoidal fuzzy matrices. Then the positive fuzzy solution to the PTrFFSME $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}$ in LR form exists and it is unique if the following conditions are satisfied: $\begin{array}{rl}If{R}_1& =\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)thendet\left(R_1\right)\ne 0,\\ If{R}_2& =\left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)thendet\left(R_2\right)\ne 0.\end{array}$

Proof:

1. Let $\tilde{A}$, $\tilde{B}$ are $\tilde{C}$ are non-negative matrices, and $K^{-1}$ and $L^{-1}$ exists. The PTrFFSME $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}$ can be written as a system of equations in Equation (3.8).

$If{R}_1=\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)$ then the matrix equation $S_1={R_1}^{-1}\cdot T_1$ has a unique solution only if $det\left(R_1\right)\ne 0$.

• (II) Similarly, $If{R}_2=\left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)$ then the matrix equation $S_2={R_2}^{-1}\cdot T_2$ has a unique solution only if $det\left(R_2\right)\ne 0$.

The following Corollary 3.2 shows the equivalency between the positive solution of the system of SME in Equation (3.1) and the positive fuzzy solution to the PTrFFSME.

Corollary 3.2:

Suppose that $\tilde{A}=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right)$, $\tilde{B}=\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right)$ and $\tilde{C}=\left(c_{ij},g_{ij},h_{ij},f_{ij}\right)$ are three positive trapezoidal fuzzy matrices. Then the positive fuzzy solution to the PTrFFSME $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}$ in LR form and the solution to the system of SME in Equation (3.1) are equivalent if the following conditions are satisfied:

I)	$det\left(R_1\right)\ne 0and{R_1}^{-1}={\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)}^{-1}>0$,
II)	$\begin{array}{l}det\left(R_2\right)\ne 0{R_2}^{-1}={\left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)}^{-1}>0\\ \text{and}\left(\begin{array}{c}Vec\left(h_{ij}\right)\\ Vec\left(f_{ij}\right)\end{array}\right)>\left(\begin{array}{cc}{I}_n\otimes {\alpha }_{ij}& {{\delta }_{ij}}_1^T\otimes I_m\\ {{\gamma }_{ij}}_1^T\otimes I_m& I_n\otimes {\text{β}}_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right),\end{array}$,
III)	$y_{ij}-x_{ij}\ge 0$ and $x_{ij}-z_{ij}\ge 0$.

Proof:

• I) Let $\tilde{A}$, $\tilde{B}$ and $\tilde{C}$ are non-negative matrices. The PTrFFSME can be written as a system of SME in Equation (3.1) by Theorem 3.1. By applying the Vec-operator and Kronecker product to the system of SME, the following system is obtained $\{\begin{array}{c}{S}_1={R_1}^{-1}\cdot T_1,\\ S_2={R_2}^{-1}\cdot T_2.\end{array}$

Therefore, if $det\left(R_1\right)\ne 0,{R_1}^{-1}={\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)}^{-1}>0$, and since

$T_1=\left(\begin{array}{c}Vec\left(c_{ij}\right)\\ Vec\left(g_{ij}\right)\end{array}\right)>0$ then $\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right)>0.$

• II) Similarly, if $det\left(R_2\right)\ne 0,{R_2}^{-1}={\left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)}^{-1}>0$,

and if $\left(\begin{array}{c}Vec\left(h_{ij}\right)\\ Vec\left(f_{ij}\right)\end{array}\right)>\left(\begin{array}{cc}{I}_n\otimes {\alpha }_{ij}& {{\delta }_{ij}}_1^T\otimes I_m\\ {{\gamma }_{ij}}_1^T\otimes I_m& I_n\otimes {\text{β}}_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right),$ then

$T_2=\left(\begin{array}{c}Vec\left(h_{ij}\right)\\ Vec\left(f_{ij}\right)\end{array}\right)-\left(\begin{array}{cc}{I}_n\otimes {\alpha }_{ij}& {{\delta }_{ij}}_1^T\otimes I_m\\ {{\gamma }_{ij}}_1^T\otimes I_m& I_n\otimes {\text{β}}_{ij}\end{array}\right)\left(\begin{array}{c}Vec\left(x_{ij}\right)\\ Vec\left(y_{ij}\right)\end{array}\right)>0$ and therefore, $\left(\begin{array}{c}Vec\left(z_{ij}\right)\\ Vec\left(q_{ij}\right)\end{array}\right)>0.$

• III) So far, the solution to the SME in Equation (3.1) positive i.e. $x_{ij}\ge 0$, $y_{ij}\ge 0$, $z_{ij}\ge 0$, $q_{ij}\ge 0$. In order for this solution to be equivalent to the positive fuzzy solution to the PTrFFSME the following conditions must be satisfied $y_{ij}-x_{ij}\ge 0$ and $x_{ij}-z_{ij}\ge 0$.

