^{1}

^{1}

^{1}

^{2}

The system of linear equations plays a vital role in real life problems such as optimization, economics, and engineering. The parameters of the system of linear equations are modeled by taking the experimental or observation data. So the parameters of the system actually contain uncertainty rather than the crisp one. The uncertainties may be considered in term of interval or fuzzy numbers. In this paper, a detailed study of three solution techniques namely Classical Method, Extension Principle method and α-cuts and interval Arithmetic Method to solve the system of fuzzy linear equations has been done. Appropriate applications are given to illustrate each technique. Then we discuss the comparison of the different methods numerically and graphically.

There are many linear equation systems in many areas of science and engineering. According to Moore [

In 1965, Lotfi A. Zadeh [

A fuzzy set [

Fuzzy linear equations are similar to ordinary linear equations in classical mathematics. A fuzzy linear equation is of the form A ¯ ⋅ X ¯ = B ¯ or A ¯ + X ¯ = B ¯ or A ¯ ⋅ X ¯ + C ¯ = B ¯ where A ¯ , B ¯ and C ¯ are given fuzzy numbers and X ¯ is an unknown fuzzy number by which the equation is satisfied.

A system of fuzzy linear equations is of the form A ¯ ⋅ X ¯ = B ¯ , where A ¯ = [ a ¯ i j ] is a n × n matrix of fuzzy numbers a ¯ i j , X ¯ t = ( x ¯ 1 , x ¯ 2 , ⋯ , x ¯ n ) is an unknown n × 1 vector of fuzzy numbers x ¯ i by which the equation is satisfied and B ¯ t = ( b ¯ 1 , b ¯ 2 , ⋯ , b ¯ n ) is a n × 1 vector of fuzzy numbers b ¯ i . Now we can write corresponding n × n system for all a ¯ i j ∈ ℝ , 1 ≤ i , j ≤ n as follows:

( a ¯ 11 a ¯ 11 ⋯ a ¯ 11 a ¯ 21 a ¯ 22 ⋯ a ¯ 2 n ⋮ ⋮ ⋱ ⋮ a ¯ n 1 a ¯ n 2 ⋯ a ¯ n n ) ( x ¯ 1 x ¯ 2 ⋮ x ¯ n ) = ( b ¯ 1 b ¯ 2 ⋮ b ¯ n )

We denote the classical solution of A ¯ ⋅ X ¯ = B ¯ as X ¯ c , if it exists.

Substitute the α-cuts of a ¯ i j , x ¯ i and b ¯ i for a ¯ i j , x ¯ i and b ¯ i ( 1 ≤ i , j ≤ 3 ) respectively in the system of linear equations

a ¯ 11 x ¯ 1 + a ¯ 12 x ¯ 2 + a ¯ 13 x ¯ 3 = b ¯ 1 , (1)

a ¯ 21 x ¯ 1 + a ¯ 22 x ¯ 2 + a ¯ 23 x ¯ 3 = b ¯ 2 , (2)

a ¯ 31 x ¯ 1 + a ¯ 32 x ¯ 2 + a ¯ 33 x ¯ 3 = b ¯ 3 , (3)

After substituting the α-cuts of a ¯ i j , x ¯ i and b ¯ i for a ¯ i j , x ¯ i and b ¯ i ( 1 ≤ i , j ≤ 3 ) in the Equations (1)-(3), we get the following three interval equations ∀ α ∈ [ 0 , 1 ] ,

[ a 11 L ( α ) , a 11 U ( α ) ] ⋅ [ x 1 L ( α ) , x 1 U ( α ) ] + [ a 12 L ( α ) , a 12 U ( α ) ] ⋅ [ x 2 L ( α ) , x 2 U ( α ) ] + [ a 13 L ( α ) , a 13 U ( α ) ] ⋅ [ x 3 L ( α ) , x 3 U ( α ) ] = [ b 1 L ( α ) , b 1 U ( α ) ] (4)

[ a 21 L ( α ) , a 21 U ( α ) ] ⋅ [ x 1 L ( α ) , x 1 U ( α ) ] + [ a 22 L ( α ) , a 22 U ( α ) ] ⋅ [ x 2 L ( α ) , x 2 U ( α ) ] + [ a 23 L ( α ) , a 23 U ( α ) ] ⋅ [ x 3 L ( α ) , x 3 U ( α ) ] = [ b 2 L ( α ) , b 2 U ( α ) ] (5)

[ a 31 L ( α ) , a 31 U ( α ) ] ⋅ [ x 1 L ( α ) , x 1 U ( α ) ] + [ a 32 L ( α ) , a 32 U ( α ) ] ⋅ [ x 2 L ( α ) , x 2 U ( α ) ] + [ a 33 L ( α ) , a 33 U ( α ) ] ⋅ [ x 3 L ( α ) , x 3 U ( α ) ] = [ b 3 L ( α ) , b 3 U ( α ) ] (6)

We now need to simplify these equations.

Assuming that all the a ¯ i j and b ¯ i are triangular fuzzy numbers and put α = 1 in Equations (4)-(6). Then we obtain the crisp linear system of equations

a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1

a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2

a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3

The sign of the solutions x 1 , x 2 and x 3 determines the sign of the unknown fuzzy numbers x ¯ 1 , x ¯ 2 and x ¯ 3 . Let us assume for this discussion that all the a ¯ i j > 0 and all the b ¯ i > 0 , so that we try for x ¯ i > 0 , i = 1 , 2 , 3 . We get from Equations (4)-(6)

[ a 11 L ( α ) ⋅ x 1 L ( α ) , a 11 U ( α ) ⋅ x 1 U ( α ) ] + [ a 12 L ( α ) ⋅ x 2 L ( α ) , a 12 U ( α ) ⋅ x 2 U ( α ) ] + [ a 13 L ( α ) ⋅ x 3 L ( α ) , a 13 U ( α ) ⋅ x 3 U ( α ) ] = [ b 1 L ( α ) , b 1 U ( α ) ] (7)

