113 
 

 American Scientific Research Journal for Engineering, Technology,  and Sciences  (ASRJETS) 
ISSN (Print) 2313-4410, ISSN (Online) 2313-4402 

© Global Society of Scientific Research and Researchers  

http://asrjetsjournal.org/  
 

Alternative Multiplying Triangular Fuzzy Number and 

Applied in Fully Fuzzy Linear System 

Zulya Desmitaa*, Mashadib  

aDepartment of Mathematics, University of Riau, Pekanbaru 28293, Indonesia 
bDepartment of Mathematics, University of Riau, Pekanbaru 28293, Indonesia 

aEmail: zulya.desmita7317@grad.unri.ac.id   
bEmail: mash-math@unri.ac.id 

Abstract 

In this paper, a new concept of arithmetic fuzzy number will be introduced using the broad area concept 

triangular fuzzy number so that we will get the form of multiplying fuzzy number in some cases. New arithmetic 

concept fuzzy number will be applied to solve the fully fuzzy linear system using Gauss Seidel method and the 

solution obtained is a single solution. 

Keywords: Arithmetic Fuzzy Number, Fully Fuzzy Linear System , Triangular Fuzzy Number. 

2010 Mathematics Subject Classification: 94D05, 08A72, 15B15 

1. Introduction 

System linear is one part of linear algebra which is studied in mathematics. The form of the matrix equation the 

system linear is 𝑨𝑨𝑨𝑨 = 𝒃𝒃 with all entries real number. Along with the development of mathematics, the system 

linear is not only used in real number, but variable and constant in system linear can be fuzzy number. Fuzzy in 

introduced by Lothfi A.Zadeh (1965). The system linear using fuzzy number consists of the fuzzy linear system 

of equation, the fully fuzzy linear system of equation [1, 2 and 3], and the dual fully fuzzy linear system of 

equation [4, 5, 6, and 7]. 

Many methods are to solve the fully fuzzy linear system of equation that has been discussed, using the 

decomposition [8, 9, and 10], using the direct method and iteration method [11, 12, and 13], Huang’s method 

[14], using LU factorization of the coefficient matrix [4 and 6], and using cramers rules [15]. In this paper, the 

system linear to be discussed is the fully fuzzy linear system of equation. Dehghan and Hashemi (2006) [8] the 

fully fuzzy linear system of equation is a system linear with elements from the matrix and the vector is in the 

------------------------------------------------------------------------ 

* Corresponding author.  

E-mail address: zulya.desmita7317@grad.unri.ac.id 

http://asrjetsjournal.org/
mailto:zulya.desmita7317@grad.unri.ac.id.m


American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

114 
 

form fuzzy number. The fully fuzzy linear system of equation  𝑨𝑨� ⊗ 𝑨𝑨� = 𝒃𝒃�  where  𝑨𝑨�  is matrix with fuzzy 

number, 𝑨𝑨�  and  𝒃𝒃�  are vector with fuzzy number. In this paper, new definition of positive fuzzy number and 

negative fuzzy number will be given in some cases using broad comparison, then by defining new fuzzy 

number, algebra will be constructed from fuzzy number in fully fuzzy linear system of equation. Fuzzy number 

to be used in this paper is triangular fuzzy number and to solve the fully fuzzy linear system of equation using 

the Gauss Seidel method. 

2. The Basic Definition 

Some of  the basic definitions of fuzzy number have been in Ming Ma (2000) [16]. 

Definition 2.1 A fuzzy number is a fuzzy set  𝑎𝑎�:𝑅𝑅 → [0,1] which satisfies: 

1. 𝑎𝑎�  is upper semicontinuous ; 

2. 𝑎𝑎�(𝑥𝑥) = 0 outside some interval [0,1]; 

3. There are real numbers 𝑎𝑎, 𝑏𝑏 in [𝑐𝑐,𝑑𝑑] for which, 

(i) 𝑎𝑎�(𝑥𝑥) is monotonic increasing on [𝑐𝑐, 𝑎𝑎]; 

(ii) 𝑎𝑎�(𝑥𝑥) is monotonic decreasing on [𝑏𝑏,𝑑𝑑]; 

(iii) 𝑎𝑎�(𝑥𝑥) = 1, for  𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏. 

 

An equivalent parametric definition is given in Friedman (1998) [17]. 

Definition 2.2 A fuzzy number  𝑎𝑎� is a pair (𝑎𝑎(𝑟𝑟), 𝑎𝑎(𝑟𝑟)) of functions  𝑎𝑎(𝑟𝑟), 𝑎𝑎(𝑟𝑟); 0 ≤ 𝑟𝑟 ≤ 1 which satisfy the 

following requirements: 

1. 𝑎𝑎(𝑟𝑟) is a bounded left continuous nondecreasing functions over [0,1] ; 

2. 𝑎𝑎(𝑟𝑟) is a bounded left continuous nonincreasing functions over [0,1]; 

3. 𝑎𝑎(𝑟𝑟) ≤ 𝑎𝑎(𝑟𝑟),0 ≤ 𝑟𝑟 ≤ 1. 