Now, the feasibility of the positive solution to the PTrFFSME in LR form is discussed.

3.1. Feasibility of the Positive Solution to the PTrFFSME in LR Form

Let, $\tilde{A}=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right)\ge 0$, $\tilde{B}=\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right)\ge 0$ and $\left(\begin{array}{cc}{I}_n\otimes m_{ij}& -{b_{ij}}_1^T\otimes I_m\\ -{a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)$and $\left(\begin{array}{cc}{I}_n\otimes m_{ij}& {b_{ij}}_1^T\otimes I_m\\ {a_{ij}}_1^T\otimes I_m& I_n\otimes n_{ij}\end{array}\right)$ be a non-singular matrix, then the PTrFFSME has a positive fuzzy solution if:

1. $x_{ij}\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$

2. $y_{ij}\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$

3. $z_{ij}\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$

4. $q_{ij}\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$

5. $y_{ij}-x_{ij}\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$

6. $x_{ij}-z_{ij}\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le n,m\right\}.$

4. Numerical Examples

In this section, the proposed methods are illustrated by solving the following example.

Example 4.1:

Consider the $2\times 2$ TrFFSME: $\begin{array}{rl}& \left(\begin{array}{cc}\left(30,31,1,1\right)& \left(35,36,1,1\right)\\ \left(32,35,1,2\right)& \left(30,31,1,1\right)\end{array}\right)\cdot \left(\begin{array}{cc}{\tilde{x}}_{11}& {\tilde{x}}_{12}\\ {\tilde{x}}_{21}& {\tilde{x}}_{22}\end{array}\right)-\left(\begin{array}{cc}{\tilde{x}}_{11}& {\tilde{x}}_{12}\\ {\tilde{x}}_{21}& {\tilde{x}}_{22}\end{array}\right)\cdot \left(\begin{array}{cc}\left(2,3,1,2\right)& \left(2,3,1,2\right)\\ \left(2,4,1,1\right)& \left(3,4,1,1\right)\end{array}\right)\\ & =\left(\begin{array}{cc}\left(190,283,159,130\right)& \left(190,279,189,97\right)\\ \left(190,287,153,125\right)& \left(190,284,185,96\right)\end{array}\right).\end{array}$ where, $\tilde{X}=\left(\begin{array}{cc}{\tilde{x}}_{11}& {\tilde{x}}_{12}\\ {\tilde{x}}_{21}& {\tilde{x}}_{22}\end{array}\right)$

and ${\tilde{x}}_{ij}=(w_{ij},y_{ij},z_{ij},q_{ij}),\mathrm{\forall }\left\{1\le i,j\le 2\right\}$

Solution:

The three proposed methods are applied to obtain the fuzzy solution $\tilde{X}=\left(\begin{array}{cc}{\tilde{x}}_{11}& {\tilde{x}}_{12}\\ {\tilde{x}}_{21}& {\tilde{x}}_{22}\end{array}\right)$ as follows:

Method 1: Generalized BSM for solving PTrFFSME.

Step 1:

Given $\tilde{A}=\left(\begin{array}{cc}\left(30,31,1,1\right)& \left(35,36,1,1\right)\\ \left(32,35,1,2\right)& \left(30,31,1,1\right)\end{array}\right)$, we can obtain the following: $\begin{array}{l}{m}_{ij}=\left(\begin{array}{cc}{m}_{11}& m_{12}\\ m_{21}& m_{22}\end{array}\right)=\left(\begin{array}{cc}30& 35\\ 32& 30\end{array}\right),n_{ij}=\left(\begin{array}{cc}{n}_{11}& n_{12}\\ n_{21}& n_{22}\end{array}\right)=\left(\begin{array}{cc}31& 36\\ 35& 31\end{array}\right),\\ {\alpha }_{ij}=\left(\begin{array}{cc}{\alpha }_{11}& {\alpha }_{12}\\ {\alpha }_{21}& {\alpha }_{22}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\text{, and }{\text{β}}_{ij}=\left(\begin{array}{cc}{\text{β}}_{11}& {\text{β}}_{12}\\ {\text{β}}_{21}& {\text{β}}_{22}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right),\end{array}$