[ a 21 L ( α ) ⋅ x 1 L ( α ) , a 21 U ( α ) ⋅ x 1 U ( α ) ] + [ a 22 L ( α ) ⋅ x 2 L ( α ) , a 22 U ( α ) ⋅ x 2 U ( α ) ] + [ a 23 L ( α ) ⋅ x 3 L ( α ) , a 23 U ( α ) ⋅ x 3 U ( α ) ] = [ b 2 L ( α ) , b 2 U ( α ) ] (8)

[ a 31 L ( α ) ⋅ x 1 L ( α ) , a 31 U ( α ) ⋅ x 1 U ( α ) ] + [ a 32 L ( α ) ⋅ x 2 L ( α ) , a 32 U ( α ) ⋅ x 2 U ( α ) ] + [ a 33 L ( α ) ⋅ x 3 L ( α ) , a 33 U ( α ) ⋅ x 3 U ( α ) ] = [ b 3 L ( α ) , b 3 U ( α ) ] (9)

which yields a 6 × 6 crisp system of linear equations as below

a 11 L ( α ) ⋅ x 1 L ( α ) + a 12 L ( α ) ⋅ x 2 L ( α ) + a 13 L ( α ) ⋅ x 3 L ( α ) = b 1 L ( α ) (10)

a 21 L ( α ) ⋅ x 1 L ( α ) + a 22 L ( α ) ⋅ x 2 L ( α ) + a 23 L ( α ) ⋅ x 3 L ( α ) = b 2 L ( α ) (11)

a 31 L ( α ) ⋅ x 1 L ( α ) + a 32 L ( α ) ⋅ x 2 L ( α ) + a 33 L ( α ) ⋅ x 3 L ( α ) = b 3 L ( α ) (12)

a 11 U ( α ) ⋅ x 1 U ( α ) + a 12 U ( α ) ⋅ x 2 U ( α ) + a 13 U ( α ) ⋅ x 3 U ( α ) = b 1 U ( α ) (13)

a 21 U ( α ) ⋅ x 1 U ( α ) + a 22 U ( α ) ⋅ x 2 U ( α ) + a 23 U ( α ) ⋅ x 3 U ( α ) = b 2 U ( α ) (14)

a 31 U ( α ) ⋅ x 1 U ( α ) + a 32 U ( α ) ⋅ x 2 U ( α ) + a 33 U ( α ) ⋅ x 3 U ( α ) = b 3 U ( α ) (15)

We solve this system for x i L ( α ) and x i U ( α ) , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] .

Using matrix notation this system can be written as

[ a 11 L ( α ) a 12 L ( α ) a 13 L ( α ) 0 0 0 a 21 L ( α ) a 22 L ( α ) a 23 L ( α ) 0 0 0 a 31 L ( α ) a 32 L ( α ) a 33 L ( α ) 0 0 0 0 0 0 a 11 U ( α ) a 12 U ( α ) a 13 U ( α ) 0 0 0 a 21 U ( α ) a 22 U ( α ) a 23 U ( α ) 0 0 0 a 31 U ( α ) a 32 U ( α ) a 33 U ( α ) ] ⋅ [ x 1 L ( α ) x 2 L ( α ) x 3 L ( α ) x 1 U ( α ) x 2 U ( α ) x 3 U ( α ) ] = [ b 1 L ( α ) b 2 L ( α ) b 3 L ( α ) b 1 U ( α ) b 2 U ( α ) b 3 U ( α ) ] (16)

Using

W = [ a 11 L ( α ) a 12 L ( α ) a 13 L ( α ) 0 0 0 a 21 L ( α ) a 22 L ( α ) a 23 L ( α ) 0 0 0 a 31 L ( α ) a 32 L ( α ) a 33 L ( α ) 0 0 0 0 0 0 a 11 U ( α ) a 12 U ( α ) a 13 U ( α ) 0 0 0 a 21 U ( α ) a 22 U ( α ) a 23 U ( α ) 0 0 0 a 31 U ( α ) a 32 U ( α ) a 33 U ( α ) ] , S = [ x 1 L ( α ) x 2 L ( α ) x 3 L ( α ) x 1 U ( α ) x 2 U ( α ) x 3 U ( α ) ] and V = [ b 1 L ( α ) b 2 L ( α ) b 3 L ( α ) b 1 U ( α ) b 2 U ( α ) b 3 U ( α ) ] , (17)

we get a crisp system of the form W ⋅ S = V .

For obtaining the fuzzy solution for the fully fuzzy linear system of equations, the necessary condition is that the coefficient matrix of the converted crisp system is invertible ∀ α ∈ [ 0 , 1 ] .

The system (5.2.29) can be partitioned into two system as

[ a 11 L ( α ) a 12 L ( α ) a 13 L ( α ) a 21 L ( α ) a 22 L ( α ) a 23 L ( α ) a 31 L ( α ) a 32 L ( α ) a 33 L ( α ) ] ⋅ [ x 1 L ( α ) x 2 L ( α ) x 3 L ( α ) ] = [ b 1 L ( α ) b 2 L ( α ) b 3 L ( α ) ]

and

[ a 11 U ( α ) a 12 U ( α ) a 13 U ( α ) a 21 U ( α ) a 22 U ( α ) a 23 U ( α ) a 31 U ( α ) a 32 U ( α ) a 33 U ( α ) ] ⋅ [ x 1 U ( α ) x 2 U ( α ) x 3 U ( α ) ] = [ b 1 U ( α ) b 2 U ( α ) b 3 U ( α ) ]

That is, W L ⋅ S L = V L and W U ⋅ S U = V U . The solution of these crisp systems determines x i L ( α ) and x i U ( α ) , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] , which are used to reconstruct the components of 3 × 1 fuzzy vector X ¯ .

After solving for the x i L ( α ) and x i U ( α ) , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] we check to see if the intervals x ¯ i = [ x i L ( α ) , x i U ( α ) ] , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] define continuous fuzzy numbers for i = 1 , 2 . What is needed is:

1) ∂ ∂ α ( x i L ( α ) ) > 0 ,

2) ∂ ∂ α ( x i U ( α ) ) < 0 , and

3) x i L ( 1 ) ≤ x i U ( 1 ) for i = 1 , 2 , 3 (equality for triangular shaped fuzzy numbers).