3. Triangular Fuzzy Number 

In this section, we discuss the concept triangular fuzzy number, new definitions for positive triangular fuzzy 

number and negative triangular fuzzy number, and arithmetic algebraic operations consisting of addition, 

subtraction, scalar product, multiplication of two fuzzy number and inverse fuzzy number. 

In this paper, we write a fuzzy number in the form of  𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽), where 𝒂𝒂 is the center, 𝜶𝜶 is the left width, 

and 𝜷𝜷 is the right width. For arbitrary fuzzy number  𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽), the membership function is of the form: 

𝜇𝜇𝑎𝑎�(𝑥𝑥) =

⎩
⎪
⎨

⎪
⎧1 −

𝑎𝑎 − 𝑥𝑥
𝛼𝛼

, 𝑎𝑎 − 𝛼𝛼 ≤ 𝑥𝑥 ≤ 𝑎𝑎,

1 −
𝑥𝑥 − 𝑎𝑎
𝛽𝛽

, 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑎𝑎 + 𝛽𝛽,

0, 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑟𝑟𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒.

 

On the other hand, a parametric fuzzy number  𝑎𝑎� = �𝑎𝑎(𝑟𝑟), 𝑎𝑎(𝑟𝑟)�  can be represented as: 

𝑎𝑎(𝑟𝑟) = 𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼  and   𝑎𝑎(𝑟𝑟) = 𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽 

 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

115 
 

3.1. Triangular Fuzzy Number Positive and Negative 

Triangular fuzzy number  𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) is said to be positive or negative: 
1. If  𝑎𝑎 − 𝛼𝛼 ≥ 0, then 𝑎𝑎� is said to be positive, and if  𝑎𝑎 + 𝛽𝛽 ≤ 0, then 𝑎𝑎� is said to be negative. Seen in Figure 

1: 

 

Figure 1: Triangular Fuzzy Number Potitif and Negative 

2. If  𝑎𝑎 > 0 and 𝑎𝑎 − 𝛼𝛼 < 0. Seen in Figure 2: 

 

Figure 2: Triangular Fuzzy Number  𝑎𝑎 > 0 

𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) is said to be positive if  𝑃𝑃 > 𝑄𝑄, and said to be negative if 𝑃𝑃 < 𝑄𝑄  or can be expressed in form  

𝑃𝑃 = 𝛽𝛽
2

+ 𝑎𝑎 + 𝑎𝑎2

2𝛽𝛽
> 𝛼𝛼

2
− 𝑎𝑎 − 𝑎𝑎2

2𝛽𝛽
= 𝑄𝑄  positive, and  𝑃𝑃 = 𝛽𝛽

2
+ 𝑎𝑎 + 𝑎𝑎2

2𝛽𝛽
< 𝛼𝛼

2
− 𝑎𝑎 − 𝑎𝑎2

2𝛽𝛽
= 𝑄𝑄  negative. 

3. If  𝑎𝑎 < 0  and  𝑎𝑎 + 𝛽𝛽 > 0. Seen in Figure 3: 

 

Figure 3: Triangular Fuzzy Number  𝑎𝑎 < 0 

𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) is said to be positive if  𝑃𝑃 > 𝑄𝑄, and said to be negative if  𝑃𝑃 < 𝑄𝑄  or can be expressed in form  

𝑃𝑃 = 𝛽𝛽
2

+ 𝑎𝑎 − 𝑎𝑎2

2𝛼𝛼
> 𝛼𝛼

2
− 𝑎𝑎 + 𝑎𝑎2

2𝛼𝛼
= 𝑄𝑄  positive, and 𝑃𝑃 = 𝛽𝛽

2
+ 𝑎𝑎 − 𝑎𝑎2

2𝛼𝛼
< 𝛼𝛼

2
− 𝑎𝑎 + 𝑎𝑎2

2𝛼𝛼
= 𝑄𝑄  negative. 

4. If 𝑎𝑎 = 0. Seen in Figure 4: 

 

Figure 4: Triangular Fuzzy Number  𝑎𝑎 = 0 

𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) is said to be positive if  𝑃𝑃 > 𝑄𝑄, and is said to be negative if  𝑃𝑃 < 𝑄𝑄  or can be expressed in 
form  𝑃𝑃 = 𝛽𝛽 > 𝛼𝛼 = 𝑄𝑄  Positive, and 𝑃𝑃 = 𝛽𝛽 < 𝛼𝛼 = 𝑄𝑄  negative. 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

116 
 

3.2. New Arithmetic Triangular Fuzzy Number 

Arithmetic algebraic operations will be given for triangular fuzzy number. For  𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) and  𝑏𝑏� = (𝑏𝑏, 𝛾𝛾, 𝛿𝛿) , 

then the parametric forms are as follows: 

𝑎𝑎� = �𝑎𝑎(𝑟𝑟), 𝑎𝑎(𝑟𝑟)� = [𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼, 𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽]  

𝑏𝑏� = �𝑏𝑏(𝑟𝑟), 𝑏𝑏(𝑟𝑟)� = [𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾, 𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿] 