Also given, $\tilde{B}=\left(\begin{array}{cc}\left(2,3,1,2\right)& \left(2,3,1,2\right)\\ \left(2,4,1,1\right)& \left(3,4,1,1\right)\end{array}\right)$, we can obtain the following: $\begin{array}{l}{a}_{ij}=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 2& 3\end{array}\right),b_{ij}=\left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right)=\left(\begin{array}{cc}3& 3\\ 4& 4\end{array}\right),\\ {\gamma }_{ij}=\left(\begin{array}{cc}{\gamma }_{11}& {\gamma }_{12}\\ {\gamma }_{21}& {\gamma }_{22}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\text{, and }{\delta }_{ij}=\left(\begin{array}{cc}{\delta }_{11}& {\delta }_{12}\\ {\delta }_{21}& {\delta }_{22}\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 1& 1\end{array}\right).\end{array}$

And, $\tilde{C}=\left(\begin{array}{cc}\left(190,283,159,130\right)& \left(190,279,189,97\right)\\ \left(190,287,153,125\right)& \left(190,284,185,96\right)\end{array}\right)$, we can obtain the following: $\begin{array}{l}{c}_{ij}=\left(\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right)=\left(\begin{array}{cc}190& 190\\ 190& 190\end{array}\right),g_{ij}=\left(\begin{array}{cc}{g}_{11}& g_{12}\\ g_{21}& g_{22}\end{array}\right)=\left(\begin{array}{cc}283& 279\\ 287& 284\end{array}\right),\\ h_{ij}=\left(\begin{array}{cc}{h}_{11}& h_{12}\\ h_{21}& h_{22}\end{array}\right)=\left(\begin{array}{cc}159& 189\\ 153& 185\end{array}\right)\text{, and }{f}_{ij}=\left(\begin{array}{cc}{f}_{11}& f_{12}\\ f_{21}& f_{22}\end{array}\right)=\left(\begin{array}{cc}130& 97\\ 125& 96\end{array}\right).\end{array}$

We decompose the following matrices by applying Definition 2.11 as follows: $\begin{array}{l}{m}_{ij}=U_1{R}_1{U}_1^T,a_{ij}=V_1{S}_1{V}_1^T,n_{ij}=U_2{R}_2{U}_2^T,b_{ij}=V_2{S}_2{V}_2^T.\end{array}$

We get: $\begin{array}{rl}{U}_1& =\left(\begin{array}{cc}0.7227& -0.691\\ 0.691& 0.7227\end{array}\right),U_1^T=\left(\begin{array}{cc}0.7227& 0.691\\ -0.691& 0.7227\end{array}\right)\text{ and }{R}_1=\left(\begin{array}{cc}63.446& 3\\ 0& -3.466\end{array}\right).\\ U_2& =\left(\begin{array}{cc}0.712& -0.702\\ 0.702& 0.712\end{array}\right),U_2^T=\left(\begin{array}{cc}0.712& 0.702\\ -0.702& 0.712\end{array}\right)\text{ and }{R}_2=\left(\begin{array}{cc}66.496& 1\\ 0& -4.496\end{array}\right).\\ V_1& =\left(\begin{array}{cc}-0.788& -0.615\\ 0.615& -0.788\end{array}\right),V_1^T=\left(\begin{array}{cc}-0.788& 0.615\\ -0.615& -0.788\end{array}\right)\text{ and }{S}_1=\left(\begin{array}{cc}0.43844& 0\\ 0.& 4.562\end{array}\right).\\ V_2& =\left(\begin{array}{cc}-0.707& -0.707\\ 0.707& -0.707\end{array}\right),V_2^T=\left(\begin{array}{cc}-0.707& 0.707\\ -0.707& -0.707\end{array}\right)\text{ and }{S}_2=\left(\begin{array}{cc}0& -1\\ 0& 7\end{array}\right).\end{array}$