We denote the extension principle solution of a system of fuzzy linear equations by X ¯ e and it always exists but may, or may not satisfy the original system of fuzzy linear equations. That is, A ¯ ⋅ X ¯ e = B ¯ may, or may not be true.

Let the components of X ¯ e are x ¯ 1 , x ¯ 2 and x ¯ 3 . In this method, we need to fuzzify the crisp solutions

x 1 = b 1 ( a 22 a 33 − a 23 a 32 ) − a 12 ( a 33 b 2 − a 23 b 3 ) + a 13 ( a 32 b 2 − a 22 b 3 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) (18)

x 2 = a 11 ( a 33 b 2 − a 23 b 3 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) (19)

x 3 = a 11 ( a 22 b 3 − a 32 b 2 ) − a 12 ( a 21 b 3 − a 31 b 2 ) + b 1 ( a 21 a 32 − a 31 a 22 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) (20)

using the extension principle.

Let

h 1 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) = x 1 ,

h 2 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) = x 2 ,

and

h 3 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) = x 3 .

To obtain the first component x ¯ 1 in X ¯ e , we substitute

a ¯ 11 , a ¯ 12 , a ¯ 13 , a ¯ 21 , a ¯ 22 , a ¯ 23 , a ¯ 31 , a ¯ 32 , a ¯ 33 , b ¯ 1 , b ¯ 2 , b ¯ 3

for

a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3

in h 1 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) and evaluate using the extension principle.

It should be noted that,

Δ = a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ≠ 0 ,

that is, the determinant of the coefficients matrix must be nonsingular and invertible.

Let α-cut of x ¯ 1 is x ¯ 1 [ α ] = [ x 1 L ( α ) , x 1 U ( α ) ] .

Then the α-cut of x ¯ 1 can be written as

x 1 L ( α ) = min { h 1 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) : a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] }

x 1 U ( α ) = max { h 1 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) : a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] }

Or,

x 1 L ( α ) = min { b 1 ( a 22 a 33 − a 23 a 32 ) − a 12 ( a 33 b 2 − a 23 b 3 ) + a 13 ( a 32 b 2 − a 22 b 3 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ; a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] } (21)

x 1 U ( α ) = max { b 1 ( a 22 a 33 − a 23 a 32 ) − a 12 ( a 33 b 2 − a 23 b 3 ) + a 13 ( a 32 b 2 − a 22 b 3 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ; a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] } (22)

for α ∈ [ 0 , 1 ] and 1 ≤ i , j ≤ 3 .

Similarly, α-cut of x ¯ 2 is x ¯ 2 [ α ] = [ x 2 L ( α ) , x 2 U ( α ) ] . Then the α-cut of x ¯ 2 is

x 2 L ( α ) = min { a 11 ( a 33 b 2 − a 23 b 3 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ; a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] } (23)

x 2 U ( α ) = max { a 11 ( a 33 b 2 − a 23 b 3 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ; a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] } (24)

for α ∈ [ 0 , 1 ] and 1 ≤ i , j ≤ 3 .

And the α-cut of x ¯ 3 is x ¯ 3 [ α ] = [ x 3 L ( α ) , x 3 U ( α ) ] . Then the α-cut of x ¯ 3 is

x 3 L ( α ) = min { a 11 ( a 22 b 3 − a 32 b 2 ) − a 12 ( a 21 b 3 − a 31 b 2 ) + b 1 ( a 21 a 32 − a 31 a 22 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ; a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] } (25)

x 3 U ( α ) = max { a 11 ( a 22 b 3 − a 32 b 2 ) − a 12 ( a 21 b 3 − a 31 b 2 ) + b 1 ( a 21 a 32 − a 31 a 22 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ; a i j ∈ a ¯ i j [ α ] , b j ∈ b ¯ j [ α ] } (26)

for α ∈ [ 0 , 1 ] and 1 ≤ i , j ≤ 3 .

After solving for the x i L ( α ) and x i U ( α ) , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] we check to see if the intervals x ¯ i = [ x i L ( α ) , x i U ( α ) ] , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] define continuous fuzzy numbers for i = 1 , 2 . What is needed is:

1) ∂ ∂ α ( x i L ( α ) ) > 0 ,

2) ∂ ∂ α ( x i U ( α ) ) < 0 , and

3) x i L ( 1 ) ≤ x i U ( 1 ) for i = 1 , 2 , 3 (equality for triangular shaped fuzzy numbers).

Now by setting X ¯ e = [ x ¯ 1 x ¯ 2 x ¯ 3 ] we can check whether A ¯ ⋅ X ¯ e = B ¯ is true or false.

If we set α = 1 , we get the crisp solution x 1 from Equation (21) and Equation (22); crisp solution x 2 from Equation (23) and Equation (24) and crisp solution x 3 from Equation (25) and Equation (26) by assuming all the a ¯ i j and b ¯ j are triangular shaped fuzzy numbers and x i L ( 1 ) = x i U ( 1 ) = x i , i = 1 , 2 , 3 .

We denote the α-cut and interval arithmetic solution of a system of fuzzy linear equations by X ¯ I and it always exists but may, or may not satisfy the original system of fuzzy linear equations. That is, A ¯ ⋅ X ¯ I = B ¯ may, or may not be true.

Let the components of X ¯ I are x ¯ 1 , x ¯ 2 and x ¯ 3 . In this method, we need to fuzzify the crisp solutions

x 1 = b 1 ( a 22 a 33 − a 23 a 32 ) − a 12 ( a 33 b 2 − a 23 b 3 ) + a 13 ( a 32 b 2 − a 22 b 3 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) (27)

x 2 = a 11 ( a 33 b 2 − a 23 b 3 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) (28)

x 3 = a 11 ( a 22 b 3 − a 32 b 2 ) − a 12 ( a 21 b 3 − a 31 b 2 ) + b 1 ( a 21 a 32 − a 31 a 22 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) (29)

using α-cut and interval arithmetic.

We substitute the α-cuts of a ¯ 11 , a ¯ 12 , a ¯ 13 , a ¯ 21 , a ¯ 22 , a ¯ 23 , a ¯ 31 , a ¯ 32 , a ¯ 33 , b ¯ 1 , b ¯ 2 , b ¯ 3 for a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 in Equations (27)-(29) and the simplifying using interval arithmetic we obtain the α-cuts of x ¯ 1 , x ¯ 2 and x ¯ 3 .