Then for the arithmetic algebraic process the two triangular fuzzy numbers are as follows: 

a. Addition 

𝑎𝑎� ⊕ 𝑏𝑏� = �𝑎𝑎(𝑟𝑟) + 𝑏𝑏(𝑟𝑟), 𝑎𝑎(𝑟𝑟) + 𝑏𝑏(𝑟𝑟)� 

             = [(𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼) + (𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾), (𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽) + (𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿)] 

             = [(𝑎𝑎 + 𝑏𝑏) − (1 − 𝑟𝑟)(𝛼𝛼 + 𝛾𝛾), (𝑎𝑎 + 𝑏𝑏) + (1 − 𝑟𝑟)(𝛽𝛽 + 𝛿𝛿)] 

Transforming back into the triangular form, we have:  

𝑎𝑎� ⊕ 𝑏𝑏� = (𝑎𝑎 + 𝑏𝑏,𝛼𝛼 + 𝛾𝛾,𝛽𝛽 + 𝛿𝛿) 

b. Subtraction 

𝑎𝑎� ⊖ 𝑏𝑏� = �𝑎𝑎(𝑟𝑟) − 𝑏𝑏(𝑟𝑟), 𝑎𝑎(𝑟𝑟) − 𝑏𝑏(𝑟𝑟)� 

             = [(𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼) − (𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿), (𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽) − (𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾)] 

             = [(𝑎𝑎 − 𝑏𝑏) − (1 − 𝑟𝑟)(𝛼𝛼 + 𝛿𝛿), (𝑎𝑎 − 𝑏𝑏) + (1 − 𝑟𝑟)(𝛽𝛽 + 𝛾𝛾)] 

Transforming back into the triangular form, we have:   

𝑎𝑎� ⊖ 𝑏𝑏� = (𝑎𝑎 − 𝑏𝑏,𝛼𝛼 + 𝛿𝛿,𝛽𝛽 + 𝛾𝛾) 

c. Scalar product as  𝜆𝜆 ⊗ 𝑎𝑎� = � (𝜆𝜆𝑎𝑎, 𝜆𝜆𝛼𝛼, 𝜆𝜆𝛽𝛽) 𝜆𝜆 ≥ 0,
(𝜆𝜆𝑎𝑎,−𝜆𝜆𝛽𝛽,−𝜆𝜆𝛼𝛼) 𝜆𝜆 < 0. 

d. Multiplication  

If 𝑎𝑎� = �𝑎𝑎(𝑟𝑟), 𝑎𝑎(𝑟𝑟)� and 𝑏𝑏� = �𝑏𝑏(𝑟𝑟), 𝑏𝑏(𝑟𝑟)� are two positive fuzzy numbers, then  �̃�𝑐 = 𝑎𝑎� ⊗ 𝑏𝑏� = �𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� 

for every 𝑟𝑟 ∈ [0,1] . The following is given some cases for triangular fuzzy number to multiplication 

operations. 

(i) If  𝑎𝑎�  is positive and 𝑏𝑏� is positive, then: 

�
𝑐𝑐(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)
𝑐𝑐(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)

                                                                                        (1) 

From equation (1) : 

�𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [(𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾) − 𝑎𝑎𝑏𝑏, (𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿) − 𝑎𝑎𝑏𝑏] 

we have  �̃�𝑐 = �𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [𝑎𝑎𝑏𝑏 − (1 − 𝑟𝑟)(𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛼𝛼), 𝑎𝑎𝑏𝑏 + (1 − 𝑟𝑟)(𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛽𝛽)]        (2) 

If we let  �̃�𝑐 = (𝑐𝑐, 𝜉𝜉,𝜓𝜓), then the parametric fuzzy number is of the form:  

�̃�𝑐 = [𝑐𝑐 − (1 − 𝑟𝑟)𝜉𝜉, 𝑐𝑐 + (1 − 𝑟𝑟)𝜓𝜓]                                                                                                       (3) 

From equation (2) and (3) we have: 𝜉𝜉 = 𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛼𝛼 𝜓𝜓 = 𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛽𝛽 

So multiplication 𝑎𝑎�  positive and 𝑏𝑏�  positive can be written as  �̃�𝑐 = 𝑎𝑎� ⊗ 𝑏𝑏� = (𝑎𝑎𝑏𝑏, 𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛼𝛼, 𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛽𝛽) 

 

(ii) If  𝑎𝑎�  is positive and  𝑏𝑏�  is negative, then: 

    �
𝑐𝑐(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)
𝑐𝑐(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)

                                                                                       (4) 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

117 
 

From equation (4) : 

�𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [(𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾) − 𝑎𝑎𝑏𝑏, (𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿) − 𝑎𝑎𝑏𝑏] 

we have  �̃�𝑐 = �𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [𝑎𝑎𝑏𝑏 − (1 − 𝑟𝑟)(𝑎𝑎𝛾𝛾 − 𝑏𝑏𝛽𝛽), 𝑎𝑎𝑏𝑏 + (1 − 𝑟𝑟)(𝑎𝑎𝛿𝛿 − 𝑏𝑏𝛼𝛼)]                                        (5) 