This will be followed by obtaining $P_1$ and $P_2$ by the definition of Kronecker difference Definition 2.9 as follows: $P_1=\left(\begin{array}{c}\begin{array}{cc}{I}_n\otimes R_1& -S_2^T\otimes I_m\end{array}\\ \begin{array}{cc}-S_1^T\otimes I_m& I_n\otimes R_2\end{array}\end{array}\right),$ $P_1=\left(\begin{array}{cccccccc}63.46& 3.& 0.& 0.& 0.& 0.& 0.& 0.\\ 0.& -3.46& 0.& 0.& 0.& 0.& 0.& 0.\\ 0.& 0.& 63.46& 3.& 1.& 0.& -7.& 0.\\ 0.& 0.& 0.& -3.46& 0.& 1.& 0.& -7.\\ -0.438& 0.& 0.& 0.& 66.49& 1.& 0.& 0.\\ 0.& -0.438& 0.& 0.& 0.& -4.49& 0.& 0.\\ 0.& 0.& -4.56& 0.& 0.& 0.& 66.49& 1.\\ 0.& 0.& 0.& -4.56& 0.& 0.& 0.& -4.49\end{array}\right).$

Also, $D_1$ and $D_2$ can be computed as follows: $D_1=U_1^T{c}_{ij}{V}_2=\left(\begin{array}{cc}0.& -379.90\\ 0.& -8.50\end{array}\right)$ and $D_2=U_2^T{g}_{ij}{V}_1=\left(\begin{array}{cc}-72.68& -561.78\\ -0.565& -8.48\end{array}\right).$

Now, we can find $d_1$ and $d_2$ by applying Definition 2.7 on $D_1$, $D_2$, $W_1$ and $W_2$ as follow: $d_1=vec\left(D_1\right)\text{ and }{d}_2=vec\left(D_2\right).$ $d_1=vec\left(D_1\right)=\left(\begin{array}{c}0\\ -379.90\\ 0\\ -8.50\end{array}\right),d_2=vec\left(D_2\right)=\left(\begin{array}{c}-72.68\\ -561.78\\ -0.565\\ -8.48\end{array}\right).$

And, since $W_1=\left(\begin{array}{cc}{w}_{11}^{\left(a\right)}& w_{12}^{\left(a\right)}\\ w_{21}^{\left(a\right)}& w_{22}^{\left(a\right)}\end{array}\right)$ and $W_2=\left(\begin{array}{cc}{w}_{11}^{\left(b\right)}& w_{12}^{\left(b\right)}\\ w_{21}^{\left(b\right)}& w_{22}^{\left(b\right)}\end{array}\right)$. Applying Definition 2.7 on $W_1$ and $W_2$ gives: $w_1=vec\left(W_1\right)=\left(\begin{array}{c}{w}_{11}^{\left(a\right)}\\ w_{12}^{\left(a\right)}\\ w_{21}^{\left(a\right)}\\ w_{22}^{\left(a\right)}\end{array}\right)\text{ and }{w}_2=vec\left(W_2\right)=\left(\begin{array}{c}{w}_{11}^{\left(b\right)}\\ w_{12}^{\left(b\right)}\\ w_{21}^{\left(b\right)}\\ w_{22}^{\left(b\right)}\end{array}\right).$ Now we can solve for $w_1$ and $w_2$ as follows: $P_1{w}_1=d_1$ $\left(\begin{array}{cccccccc}63.46& 3.& 0.& 0.& 0.& 0.& 0.& 0.\\ 0.& -3.46& 0.& 0.& 0.& 0.& 0.& 0.\\ 0.& 0.& 63.46& 3.& 1.& 0.& -7.& 0.\\ 0.& 0.& 0.& -3.46& 0.& 1.& 0.& -7.\\ -0.438& 0.& 0.& 0.& 66.49& 1.& 0.& 0.\\ 0.& -0.438& 0.& 0.& 0.& -4.49& 0.& 0.\\ 0.& 0.& -4.56& 0.& 0.& 0.& 66.49& 1.\\ 0.& 0.& 0.& -4.56& 0.& 0.& 0.& -4.49\end{array}\right)\left(\begin{array}{c}{w}_{11}^{\left(a\right)}\\ w_{12}^{\left(a\right)}\\ w_{21}^{\left(a\right)}\\ w_{22}^{\left(a\right)}\\ w_{11}^{\left(b\right)}\\ w_{12}^{\left(b\right)}\\ w_{21}^{\left(b\right)}\\ w_{22}^{\left(b\right)}\end{array}\right)=\left(\begin{array}{c}0\\ -379.90\\ 0\\ -8.50\\ -72.68\\ -561.78\\ -0.565\\ -8.48\end{array}\right).$ Gaussian elimination and back substitution are applied to obtain $W_1$ and $W_2$. $W_1=\left(\begin{array}{cc}7.48& -6.11\\ 67.62& -74.31\end{array}\right)$ and $W_2=\left(\begin{array}{cc}-0.05& -9.0\\ -0.134& 0.52\end{array}\right).$