It should be noted that

Δ = a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) ≠ 0 ,

that is, the determinant of the coefficients matrix must be nonsingular and invertible.

Let α-cut of x ¯ 1 is x ¯ 1 [ α ] = [ x 1 L ( α ) , x 1 U ( α ) ] .

To find the α-cut of x ¯ 1 we substitute

a ¯ 11 , a ¯ 12 , a ¯ 13 , a ¯ 21 , a ¯ 22 , a ¯ 23 , a ¯ 31 , a ¯ 32 , a ¯ 33 , b ¯ 1 , b ¯ 2 , b ¯ 3

for a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 in Equation (27). Then we get,

x ¯ 1 [ α ] = b ¯ 1 [ α ] ( a ¯ 22 [ α ] a ¯ 33 [ α ] − a ¯ 23 [ α ] a ¯ 32 [ α ] ) − a ¯ 12 [ α ] ( a ¯ 33 [ α ] b ¯ 2 [ α ] − a ¯ 23 [ α ] b ¯ 3 [ α ] ) + a ¯ 13 [ α ] ( a ¯ 32 [ α ] b ¯ 2 [ α ] − a ¯ 22 [ α ] b ¯ 3 [ α ] ) a ¯ 11 [ α ] ( a ¯ 22 [ α ] a ¯ 33 [ α ] − a ¯ 32 [ α ] a ¯ 23 [ α ] ) − a ¯ 12 [ α ] ( a ¯ 21 [ α ] a ¯ 33 [ α ] − a ¯ 31 [ α ] a ¯ 23 [ α ] ) + a ¯ 13 [ α ] ( a ¯ 21 [ α ] a ¯ 32 [ α ] − a ¯ 31 [ α ] a ¯ 22 [ α ] )

Let us assume that all a ¯ i j > 0 and all the b ¯ j > 0 . Then by simplifying and using the interval arithmetic we get,

x 1 L ( α ) = ( b 1 L a 22 L a 33 L − b 1 U a 23 U a 32 U ) − ( a 12 L a 33 L b 2 L − a 12 U a 23 U b 3 U ) + ( a 13 L a 32 L b 2 L − a 13 U a 22 U b 3 U ) ( a 11 U a 22 U a 33 U − a 11 L a 32 L a 23 L ) − ( a 12 U a 21 U a 33 U − a 12 L a 31 L a 23 L ) + ( a 13 U a 21 U a 32 U − a 13 L a 31 L a 22 L ) (30)

And

x 1 U ( α ) = ( b 1 U a 22 U a 33 U − b 1 L a 23 L a 32 L ) − ( a 12 U a 33 U b 2 U − a 12 L a 23 L b 3 L ) + ( a 13 U a 32 U b 2 U − a 13 L a 22 L b 3 L ) ( a 11 L a 32 L a 23 L − a 11 U a 22 U a 33 U ) − ( a 12 L a 31 L a 23 L − a 12 U a 21 U a 33 U ) + ( a 13 L a 31 L a 22 L − a 13 U a 21 U a 32 U ) (31)

Again let α-cut of x ¯ 2 is x ¯ 2 [ α ] = [ x 2 L ( α ) , x 2 U ( α ) ] . To find the α-cut of x ¯ 2 we substtute a ¯ 11 , a ¯ 12 , a ¯ 13 , a ¯ 21 , a ¯ 22 , a ¯ 23 , a ¯ 31 , a ¯ 32 , a ¯ 33 , b ¯ 1 , b ¯ 2 , b ¯ 3 for

a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3

in Equations (28). Then we get,

x ¯ 2 [ α ] = a ¯ 11 [ α ] ( a ¯ 33 [ α ] b ¯ 2 [ α ] − a ¯ 23 [ α ] b ¯ 3 [ α ] ) − b ¯ 1 [ α ] ( a ¯ 21 [ α ] a ¯ 33 [ α ] − a ¯ 31 [ α ] a ¯ 23 [ α ] ) + a ¯ 13 [ α ] ( a ¯ 21 [ α ] b ¯ 3 [ α ] − a ¯ 31 [ α ] b ¯ 2 [ α ] ) a ¯ 11 [ α ] ( a ¯ 22 [ α ] a ¯ 33 [ α ] − a ¯ 32 [ α ] a ¯ 23 [ α ] ) − a ¯ 12 [ α ] ( a ¯ 21 [ α ] a ¯ 33 [ α ] − a ¯ 31 [ α ] a ¯ 23 [ α ] ) + a ¯ 13 [ α ] ( a ¯ 21 [ α ] a ¯ 32 [ α ] − a ¯ 31 [ α ] a ¯ 22 [ α ] )

Let us assume that all a ¯ i j > 0 and all the b ¯ j > 0 . Then by simplifying and using the interval arithmetic we get,

x 2 L ( α ) = ( a 11 L a 33 L b 2 L − a 11 U a 23 U b 3 U ) − ( a 21 L a 33 L b 1 L − a 31 U a 23 U b 1 U ) + ( a 13 L a 21 L b 3 L − a 13 U a 31 U b 2 U ) ( a 11 U a 22 U a 33 U − a 11 L a 32 L a 23 L ) − ( a 12 U a 21 U a 33 U − a 12 L a 31 L a 23 L ) + ( a 13 U a 21 U a 32 U − a 13 L a 31 L a 22 L ) (32)

And

x 2 U ( α ) = ( a 11 U a 23 U b 3 U − a 11 L a 33 L b 2 L ) − ( a 31 U a 23 U b 1 U − a 21 L a 33 L b 1 L ) + ( a 13 U a 31 U b 2 U − a 13 L a 21 L b 3 L ) ( a 11 L a 32 L a 23 L − a 11 U a 22 U a 33 U ) − ( a 12 L a 31 L a 23 L − a 12 U a 21 U a 33 U ) + ( a 13 L a 31 L a 22 L − a 13 U a 21 U a 32 U ) (33)