From equation (5) and (3) we have: 𝜉𝜉 = 𝑎𝑎𝛾𝛾 − 𝑏𝑏𝛽𝛽 𝜓𝜓 = 𝑎𝑎𝛿𝛿 − 𝑏𝑏𝛼𝛼 

So multiplication 𝑎𝑎� positive and  𝑏𝑏� negative can be written as  �̃�𝑐 = 𝑎𝑎� ⊗ 𝑏𝑏� = (𝑎𝑎𝑏𝑏, 𝑎𝑎𝛾𝛾 − 𝑏𝑏𝛽𝛽, 𝑎𝑎𝛿𝛿 − 𝑏𝑏𝛼𝛼) 

 

(iii) If  𝑎𝑎�  is negative and 𝑏𝑏� is positive,then: 

�𝑐𝑐
(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)
𝑐𝑐(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)

                                                                                        (6) 

From equation (6) :  

�𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [(𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿) − 𝑎𝑎𝑏𝑏, (𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾) − 𝑎𝑎𝑏𝑏] 

we have  �̃�𝑐 = �𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [𝑎𝑎𝑏𝑏 − (1 − 𝑟𝑟)(−𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛼𝛼), 𝑎𝑎𝑏𝑏 + (1 − 𝑟𝑟)(−𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛽𝛽)]                                  (7) 

From equation (7) and (3) we have: 𝜉𝜉 = −𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛼𝛼 𝜓𝜓 = −𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛽𝛽 

So multiplication 𝑎𝑎� negative and 𝑏𝑏� positive can be written as  �̃�𝑐 = 𝑎𝑎� ⊗ 𝑏𝑏� = (𝑎𝑎𝑏𝑏,−𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛼𝛼,−𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛽𝛽) 

 

(iv) If  𝑎𝑎� is negative and 𝑏𝑏� is negative, then: 

�𝑐𝑐
(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)
𝑐𝑐(𝑟𝑟) = 𝑎𝑎(𝑟𝑟)𝑏𝑏(1) + 𝑎𝑎(1)𝑏𝑏(𝑟𝑟) − 𝑎𝑎(1)𝑏𝑏(1)

                                                                                        (8) 

From equation (8) : 

�𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [(𝑎𝑎 + (1 − 𝑟𝑟)𝛽𝛽)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 + (1 − 𝑟𝑟)𝛿𝛿) − 𝑎𝑎𝑏𝑏, (𝑎𝑎 − (1 − 𝑟𝑟)𝛼𝛼)𝑏𝑏 + 𝑎𝑎(𝑏𝑏 − (1 − 𝑟𝑟)𝛾𝛾) − 𝑎𝑎𝑏𝑏] 

We have  �̃�𝑐 = �𝑐𝑐(𝑟𝑟), 𝑐𝑐(𝑟𝑟)� = [𝑎𝑎𝑏𝑏 − (1 − 𝑟𝑟)(−𝑎𝑎𝛿𝛿 − 𝑏𝑏𝛽𝛽), 𝑎𝑎𝑏𝑏 + (1 − 𝑟𝑟)(−𝑎𝑎𝛾𝛾 − 𝑏𝑏𝛼𝛼)]                                  (9) 

From equation (9) and (3) we have: 𝜉𝜉 = −𝑎𝑎𝛿𝛿 − 𝑏𝑏𝛽𝛽 𝜓𝜓 = −𝑎𝑎𝛾𝛾 − 𝑏𝑏𝛼𝛼 

So multiplication 𝑎𝑎� negative and 𝑏𝑏� negative can be written as  �̃�𝑐 = 𝑎𝑎� ⊗ 𝑏𝑏� = (𝑎𝑎𝑏𝑏,−(𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛽𝛽),−(𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛼𝛼) 

 

e. Inverse 

The identity element for triangular fuzzy number is: 

𝐼𝐼 = (1,0,0) 
where 𝐼𝐼 = (1,0,0) is positive. Let two fuzzy number 𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) and 𝑏𝑏� = (𝑏𝑏, 𝛾𝛾, 𝛿𝛿) have inverse: 

𝑏𝑏� =
1
𝑎𝑎�

 

will be indicated  𝑎𝑎� ⊗ 𝑏𝑏� = (1,0,0). Fuzzy number 𝑎𝑎� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) the condition for having inverse is 𝑎𝑎 ≠ 0. 

Therefore, inverse for triangular fuzzy number consist of two cases, as follows: 

(i) If  𝑎𝑎� > 0 and  𝑏𝑏� > 0, 𝑎𝑎� ⊗ 𝑏𝑏� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) ⊗ (𝑏𝑏, 𝛾𝛾, 𝛿𝛿) = (𝑎𝑎𝑏𝑏, 𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛼𝛼, 𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛽𝛽) = (1,0,0) 

(ii) If  𝑎𝑎� < 0 and  𝑏𝑏� < 0, 𝑎𝑎� ⊗ 𝑏𝑏� = (𝑎𝑎,𝛼𝛼,𝛽𝛽) ⊗ (𝑏𝑏, 𝛾𝛾, 𝛿𝛿) = (𝑎𝑎𝑏𝑏,−(𝑎𝑎𝛿𝛿 + 𝑏𝑏𝛽𝛽),−(𝑎𝑎𝛾𝛾 + 𝑏𝑏𝛼𝛼) = (1,0,0) 