Step 2: We compute $x_{ij}$ and $y_{ij}$ as follows: $x_{ij}=U_2{W}_1{V}_1^T=\left(\begin{array}{cc}0.712& -0.702\\ 0.702& 0.712\end{array}\right)\left(\begin{array}{cc}-0.745& -6.99\\ 0.299& 1.349\end{array}\right)\left(\begin{array}{cc}-0.788& 0.615\\ -0.615& -0.788\end{array}\right)$ Thus, $x_{ij}=\left(\begin{array}{cc}4& 4\\ 3& 3\end{array}\right).$ $y_{ij}=U_1{W}_2{V}_2^T=\left(\begin{array}{cc}0.7227& -0.691\\ 0.691& 0.7227\end{array}\right)\left(\begin{array}{cc}-0.05& -9.0\\ -0.134& 0.52\end{array}\right)\left(\begin{array}{cc}-0.707& 0.707\\ -0.707& -0.707\end{array}\right).$

Thus, $y_{ij}=\left(\begin{array}{cc}5& 5\\ 4& 4\end{array}\right).$

Step 3: The values of $x_{ij}$ and $y_{ij}$ are used to compute $h_1^{\alpha }$ and $f_1^{\alpha }$ as follows: $h_1^{\alpha }=h_{ij}-{\alpha }_{ij}{x}_{ij}-x_{ij}{\gamma }_{ij}=\left(\begin{array}{cc}137& 167\\ 134& 166\end{array}\right),$ $f_1^{\alpha }=f_{ij}-{\text{β}}_{ij}{y}_{ij}-y_{ij}{\delta }_{ij}=\left(\begin{array}{cc}113& 80\\ 105& 76\end{array}\right).$

The values of $h_1^{\alpha }$ and $f_1^{\alpha }$ are substituted in the following equations. $\{\begin{array}{c}{m}_{ij}{z}_{ij}+q_{ij}{b}_{ij}=h_1^{\alpha },\\ n_{ij}{q}_{ij}+z_{ij}{a}_{ij}=f_1^{\alpha }.\end{array}$

Step 4: Since the obtained equations have exactly the same structure as the first two equations in Equation (3.1), $z_{ij}$ and $q_{ij}$ can be computed similar to $x_{ij}$ and $y_{ij}$. Thus, $z_{ij}=\left(\begin{array}{cc}2& 3\\ 2& 2\end{array}\right)$ and $q_{ij}=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right).$

Step 5: The solution $\tilde{X}$ of the given PTrFFSME is: $\tilde{X}=\left(\begin{array}{cc}\left(◻\begin{array}{c}4,5,2,1\end{array}◻\right)& \left(◻\text{4, 5, 3,1}◻\right)\\ \left(◻\begin{array}{c}3,4,2,2\end{array}◻\right)& \left(◻\begin{array}{c}3,4,2,1\end{array}◻\right)\end{array}\right).$ Method 2: MVM for solving PTrFFSME.

Step 1: Since $\tilde{A}=\left(m_{ij},n_{ij},{\alpha }_{ij},{\text{β}}_{ij}\right),\mathrm{\forall }\left\{1\le i,j\le 2\right\},$ $\tilde{B}=\left(a_{ij},b_{ij},{\gamma }_{ij},{\delta }_{ij}\right),\mathrm{\forall }\left\{1\le i,j\le 2\right\}$, $\tilde{X}=\left(x_{ij},y_{ij},z_{ij},q_{ij}\right),\mathrm{\forall }\left\{1\le i,j\le 2\right\}\text{ and }\tilde{C}=\left(c_{ij},g_{ij},h_{ij},f_{ij}\right),\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

We can obtain the following: $m_{ij}=\left(\begin{array}{cc}30& 35\\ 32& 30\end{array}\right),n_{ij}=\left(\begin{array}{cc}31& 36\\ 35& 31\end{array}\right),{\alpha }_{ij}=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),{\text{β}}_{ij}=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right),$ and $a_{ij}=\left(\begin{array}{cc}2& 2\\ 2& 3\end{array}\right),b_{ij}=\left(\begin{array}{cc}3& 3\\ 4& 4\end{array}\right),{\gamma }_{ij}=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),{\delta }_{ij}=\left(\begin{array}{cc}2& 2\\ 1& 1\end{array}\right),$ and $c_{ij}=\left(\begin{array}{cc}190& 190\\ 190& 190\end{array}\right),g_{ij}=\left(\begin{array}{cc}283& 279\\ 287& 284\end{array}\right),h_{ij}=\left(\begin{array}{cc}159& 189\\ 153& 185\end{array}\right),f_{ij}=\left(\begin{array}{cc}130& 97\\ 125& 96\end{array}\right).$