And finally, let α-cut of x ¯ 3 is x ¯ 3 [ α ] = [ x 3 L ( α ) , x 3 U ( α ) ] . To find the α-cut of x ¯ 3 we substitute a ¯ 11 , a ¯ 12 , a ¯ 13 , a ¯ 21 , a ¯ 22 , a ¯ 23 , a ¯ 31 , a ¯ 32 , a ¯ 33 , b ¯ 1 , b ¯ 2 , b ¯ 3 for

a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3

in Equations (29). Then we get,

x ¯ 3 [ α ] = a ¯ 11 [ α ] ( a ¯ 22 [ α ] b ¯ 3 [ α ] − a ¯ 32 [ α ] b ¯ 2 [ α ] ) − a ¯ 12 [ α ] ( a ¯ 21 [ α ] b ¯ 3 [ α ] − a ¯ 31 [ α ] b ¯ 2 [ α ] ) + b ¯ 1 [ α ] ( a ¯ 21 [ α ] a ¯ 32 [ α ] − a ¯ 31 [ α ] a ¯ 22 [ α ] ) a ¯ 11 [ α ] ( a ¯ 22 [ α ] a ¯ 33 [ α ] − a ¯ 32 [ α ] a ¯ 23 [ α ] ) − a ¯ 12 [ α ] ( a ¯ 21 [ α ] a ¯ 33 [ α ] − a ¯ 31 [ α ] a ¯ 23 [ α ] ) + a ¯ 13 [ α ] ( a ¯ 21 [ α ] a ¯ 32 [ α ] − a ¯ 31 [ α ] a ¯ 22 [ α ] )

Let us assume that all a ¯ i j > 0 and all the b ¯ j > 0 . Then by simplifying and using the interval arithmetic we get,

x 3 L ( α ) = ( a 11 L a 22 L b 3 L − a 11 U a 32 U b 2 U ) − ( a 12 L a 21 L b 3 L − a 12 U a 31 U b 2 U ) + ( a 21 L a 32 L b 1 L − a 31 U a 22 U b 1 U ) ( a 11 U a 22 U a 33 U − a 11 L a 32 L a 23 L ) − ( a 12 U a 21 U a 33 U − a 12 L a 31 L a 23 L ) + ( a 13 U a 21 U a 32 U − a 13 L a 31 L a 22 L ) (34)

And

x 3 U ( α ) = ( a 11 U a 32 U b 2 U − a 11 L a 22 L b 3 L ) − ( a 12 U a 31 U b 2 U − a 12 L a 21 L b 3 L ) + ( a 31 U a 22 U b 1 U − a 21 L a 32 L b 1 L ) ( a 11 L a 32 L a 23 L − a 11 U a 22 U a 33 U ) − ( a 12 L a 31 L a 23 L − a 12 U a 21 U a 33 U ) + ( a 13 L a 31 L a 22 L − a 13 U a 21 U a 32 U ) (35)

After solving for the x i L ( α ) and x i U ( α ) , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] we check to see if the intervals x ¯ i = [ x i L ( α ) , x i U ( α ) ] , i = 1 , 2 , 3 , α ∈ [ 0 , 1 ] define continuous fuzzy numbers for i = 1 , 2 . What is needed is:

1) ∂ ∂ α ( x i L ( α ) ) > 0 ,

2) ∂ ∂ α ( x i U ( α ) ) < 0 , and

3) x i L ( 1 ) ≤ x i U ( 1 ) for i = 1 , 2 , 3 (equality for triangular shaped fuzzy numbers).

Now by setting X ¯ I = [ x ¯ 1 x ¯ 2 x ¯ 3 ] we can check whether A ¯ ⋅ X ¯ I = B ¯ is true or false.

If we set α = 1 , we get the crisp solution x 1 from Equation (30) and Equation (31); crisp solution x 2 from Equation (32) and Equation (33) and crisp solution x 3 from Equation (34) and Equation (35) by assuming all the a ¯ i j and b ¯ j are triangular shaped fuzzy numbers and x i L ( 1 ) = x i U ( 1 ) = x i , i = 1 , 2 , 3 .

Consider the system of fuzzy linear equation in matrix form

[ ( 1 / 2 / 3 ) 0 0 0 ( 3 / 4 / 5 ) 0 0 0 ( 6 / 8 / 12 ) ] [ x ¯ 1 x ¯ 2 x ¯ 3 ] = [ ( − 1 / 1 / 2 ) ( 1 / 2 / 3 ) ( 2 / 5 / 8 ) ]

We solve this fuzzy matrix equation using the classical method.

The above system can be written as

( 1 / 2 / 3 ) x ¯ 1 = ( − 1 / 1 / 2 ) ( 3 / 4 / 5 ) x ¯ 2 = ( 1 / 2 / 3 ) ( 6 / 8 / 12 ) x ¯ 3 = ( 2 / 5 / 8 ) } (36)

Here a ¯ 11 = ( 1 / 2 / 3 ) , a ¯ 22 = ( 3 / 4 / 5 ) , and a ¯ 33 = ( 6 / 8 / 12 ) .

Also, we have, b ¯ 1 = ( − 1 / 1 / 2 ) , b ¯ 2 = ( 1 / 2 / 3 ) and b ¯ 3 = ( 2 / 5 / 8 ) .

Then the α-cuts are:

a ¯ 11 [ α ] = [ a 11 L ( α ) , a 11 U ( α ) ] = [ 1 + α , 3 − α ] ,

a ¯ 22 [ α ] = [ a 22 L ( α ) , a 22 U ( α ) ] = [ 3 + α , 5 − α ] ,

a ¯ 33 [ α ] = [ a 33 L ( α ) , a 33 U ( α ) ] = [ 6 + 2 α , 12 − 4 α ] ,

b ¯ 1 [ α ] = [ b 1 L ( α ) , b 1 U ( α ) ] = [ − 1 + 2 α , 2 − α ] ,

b ¯ 2 [ α ] = [ b 2 L ( α ) , a 2 U ( α ) ] = [ 1 + α , 3 − α ] ,

b ¯ 3 [ α ] = [ b 3 L ( α ) , a 3 U ( α ) ] = [ 2 + 3 α , 8 − 3 α ] , ∀ α ∈ [ 0 , 1 ] .