So the inverse for triangular fuzzy number is: 

𝑏𝑏� =
1
𝑎𝑎�

= �
1
𝑎𝑎

,
𝛽𝛽
𝑎𝑎2

,
𝛼𝛼
𝑎𝑎2
� 

 
 
 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

118 
 

4. Solving Fully Fuzzy Linear System of Equation  

Let the fully fuzzy linear system be as follows: 

 (𝑎𝑎�11⨂ 𝑥𝑥�1) ⊕ (𝑎𝑎�12⨂ 𝑥𝑥�2) ⊕ …⊕ (𝑎𝑎�1𝑛𝑛⨂ 𝑥𝑥�𝑛𝑛)  = 𝑏𝑏�1 

 (𝑎𝑎�21⨂ 𝑥𝑥�1) ⊕ (𝑎𝑎�22⨂ 𝑥𝑥�2) ⊕ …⊕ (𝑎𝑎�2𝑛𝑛⨂ 𝑥𝑥�𝑛𝑛)  = 𝑏𝑏�2           

           ⋮                        ⋮                 ⋮                 ⋮ 
 (𝑎𝑎�𝑛𝑛1⨂ 𝑥𝑥�1) ⊕ (𝑎𝑎�𝑛𝑛2⨂ 𝑥𝑥�2) ⊕ …⊕ (𝑎𝑎�𝑛𝑛𝑛𝑛⨂ 𝑥𝑥�𝑛𝑛) = 𝑏𝑏�𝑛𝑛 

The matrix form the fully fuzzy linear system of equation is 𝑨𝑨� ⊗ 𝑨𝑨� = 𝒃𝒃�, where 𝑨𝑨� = (𝒂𝒂�𝒊𝒊𝒊𝒊) = (𝑎𝑎𝑖𝑖𝑖𝑖 ,𝛼𝛼𝑖𝑖𝑖𝑖 ,𝛽𝛽𝑖𝑖𝑖𝑖) is a 

fuzzy matrix 𝑛𝑛 × 𝑛𝑛, 𝑨𝑨� = (𝑥𝑥�1, … , 𝑥𝑥�𝑛𝑛)  and  𝒃𝒃� = (𝑏𝑏�1, … , 𝑏𝑏�𝑛𝑛)  are fuzzy vectors 𝑛𝑛 × 1. A matrix  𝑨𝑨� = (𝒂𝒂�𝒊𝒊𝒊𝒊)  is 

called a fuzzy matrix, if each element of  𝑨𝑨�  is a fuzzy number. We may represent fuzzy matrix  𝑨𝑨� = �𝒂𝒂�𝒊𝒊𝒊𝒊� 𝑛𝑛×𝑛𝑛  

that  �𝒂𝒂�𝒊𝒊𝒊𝒊� = �𝑎𝑎𝑖𝑖𝑖𝑖 ,𝛼𝛼𝑖𝑖𝑖𝑖 ,𝛽𝛽𝑖𝑖𝑖𝑖�  with new notation  𝑨𝑨� = (𝐴𝐴,𝑀𝑀,𝑁𝑁), where 𝐴𝐴 = �𝑎𝑎𝑖𝑖𝑖𝑖�, 𝑀𝑀 = �𝛼𝛼𝑖𝑖𝑖𝑖�, and  𝑁𝑁 = �𝛽𝛽𝑖𝑖𝑖𝑖�  are 

three 𝑛𝑛 𝑥𝑥 𝑛𝑛 crips matrices. 

Next, to get a solution the fully fuzzy linear system of equation  𝑨𝑨� ⊗ 𝑨𝑨� = 𝒃𝒃� , where 𝑨𝑨� = (𝐴𝐴,𝑀𝑀,𝑁𝑁) , 𝑨𝑨� =

(𝑥𝑥,𝑦𝑦, 𝑧𝑧), and  𝒃𝒃� = (𝑏𝑏,𝑔𝑔, ℎ) so that it is obtained: 

(𝐴𝐴,𝑀𝑀,𝑁𝑁) ⊗ (𝑥𝑥,𝑦𝑦, 𝑧𝑧) = (𝑏𝑏,𝑔𝑔,ℎ) 

by using algebra multiplication of two fuzzy number for 𝑨𝑨� > 𝟎𝟎, 𝒃𝒃� > 𝟎𝟎, and  𝑨𝑨� > 𝟎𝟎 the formula that applied is 

(𝐴𝐴𝑥𝑥,𝐴𝐴𝑦𝑦 + 𝑀𝑀𝑥𝑥,𝐴𝐴𝑧𝑧 + 𝑁𝑁𝑥𝑥) = (𝑏𝑏,𝑔𝑔, ℎ). Therefore, it can be concluded that: 

𝐴𝐴𝑥𝑥 = 𝑏𝑏          

𝐴𝐴𝑦𝑦 + 𝑀𝑀𝑥𝑥 = 𝑔𝑔                         (11) 