Step 2: Since $R_1\cdot S_1=T_1$, we compute $R_1$, $T_1$ and solve for $S_1$ as follows: (4.1) $\begin{array}{rl}{R}_1& =\left(\begin{array}{cccccccc}30& 35& 0& 0& -3& 0& -4& 0\\ 32& 30& 0& 0& 0& -3& 0& -4\\ 0& 0& 30& 35& -3& 0& -4& 0\\ 0& 0& 32& 30& 0& -3& 0& -4\\ 31& 36& 0& 0& -2& 0& -2& 0\\ 35& 31& 0& 0& 0& -2& 0& -2\\ 0& 0& 31& 36& -2& 0& -3& 0\\ 0& 0& 35& 31& 0& -2& 0& -3\end{array}\right),S_1=\left(\begin{array}{c}{x}_{11}\\ x_{12}\\ x_{21}\\ x_{22}\\ y_{11}\\ y_{12}\\ y_{21}\\ y_{22}\end{array}\right)\text{ and }{T}_1=\left(\begin{array}{c}190\\ 190\\ 190\\ 190\\ 283\\ 287\\ 279\\ 284\end{array}\right).\\ & \left(\begin{array}{cccccccc}30& 35& 0& 0& -3& 0& -4& 0\\ 32& 30& 0& 0& 0& -3& 0& -4\\ 0& 0& 30& 35& -3& 0& -4& 0\\ 0& 0& 32& 30& 0& -3& 0& -4\\ 31& 36& 0& 0& -2& 0& -2& 0\\ 35& 31& 0& 0& 0& -2& 0& -2\\ 0& 0& 31& 36& -2& 0& -3& 0\\ 0& 0& 35& 31& 0& -2& 0& -3\end{array}\right)\left(\begin{array}{c}{x}_{11}\\ x_{12}\\ x_{21}\\ x_{22}\\ y_{11}\\ y_{12}\\ y_{21}\\ y_{22}\end{array}\right)=\left(\begin{array}{c}190\\ 190\\ 190\\ 190\\ 283\\ 287\\ 279\\ 284\end{array}\right).\end{array}$(4.1)

Multiplying both sides of Equation (4.1) by ${R_1}^{-1}$ we get: $S_1=\left(\begin{array}{c}{x}_{11}\\ x_{12}\\ x_{21}\\ x_{22}\\ y_{11}\\ y_{12}\\ y_{21}\\ y_{22}\end{array}\right)=\left(\begin{array}{c}4\\ 4\\ 3\\ 3\\ 5\\ 5\\ 4\\ 4\end{array}\right).$ Thus, $x_{ij}=\left(\begin{array}{cc}4& 4\\ 3& 3\end{array}\right)\text{ and }{y}_{ij}=\left(\begin{array}{cc}5& 5\\ 4& 4\end{array}\right).$

The values of $Vec\left(x_{ij}\right)$ and $Vec\left(y_{ij}\right)$ is substituted in Equation (3.13) to compute $Vec\left(z_{ij}\right)$ and $Vec\left(q_{ij}\right)$ as follows:

Step 3: We also compute, $R_2=\left(\begin{array}{cccccccc}30& 35& 0& 0& 3& 0& 4& 0\\ 32& 30& 0& 0& 0& 3& 0& 4\\ 0& 0& 30& 35& 3& 0& 4& 0\\ 0& 0& 32& 30& 0& 3& 0& 4\\ 2& 0& 2& 0& 31& 36& 0& 0\\ 0& 2& 0& 2& 35& 31& 0& 0\\ 2& 0& 3& 0& 0& 0& 31& 36\\ 0& 2& 0& 3& 0& 0& 35& 31\end{array}\right)\text{ and }{T}_2=\left(\begin{array}{c}137\\ 134\\ 167\\ 166\\ 113\\ 105\\ 80\\ 76\end{array}\right).$ And since, (4.2) $R_2\cdot S_2=T_2\left(\begin{array}{cccccccc}30& 35& 0& 0& 3& 0& 4& 0\\ 32& 30& 0& 0& 0& 3& 0& 4\\ 0& 0& 30& 35& 3& 0& 4& 0\\ 0& 0& 32& 30& 0& 3& 0& 4\\ 2& 0& 2& 0& 31& 36& 0& 0\\ 0& 2& 0& 2& 35& 31& 0& 0\\ 2& 0& 3& 0& 0& 0& 31& 36\\ 0& 2& 0& 3& 0& 0& 35& 31\end{array}\right)\left(\begin{array}{c}{z}_{11}\\ z_{12}\\ z_{21}\\ z_{22}\\ q_{11}\\ q_{12}\\ q_{21}\\ q_{22}\end{array}\right)=\left(\begin{array}{c}137\\ 134\\ 167\\ 166\\ 113\\ 105\\ 80\\ 76\end{array}\right).$(4.2)