Now substituting the α-cuts of a ¯ 11 , a ¯ 22 , a ¯ 33 , b ¯ 1 , b ¯ 2 and b ¯ 3 for a ¯ 11 , a ¯ 22 , a ¯ 33 , b ¯ 1 , b ¯ 2 and b ¯ 3 in the system (36) and we get

[ 1 + α , 3 − α ] [ x 1 L ( α ) , x 1 U ( α ) ] = [ − 1 + 2 α , 2 − α ] [ 3 + α , 5 − α ] [ x 2 L ( α ) , x 2 U ( α ) ] = [ 1 + α , 3 − α ] [ 6 + 2 α , 12 − 4 α ] [ x 3 L ( α ) , x 3 U ( α ) ] = [ 2 + 3 α , 8 − 3 α ] } (37)

If we put α = 1 , we get the crisp solutions x 1 = 1 / 2 , x 2 = 1 / 2 and x 3 = 5 / 8 . So we assume we can get a solution with x ¯ 1 > 0 , x ¯ 1 [ 1 ] = 3 / 2 , x ¯ 2 > 0 , x ¯ 2 [ 1 ] = 5 / 4 and x ¯ 3 > 0 , x ¯ 3 [ 1 ] = 5 / 8 .

From Equation (37) we get,

( 1 + α ) x 1 L ( α ) = ( − 1 + 2 α ) or x 1 L ( α ) = ( − 1 + 2 α ) ( 1 + α ) ( 3 + α ) x 2 L ( α ) = ( 1 + α ) or x 2 L ( α ) = ( 1 + α ) ( 3 + α ) ( 6 + 2 α ) x 3 L ( α ) = ( 2 + 3 α ) or x 3 L ( α ) = ( 2 + 3 α ) ( 6 + 2 α ) ( 3 − α ) x 1 U ( α ) = ( 2 − α ) or x 1 U ( α ) = ( 2 − α ) ( 3 − α ) ( 5 − α ) x 2 L ( α ) = ( 3 − α ) or x 2 U ( α ) = ( 3 − α ) ( 5 − α ) ( 12 − 4 α ) x 3 U ( α ) = ( 8 − 3 α ) or x 3 U ( α ) = ( 8 − 3 α ) ( 12 − 4 α ) } (38)

We find that,

∂ ∂ α ( x 1 L ( α ) ) = 3 ( 1 + α ) 2 > 0 ; ∂ ∂ α ( x 2 L ( α ) ) = 2 ( 1 + α ) 2 > 0 ;

∂ ∂ α ( x 3 L ( α ) ) = 14 ( 6 + 2 α ) 2 > 0 ; ∂ ∂ α ( x 1 U ( α ) ) = − 1 ( 3 − α ) 2 < 0 ;

∂ ∂ α ( x 2 U ( α ) ) = − 2 ( 5 − α ) 2 < 0 ; ∂ ∂ α ( x 3 U ( α ) ) = − 4 ( 12 − 4 α ) 2 < 0 .

That is x 1 L ( α ) , x 2 L ( α ) and x 3 L ( α ) are increasing functions of α ∈ [ 0 , 1 ] and x 1 U ( α ) , x 2 U ( α ) and x 3 U ( α ) are decreasing functions of α ∈ [ 0 , 1 ] . Also x 1 L ( 1 ) = x 1 U ( 1 ) , x 2 L ( 1 ) = x 2 U ( 1 ) and x 3 L ( 1 ) = x 3 U ( 1 ) .

Hence,

x ¯ 1 [ α ] = [ ( − 1 + 2 α ) ( 1 + α ) , ( 2 − α ) ( 3 − α ) ] , x ¯ 2 [ α ] = [ ( 1 + α ) ( 3 + α ) , ( 3 − α ) ( 5 − α ) ]

and x ¯ 3 [ α ] = [ ( 2 + 3 α ) ( 6 + 2 α ) , ( 8 − 3 α ) ( 12 − 4 α ) ] defines the α-cuts of three fuzzy numbers respectively.

Now the support of x ¯ 1 is x ¯ 1 [ 0 ] = [ ( − 1 + 2 × 0 ) ( 1 + 0 ) , ( 2 − 0 ) ( 3 − 0 ) ] = [ − 1 , 2 3 ] and modal of x ¯ 1 is x ¯ 1 [ 1 ] = [ ( − 1 + 2 × 1 ) ( 1 + 1 ) , ( 2 − 1 ) ( 3 − 1 ) ] = [ 1 2 , 1 2 ] = 1 2 ;

The support of x ¯ 2 is x ¯ 2 [ 0 ] = [ ( 1 + 0 ) ( 3 + 0 ) , ( 3 − 0 ) ( 5 − 0 ) ] = [ 1 3 , 3 5 ] and modal of x ¯ 2 is x ¯ 2 [ 1 ] = [ ( 1 + 1 ) ( 3 + 1 ) , ( 3 − 1 ) ( 5 − 1 ) ] = [ 1 2 , 1 2 ] = 1 2 ; and

The support of x ¯ 3 is x ¯ 3 [ 0 ] = [ ( 2 + 3 × 0 ) ( 6 + 2 × 0 ) , ( 8 − 3 × 0 ) ( 12 − 4 × 0 ) ] = [ 1 3 , 2 3 ] and modal of x ¯ 3 is x ¯ 3 [ 1 ] = [ ( 2 + 3 × 1 ) ( 6 + 2 × 1 ) , ( 8 − 3 × 1 ) ( 12 − 4 × 1 ) ] = [ 5 8 , 5 8 ] = 5 8 .

Therefore we can say that, the classical solution X ¯ c exists and its components are continuous triangular shaped fuzzy numbers

x ¯ 1 ≈ ( − 1 / 1 2 / 2 3 ) , x ¯ 2 ≈ ( 1 3 / 1 2 / 3 5 ) and x ¯ 3 ≈ ( 1 3 / 5 8 / 2 3 ) .

The membership function of the triangularly shaped number x ¯ 1 ≈ ( − 1 / 1 2 / 2 3 ) is

x = ( − 1 + 2 α ) ( 1 + α ) ⇒ μ 1 L ( x ) = α = ( 1 + x ) ( 2 − x ) , for − 1 ≤ x ≤ 1 2 ,

and,

x = ( 2 − α ) ( 3 − α ) = μ 1 U ( x ) = α = ( 2 − 3 x ) ( 1 − x ) , for 1 2 ≤ x ≤ 2 3 .