𝐴𝐴𝑧𝑧 + 𝑁𝑁𝑥𝑥 = ℎ                                                                                                                                

Furthermore, proving equation (11) satisfies strictly diagonal dominant with the following formula: 

|𝑎𝑎𝑖𝑖𝑖𝑖| > � �𝑎𝑎𝑖𝑖𝑖𝑖�
𝑛𝑛

𝑖𝑖=𝑖𝑖,𝑖𝑖≠𝑖𝑖

𝑒𝑒 = 1,2, … ,𝑛𝑛 

Then do the iteration process using the intial value. Gauss seidel iteration stop if tolerance has been achieved: 

�𝑥𝑥
�𝑖𝑖

(𝑗𝑗)⊖𝑥𝑥�𝑖𝑖
(𝑗𝑗−𝑖𝑖)

𝑥𝑥�𝑖𝑖
(𝑗𝑗) �                         (12) 

Example: The fully fuzzy system linear of equation with positive and positive multiplication operations as 

follows:  

( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )9.131,3.88,5.535,,1.0,1.0,5.4,,3.0,1.0,2,,2.0,1.0,2

3.109,2.76,5.434,,2.0,2.0,5.1,,4.0,1.0,4,,1.0,1.0,2
2.536,7.427,1897,,2.0,5.0,6,,5.1,5.1,12,,1,1,19

333222111

333222111

333222111

=⊗⊕⊗⊕⊗
=⊗⊕⊗⊕⊗

=⊗⊕⊗⊕⊗

zyxzyxzyx
zyxzyxzyx

zyxzyxzyx
 

 

Solution: The steps to solve the fully fuzzy linear system of equation as follow: 

Change the form of equations into the matrix  𝑨𝑨� = ( 𝐴𝐴,𝑀𝑀,𝑁𝑁) and  𝒃𝒃� = (𝑏𝑏,𝑔𝑔, ℎ) are: 

𝐴𝐴 = �
19 12 6
2 4 1.5
2 2 4.5

�          𝑀𝑀 = �
1 1.5 0.5

0.1 0.1 0.2
0.1 0.1 0.1

�         𝑁𝑁 = �
1 1.5 0.2

0.1 0.4 0.2
0.2 0.3 0.1

� 

 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

119 
 

          𝑏𝑏  = �
1897
434.5
535.5

�                       𝑔𝑔 = �
427.7
76.2
88.3

�                        ℎ = �
536.2
109.3
131.9

� 

Because 𝑨𝑨� > 𝟎𝟎, 𝒃𝒃� > 𝟎𝟎, and  𝑨𝑨� > 𝟎𝟎, the formula that applied in equation (11). Next, change the matrix into 

system linear of equation as follows: 

 𝐴𝐴𝑥𝑥 = 𝑏𝑏  

�
19 12 6
2 4 1.5
2 2 4.5

� �
𝑥𝑥1
𝑥𝑥2
𝑥𝑥3
� = �

1897
434.5
535.5

�                                                                                                                            (13) 

 𝑀𝑀𝑥𝑥 + 𝐴𝐴𝑦𝑦 = 𝑔𝑔 

�
1 1.5 0.5

0.1 0.1 0.2
0.1 0.1 0.1

�  �
𝑥𝑥1
𝑥𝑥2
𝑥𝑥3
�  + �

19 12 6
2 4 1.5
2 2 4.5

� �
𝑦𝑦1
𝑦𝑦2
𝑦𝑦3
� = �

427.7
76.2
88.3

�                                                                                  (14) 

 𝐴𝐴𝑧𝑧 + 𝑁𝑁𝑥𝑥 = ℎ 

�
19 12 6
2 4 1.5
2 2 4.5

� �
𝑧𝑧1
𝑧𝑧2
𝑧𝑧3
�  +  �

1 1.5 0.2
0.1 0.4 0.2
0.2 0.3 0.1

�  �
𝑥𝑥1
𝑥𝑥2
𝑥𝑥3
� = �

536.2
109.3
131.9

�                                                                                 (15) 

For equation (13), the equation obtained is as follows: 

19𝑥𝑥1 + 12𝑥𝑥2 + 6𝑥𝑥3    = 1897 

2𝑥𝑥1   + 4𝑥𝑥2   + 1.5𝑥𝑥3 = 434.5 

2𝑥𝑥1   + 2𝑥𝑥2   + 4.5𝑥𝑥3 = 535.5 

To get the values  𝑥𝑥1, 𝑥𝑥2 and 𝑥𝑥3  it must first be proven that equation (13) strictly diagonally dominant in the 

following: 

 |𝑎𝑎11|  ≥  |𝑎𝑎12|  + |𝑎𝑎13| →  |19|   ≥  |12| + |6|     = 19  ≥ 18 

 |𝑎𝑎22|  ≥  |𝑎𝑎21|  +  |𝑎𝑎23| →  |4|     ≥  |2|    + |1.5| = 4    ≥ 3.5 

 |𝑎𝑎33|  ≥  |𝑎𝑎31|  +  |𝑎𝑎32| →  |4.5|  ≥  |2|    + |2|    = 4.5 ≥ 4 

All equations (13) are proven to be diagonally dominant, so obtained: 