Multiplying both sides of Equation (4.2) by ${R_2}^{-1}$ we get: $S_2=\left(\begin{array}{c}{z}_{11}\\ z_{12}\\ z_{21}\\ z_{22}\\ q_{11}\\ q_{12}\\ q_{21}\\ q_{22}\end{array}\right)=\left(\begin{array}{c}2\\ 3\\ 2\\ 2\\ 1\\ 1\\ 2\\ 1\end{array}\right).$

Thus, $z_{ij}=\left(\begin{array}{cc}2& 3\\ 2& 2\end{array}\right)\text{ and }{q}_{ij}=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right).$

Step 4: The solution $\tilde{X}$ of the given TrFFSME is: $\tilde{X}=\left(\begin{array}{cc}\left(◻\begin{array}{c}4,5,2,1\end{array}◻\right)& \left(◻\text{4, 5, 3, 1}◻\right)\\ \left(◻\begin{array}{c}3,4,2,2\end{array}◻\right)& \left(◻\begin{array}{c}3,4,2,1\end{array}◻\right)\end{array}\right).$

Step 5: Feasibility of the solution

Since,

1. $x_{ij}=\left(\begin{array}{cc}4& 4\\ 3& 3\end{array}\right)\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

2. $y_{ij}=\left(\begin{array}{cc}5& 5\\ 4& 4\end{array}\right)\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

3. $z_{ij}=\left(\begin{array}{cc}2& 3\\ 2& 2\end{array}\right)\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

4. $q_{ij}=\left(\begin{array}{cc}1& 1\\ 2& 1\end{array}\right)\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

5. $y_{ij}-x_{ij}=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

6. $x_{ij}-z_{ij}=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)\ge 0$, $\mathrm{\forall }\left\{1\le i,j\le 2\right\}.$

Therefore, the solution $\tilde{X}=\left(\begin{array}{cc}\left(◻\begin{array}{c}4,5,2,1\end{array}◻\right)& \left(◻\text{4, 5, 3, 1}◻\right)\\ \left(◻\begin{array}{c}3,4,2,2\end{array}◻\right)& \left(◻\begin{array}{c}3,4,2,1\end{array}◻\right)\end{array}\right)$, is feasible.

Figure 1 shows the fuzzy solution $\tilde{X}$.

Figure 1. The fuzzy solution $\tilde{X}$ of Example 4.1.

4.1. Verification of the solution

To verify the obtained fuzzy solution, we first multiply $\tilde{A}$ and $\tilde{X}$ as follows: $\begin{array}{rl}\tilde{A}\tilde{X}& =\left(\begin{array}{cc}\left(30,31,1,1\right)& \left(35,36,1,1\right)\\ \left(32,35,1,2\right)& \left(30,31,1,1\right)\end{array}\right)\left(\begin{array}{cc}\left(◻\begin{array}{c}4,5,2,1\end{array}◻\right)& \left(◻\text{4, 5, 3, 1}◻\right)\\ \left(◻\begin{array}{c}3,4,2,2\end{array}◻\right)& \left(◻\begin{array}{c}3,4,2,1\end{array}◻\right)\end{array}\right)\\ & =\left(\begin{array}{cc}\left(\begin{array}{cccc}225& 299& 137& 112\end{array}\right)& \left(\begin{array}{cccc}225& 299& 167& 76\end{array}\right)\\ \left(\begin{array}{cccc}218& 299& 131& 111\end{array}\right)& \left(\begin{array}{cccc}218& 299& 163& 80\end{array}\right)\end{array}\right).\end{array}$