Thus the membership function of x ¯ 1 ≈ ( − 1 / 1 2 / 2 3 ) is

μ x ¯ 1 ( x ) = { ( 1 + x ) ( 2 − x ) , for − 1 ≤ x ≤ 1 2 ( 2 − 3 x ) ( 1 − x ) , for 1 2 ≤ x ≤ 2 3 0 , otherwise (39)

and its graph is shown in

The membership function of the triangularly shaped number x ¯ 2 ≈ ( 1 3 / 1 2 / 3 5 ) is

x = ( 1 + α ) ( 3 + α ) ⇒ μ 2 L ( x ) = α = ( 1 − 3 x ) ( x − 1 ) , for 1 3 ≤ x ≤ 1 2 ,

and

x = ( 3 − α ) ( 5 − α ) = μ 2 U ( x ) = α = ( 3 − 5 x ) ( 1 − x ) , for 1 2 ≤ x ≤ 3 5 .

Thus the membership function of x ¯ 2 ≈ ( 1 3 / 1 2 / 3 5 ) is

μ x ¯ 2 ( x ) = { ( 1 − 3 x ) ( x − 1 ) , for 1 3 ≤ x ≤ 1 2 ( 3 − 5 x ) ( 1 − x ) , for 1 2 ≤ x ≤ 3 5 0 , otherwise (40)

and its graph is shown in

And the membership function of the triangular shaped number x ¯ 3 ≈ ( 1 3 / 5 8 / 2 3 ) is

x = ( 2 + 3 α ) ( 6 + 2 α ) ⇒ μ 3 L ( x ) = α = ( 2 − 6 x ) ( 2 x − 3 ) , for 1 3 ≤ x ≤ 5 8 ,

and

x = ( 8 − 3 α ) ( 12 − 4 α ) = μ 3 U ( x ) = α = ( 8 − 12 x ) ( 3 − 4 x ) , for 5 8 ≤ x ≤ 2 3 .

Thus the membership function of x ¯ 3 ≈ ( 1 3 / 5 8 / 2 3 ) is

μ x ¯ 3 ( x ) = { ( 2 − 6 x ) ( 2 x − 3 ) , for 1 3 ≤ x ≤ 5 8 ( 8 − 12 x ) ( 3 − 4 x ) , for 5 8 ≤ x ≤ 2 3 0 , otherwise (41)

and its graph is shown in

Finally the graph of the classical solution

X ¯ c = [ x ¯ 1 ≈ ( − 1 / 1 2 / 2 3 ) x ¯ 2 ≈ ( 1 3 / 1 2 / 3 5 ) x ¯ 3 ≈ ( 1 3 / 5 8 / 2 3 ) ]

is shown in

Consider the system of fuzzy linear equation in matrix form

[ ( 1 / 2 / 3 ) 0 0 0 ( 3 / 4 / 5 ) 0 0 0 ( 6 / 8 / 12 ) ] [ x ¯ 1 x ¯ 2 x ¯ 3 ] = [ ( − 1 / 1 / 2 ) ( 1 / 2 / 3 ) ( 2 / 5 / 8 ) ]

We solve this fuzzy matrix equation using the extension principle method.

Here, a ¯ 11 = ( 1 / 2 / 3 ) , a ¯ 12 = 0 , a ¯ 13 = 0 , a ¯ 21 = 0 , a ¯ 22 = ( 3 / 4 / 5 ) , a ¯ 23 = 0 , a ¯ 31 = 0 , a ¯ 32 = 0 and a ¯ 33 = ( 6 / 8 / 12 ) . Also, we have, b ¯ 1 = ( − 1 / 1 / 2 ) , b ¯ 2 = ( 1 / 2 / 3 ) and b ¯ 3 = ( 2 / 5 / 8 ) .

Then the α-cuts are:

a ¯ 11 [ α ] = [ a 11 L ( α ) , a 11 U ( α ) ] = [ 1 + α , 3 − α ] ,

a ¯ 22 [ α ] = [ a 22 L ( α ) , a 22 U ( α ) ] = [ 3 + α , 5 − α ] ,

a ¯ 33 [ α ] = [ a 33 L ( α ) , a 33 U ( α ) ] = [ 6 + 2 α , 12 − 4 α ] ,

b ¯ 1 [ α ] = [ b 1 L ( α ) , b 1 U ( α ) ] = [ − 1 + 2 α , 2 − α ] ,

b ¯ 2 [ α ] = [ b 2 L ( α ) , a 2 U ( α ) ] = [ 1 + α , 3 − α ] ,

b ¯ 3 [ α ] = [ b 3 L ( α ) , a 3 U ( α ) ] = [ 2 + 3 α , 8 − 3 α ] , ∀ α ∈ [ 0 , 1 ] .

Now the crisp solutions are

x 1 = b 1 ( a 22 a 33 − a 23 a 32 ) − a 12 ( a 33 b 2 − a 23 b 3 ) + a 13 ( a 32 b 2 − a 22 b 3 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) = a 22 a 33 b 1 a 11 a 22 a 33

∴ h 1 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) = b 1 a 11 (42)

x 2 = a 11 ( a 33 b 2 − a 23 b 3 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) = a 11 a 33 b 2 a 11 a 22 a 33

∴ h 2 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) = b 2 a 22 (43)

x 3 = a 11 ( a 22 b 3 − a 32 b 2 ) − a 12 ( a 21 b 3 − a 31 b 2 ) + b 1 ( a 21 a 32 − a 31 a 22 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) = a 11 a 22 b 3 a 11 a 22 a 33

∴ h 3 ( a 11 , a 12 , a 13 , a 21 , a 22 , a 23 , a 31 , a 32 , a 33 , b 1 , b 2 , b 3 ) = b 3 a 33 (44)

Since b 1 a 11 , b 2 a 22 and b 3 a 33 are increasing functions of b 1 , b 2 and b 3 ; and decreasing functions of a 11 , a 22 and a 33 , then

min { b 1 a 11 : b 1 ∈ b ¯ 1 [ α ] , a 11 ∈ a ¯ 11 [ α ] } = b 1 L ( α ) a 11 U (α)

and max { b 1 a 11 : b 1 ∈ b ¯ 1 [ α ] , a 11 ∈ a ¯ 11 [ α ] } = b 1 U ( α ) a 11 L ( α ) , ∀ α ∈ [ 0 , 1 ] .