 𝑥𝑥1 = 1
19

 (1897 − 12𝑥𝑥2  − 6𝑥𝑥3) 

 𝑥𝑥2 = 1
4
 (434.5 − 2𝑥𝑥1 – 1.5𝑥𝑥3) 

 𝑥𝑥3 = 1
4.5

 (535.5 − 2𝑥𝑥1  − 2𝑥𝑥2) 

Then the iteration process starts with the intial value (0, 0, 0), which is as follows: 

First iteration 

 𝑥𝑥1(1) = 1
19

 (1897 – 12(0) – 6(0)) =  1
19

 (1897) = 99.8421 

 𝑥𝑥2(1) = 1
4
 (434.5 – 2(99.8421) – 1.5(0))  =  1

4
 (234.8158) = 58.7039  

 𝑥𝑥3(1) = 1
4.5

 (535.5 – 2(99.8421) – 2(58.7039)) =  1
4.5

 (218.408) = 48.5351 

Second iteration 

 𝑥𝑥1(2) = 1
19

 (1897 – 12(58.7039) – 6(48.5351)) =  1
19 

 (901.3426) = 47.4390 

 𝑥𝑥2(2) = 1
4
 (434.5 – 2(47.4390) – 1.5(48.5351) =  1

4 
 (266.8194) = 66.7048 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

120 
 

 𝑥𝑥3(2) = 1
4.5

 (535.5 – 2(47.4390) – 2(66.7048)) =  1
4.5 

 (307.2124) = 68.2694 

Based on equation (12), the iteration process until 10th iteration is obtained the value  𝑥𝑥1  = 37.0021, 𝑥𝑥2  = 

61.9987, 𝑥𝑥3 = 74.9996.  

For equation (14), the equation obtained is as follows: 

𝑥𝑥1        + 1.5𝑥𝑥2  + 0.5𝑥𝑥3 + 19𝑦𝑦1  + 12𝑦𝑦2  + 6𝑦𝑦3   = 427.7 

0.1 𝑥𝑥1 + 0.1𝑥𝑥2  + 0.2𝑥𝑥3 + 2𝑦𝑦1    + 4𝑦𝑦2    + 1.5𝑦𝑦3 = 76.2 

0.1 𝑥𝑥1 + 0.1𝑥𝑥2  + 0.1𝑥𝑥3 + 2𝑦𝑦1    + 2𝑦𝑦2    + 4.5𝑦𝑦3 = 88.3 

Because the value 𝑥𝑥1, 𝑥𝑥2, 𝑥𝑥3  has been obtained, substitution value is put into equation (14) so that the new 

equation is obtained as follows: 

19𝑦𝑦1  + 12𝑦𝑦2 + 6𝑦𝑦3     = 260.2 

2𝑦𝑦1    + 4𝑦𝑦2    + 1.5𝑦𝑦3 = 51.3 

2𝑦𝑦1    + 2𝑦𝑦2    + 4.5𝑦𝑦3 = 70.9 

To get the values 𝑦𝑦1,𝑦𝑦2 and 𝑦𝑦3, it must first be proven that equation (14) strictly diagonally dominant in the 

following: 

 |𝑎𝑎11|  ≥  |𝑎𝑎12|  + |𝑎𝑎13| →  |19|   ≥  |12|  + |6|     = 19  ≥ 18 

 |𝑎𝑎22|  ≥  |𝑎𝑎21|  +  |𝑎𝑎23| →  |4|     ≥  |2|     + |1.5| = 4    ≥ 3.5 

 |𝑎𝑎33|  ≥  |𝑎𝑎31|  +  |𝑎𝑎32| →  |4.5|  ≥  |2|     + |2|    = 4.5 ≥ 4 

All equations (14) are proven to be diagonally dominant, so obtained: 

𝑦𝑦1 = 1
19

 (260.2 − 12𝑦𝑦2 − 6𝑦𝑦3) 

𝑦𝑦2 = 1
4
 (51.3 − 2𝑦𝑦1 – 1.5𝑦𝑦3) 

𝑦𝑦3 = 1
4.5

 (70.9 − 2𝑦𝑦1 − 2𝑦𝑦2) 

Then the iteration process starts with the intial value (0, 0, 0), which is as follows: 

First iteration 

𝑦𝑦1(1) = 1
19

 (260.2 – 12(0) – 6(0)) =  1
19

 (260.2) = 13.6947  

𝑦𝑦2(1) = 1
4
 (51.3 – 2(13.6947) – 1.5(0)) =  1

4
 (23.9106) = 5.9776 

𝑦𝑦3(1) = 1
4.5

 (70.9 – 2(13.6947) – 2(5.9776)) =  1
4.5

(31.5554) = 7.0123 

Second iteration 

𝑦𝑦1(2) = 1
19

 (260.2 – 12(5.9776) – 6(7.0123)) =  1
19

 (146.395) = 7.705   

𝑦𝑦2(2) = 1
4
 (51.3 – 2(7.705) – 1.5(7.0123)) =  1

4
 (25.3716) = 6.3429 



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

121 
 

𝑦𝑦3(2) = 1
4.5

 (70.9 – 2(7.705) – 2(6.3429)) =  1
4.5

 (42.8042) = 9.5120 

Based on equation (12), the iteration process until 10th iteration is obtained the value 𝑦𝑦1  = 7.00029, 𝑦𝑦2  = 
5.49987, 𝑦𝑦3 = 10.1998.  