We also multiply $\tilde{X}$ and $\tilde{B}$ as follows: $\begin{array}{rl}\tilde{X}\tilde{B}& =\left(\begin{array}{cc}\left(◻\begin{array}{c}4,5,2,1\end{array}◻\right)& \left(◻\text{4, 5, 3, 1}◻\right)\\ \left(◻\begin{array}{c}3,4,2,2\end{array}◻\right)& \left(◻\begin{array}{c}3,4,2,1\end{array}◻\right)\end{array}\right)\left(\begin{array}{cc}\left(2,3,1,2\right)& \left(2,3,1,2\right)\\ \left(2,4,1,1\right)& \left(3,4,1,1\right)\end{array}\right)\\ & =\left(\begin{array}{cc}\left(\begin{array}{cccc}16& 35& 18& 22\end{array}\right)& \left(\begin{array}{cccc}20& 35& 21& 22\end{array}\right)\\ \left(\begin{array}{cccc}12& 28& 14& 22\end{array}\right)& \left(\begin{array}{cccc}15& 28& 16& 22\end{array}\right)\end{array}\right).\end{array}$

Therefore, $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\left(\begin{array}{cc}\left(190,283,159,130\right)& \left(190,279,189,97\right)\\ \left(190,287,153,125\right)& \left(190,284,185,96\right)\end{array}\right).$

The value of $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}$ is exactly equal to the constant matrix $\tilde{C}$.

5 Conclusion

In this paper, the solution to the PTrFFSME $\tilde{A}\tilde{X}-\tilde{X}\tilde{B}=\tilde{C}$ is obtained analytically. The positive fuzzy solution is obtained by two analytical methods, the generalized BSM, in addition to MVM. The existence and uniqueness of the positive fuzzy are discussed. In terms of accuracy, the two methods are obtained the positive fuzzy solution to the PTrFFSME. In addition, the methods can also be applied to TFFSME. Both methods can be applied for large TrFFSME using Matlab or Mathematica. The main limitation of the MVM is that it needs a long computational time and, therefore, large memory storage in order to find the fuzzy solution. In future work, a further modification to the existing arithmetic fuzzy operations is needed in order to reduce the complexity of the obtained system of SME. In addition, optimization techniques need to be developed to overcome the limitation of the developed methods. As future research, the ideas presented in this paper will be modified and applied to TrFFSME with near zero trapezoidal fuzzy numbers.

Acknowledgements

We would like to express our sincere thanks to Professor Mohammed Abdel Latif Ramadan for his valuable and constructive suggestions in the planning and development of this study.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

This research was supported by the Ministry of Higher Education (MoHE) of Malaysia through Fundamental Research Grant Scheme (FRGS/1/2018/STG06/UUM/02/12). We also want to thank the Universiti Utara Malaysia (UUM) for the facilities provided.

Notes on contributors

Ahmed Abdel Aziz Elsayed

Dr. Ahmed Abdel Aziz Elsayed works at the Institute of Applied Technology, Abu Dhabi campus, United Arab Emirates (UAE). He obtained his bachelor's degree in Mathematics and Education in 2007, a Master's in Mathematics in 2018 from the American University in Sharjah, UAE, and a PhD in Mathematics in 2022 from the School of Quantitative Sciences, Universiti Utara Malaysia (UUM). His current research interests include Fuzzy Logic, Numerical Analysis and Linear Algebra.

Nazihah Ahmad

Dr. Nazihah Ahmad works at the School of Quantitative Sciences, Universiti Utara Malaysia (UUM). She obtained her bachelor's degree in Mathematics with honours in 2001, a Master's in Mathematics in 2002, and a PhD in Mathematics in 2009. Her current research interests include fuzzy mathematics, topology, and mathematical modelling in medicine, healthcare, finance, and psychology. She has been awarded several research grants by the Ministry of Education Malaysia and UUM as a principal and co-investigator. The quality of her work has been published in international journals (indexed by SCOPUS or Web of Science), conference proceedings, books and book chapter. She has been teaching mathematics for more than 20 years at the undergraduate and master's level. She has also supervised master and PhD students. Her expertise was acknowledged by her academic peers when she was appointed as an external/internal examiner, article reviewer and external assessor. At national level, she has involved in revising the mathematics syllabus based on the Secondary School Standard Curriculum and in organizing STEM service-learning activities to inspire and strengthen STEM education among young people.

Ghassan Malkawi

Dr. Ghassan Malkawi has been on the mathematics faculty since 2003 at the Higher College of Technology (HCT) in the United Arab Emirates. Where he holds the position of Assistant Professor in the Division of Engineering. Malkawi completed his doctorate from the Universiti Utara Malaysia. His master's degree in applied mathematics from the University of Jordan. Fuzzy linear systems are the primary area of interest for Dr. Malkawi's research.