For equation (15), the equation obtained is as follows: 
19𝑧𝑧1 + 12𝑧𝑧2  + 6𝑧𝑧3   + 𝑥𝑥1        + 1.5𝑥𝑥2  +0.2𝑥𝑥3  = 536.2 

2 𝑧𝑧1  + 4𝑧𝑧2    + 1.5𝑧𝑧3 +0.1𝑥𝑥1  + 0.4𝑥𝑥2   + 0.2𝑥𝑥3 = 109.3 

2 𝑧𝑧1  + 2𝑧𝑧2    + 4.5𝑧𝑧3 + 0.2𝑥𝑥1 + 0.3𝑥𝑥2   + 0.1𝑥𝑥3 = 131.9 

Because the value of 𝑥𝑥1, 𝑥𝑥2, 𝑥𝑥3 has been obtained, substitution value is put into equation (15) so that the new 
equation is obtained as follows: 

19𝑧𝑧1 + 12𝑧𝑧2  + 6𝑧𝑧3    = 391.2 

2 𝑧𝑧1  + 4𝑧𝑧2    + 1.5𝑧𝑧3 = 65.8 

2 𝑧𝑧1  + 2𝑧𝑧2    + 4.5𝑧𝑧3 = 98.4 

To get the values 𝑧𝑧1, 𝑧𝑧2 and 𝑧𝑧3 , it must first be proven that equation (15) strictly diagonally dominant in the 

following: 

 |𝑎𝑎11|  ≥  |𝑎𝑎12|  + |𝑎𝑎13| →  |19|  ≥  |12| + |6|     = 19  ≥ 18 

 |𝑎𝑎22|  ≥  |𝑎𝑎21|  +  |𝑎𝑎23| →  |4|    ≥  |2|    + |1.5| = 4    ≥ 3.5 

 |𝑎𝑎33|  ≥  |𝑎𝑎31|  +  |𝑎𝑎32| →  |4.5| ≥  |2|    + |2|    = 4.5 ≥ 4 

All equations (15) are proven to be diagonally dominant, so obtained: 

𝑧𝑧1 = 1
19

 (391.2 − 12𝑧𝑧1 − 6𝑧𝑧3) 

𝑧𝑧2 = 1
4
 (65.8 − 2𝑧𝑧1 – 1.5𝑧𝑧3) 

𝑧𝑧3 = 1
4.5

 (98.4 − 2𝑧𝑧1 − 2𝑧𝑧2) 

Then the iteration process starts with the intial value (0, 0, 0), which is as follows: 

First iteration 
 𝑧𝑧1(1) = 1

19
 (391.2 – 12(0) – 6(0)) =  1

19
 (391.2) = 20.5895 

 𝑧𝑧2(1) = 1
4
 (65.8 – 2(20.5895) – 1.5(0)) =  1

4
 (24.621) = 6.1553  

𝑧𝑧3(1) = 1
4.5

 (98.4 – 2(20.5895) – 2(6.1553)) =  1
4.5

 (44.9104) = 9.9801         

Second iteration 
 𝑧𝑧1(2) = 1

19
 (391.2 – 12(6.1553) – 6(9.9801)) =  1

19
 (257.4558) = 13.5503 

 𝑧𝑧2(2) = 1
4
 (65.8 – 2(13.5503) – 1.5(9.9801)) =  1

4
 (23.7292) = 5.9323 

 𝑧𝑧3(2) = 1
4.5

 (98.4 – 2(13.5503) – 2(5.9323)) =  1
4.5

 (59.4348) = 13.2077    

Based on equation (12), the iteration process until 13th iteration is obtained the value  𝑧𝑧1 = 13.3016, 𝑧𝑧2 = 4.5794, 
𝑧𝑧3 = 13.9195.  



American Scientific Research Journal for Engineering, Technology, and Sciences (ASRJETS) (2019) Volume 56, No  1, pp 113-123 

 

122 
 

5. Conclusion 

In this paper, it can be concluded that the fully fuzzy linear system of equation  𝑨𝑨� ⊗ 𝑨𝑨� = 𝒃𝒃�  solved by 

changing it into (𝐴𝐴,𝑀𝑀,𝑁𝑁) ⊗ (𝑥𝑥,𝑦𝑦, 𝑧𝑧) = (𝑏𝑏,𝑔𝑔, ℎ), so we will get three system linear of equation. Based on the 

results from the example, the value obtained with relatively small errors are as follow: 

𝑥𝑥� = �
𝑥𝑥�1
𝑥𝑥�2
𝑥𝑥�3
� = �

37,7,13.4
62,5.5,4.6
75,10.2,14

� 

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