∴ x ¯ 1 [ α ] = [ b 1 L ( α ) a 11 U ( α ) , b 1 U ( α ) a 11 L ( α ) ] = [ − 1 + 2 α 3 − α , 2 − α 1 + α ]

Similarly,

x ¯ 2 [ α ] = [ b 2 L ( α ) a 22 U ( α ) , b 2 U ( α ) a 22 L ( α ) ] = [ 1 + α 5 − α , 3 − α 3 + α ] .

Also,

x ¯ 3 [ α ] = [ b 3 L ( α ) a 33 U ( α ) , b 3 U ( α ) a 33 L ( α ) ] = [ 2 + 3 α 12 − 4 α , 8 − 3 α 6 + 2 α ] .

Here,

x 1 L ( α ) = − 1 + 2 α 3 − α , x 2 L ( α ) = 1 + α 5 − α , x 3 L ( α ) = 2 + 3 α 12 − 4 α , x 1 U ( α ) = 2 − α 1 + α , x 2 U ( α ) = 3 − α 3 + α , x 3 U ( α ) = 8 − 3 α 6 + 2 α .

We find that,

∂ ∂ α ( x 1 L ( α ) ) = 5 ( 3 − α ) 2 > 0 ; ∂ ∂ α ( x 2 L ( α ) ) = 6 ( 5 − α ) 2 > 0 ;

∂ ∂ α ( x 3 L ( α ) ) = 44 ( 12 − 4 α ) 2 > 0 ; ∂ ∂ α ( x 1 U ( α ) ) = − 3 ( 1 + α ) 2 < 0 ;

∂ ∂ α ( x 2 U ( α ) ) = − 6 ( 3 + α ) 2 < 0 ; ∂ ∂ α ( x 3 U ( α ) ) = − 34 ( 6 + 2 α ) 2 < 0 .

That is, x 1 L ( α ) , x 2 L ( α ) and x 3 L ( α ) are increasing functions of α ∈ [ 0 , 1 ] and x 1 U ( α ) , x 2 U ( α ) and x 3 U ( α ) are decreasing functions of α ∈ [ 0 , 1 ] . Also x 1 L ( 1 ) = x 1 U ( 1 ) , x 2 L ( 1 ) = x 2 U ( 1 ) and x 3 L ( 1 ) = x 3 U ( 1 ) .

Hence,

x ¯ 1 [ α ] = [ − 1 + 2 α 3 − α , 2 − α 1 + α ] , x ¯ 2 [ α ] = [ 1 + α 5 − α , 3 − α 3 + α ] and x ¯ 3 [ α ] = [ 2 + 3 α 12 − 4 α , 8 − 3 α 6 + 2 α ]

define the α-cuts of three fuzzy numbers respectively.

Now the support of x ¯ 1 is x ¯ 1 [ 0 ] = [ − 1 + 2 × 0 3 − 0 , 2 − 0 1 + 0 ] = [ − 1 3 , 2 ] and modal of x ¯ 1 is x ¯ 1 [ 1 ] = [ − 1 + 2 × 1 3 − 1 , 2 − 1 1 + 1 ] = [ 1 2 , 1 2 ] = 1 2 ;

The support of x ¯ 2 is x ¯ 2 [ 0 ] = [ 1 + 0 5 − 0 , 3 − 0 3 + 0 ] = [ 1 5 , 1 ] and modal of x ¯ 2 is

x ¯ 2 [ 1 ] = [ 1 + 1 5 − 1 , 3 − 1 3 + 1 ] = [ 1 2 , 1 2 ] = 1 2 ;

and the support of x ¯ 3 is x ¯ 3 [ 0 ] = [ 2 + 3 × 0 12 − 4 × 0 , 8 − 3 × 0 6 + 2 × 0 ] = [ 1 6 , 4 3 ] and modal of x ¯ 3 is x ¯ 3 [ 1 ] = [ 2 + 3 × 1 12 − 4 × 1 , 8 − 3 × 1 6 + 2 × 1 ] = [ 5 8 , 5 8 ] = 5 8 .

Therefore we can say that, the extension principle solution X ¯ e exists and its components are continuous triangular shaped fuzzy numbers x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) , x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) and x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) .

The membership function of the triangularly shaped number x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) , is

x = ( − 1 + 2 α ) ( 3 − α ) ⇒ μ 1 L ( x ) = α = ( 1 + 3 x ) ( 2 + x ) , for − 1 3 ≤ x ≤ 1 2 ,

and

x = 2 − α 1 + α ⇒ μ 1 U ( x ) = α = ( 2 − x ) ( 1 + x ) , for 1 2 ≤ x ≤ 2 .

Thus the membership function of x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) is

μ x ¯ 1 ( x ) = { ( 1 + 3 x ) ( 2 + x ) , for − 1 3 ≤ x ≤ 1 2 ( 2 − x ) ( 1 + x ) , for 1 2 ≤ x ≤ 2 0 , otherwise (45)

and its graph is shown in

The membership function of the triangularly shaped number x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) is

x = ( 1 + α ) ( 5 − α ) ⇒ μ 2 L ( x ) = α = ( − 1 + 5 x ) ( 1 + x ) , for 1 5 ≤ x ≤ 1 2 ,

and

x = ( 3 − α ) ( 3 + α ) ⇒ μ 2 U ( x ) = α = ( 3 − 3 x ) ( 1 + x ) , for 1 2 ≤ x ≤ 1 .

Thus the membership function of x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) is

μ x ¯ 2 ( x ) = { ( − 1 + 5 x ) ( 1 + x ) , for 1 5 ≤ x ≤ 1 2 ( 3 − 3 x ) ( 1 + x ) , for 1 2 ≤ x ≤ 1 0 , otherwise (46)

and its graph is shown in

And the membership function of the triangularly shaped number x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) is

x = 2 + 3 α 12 − 4 α ⇒ μ 3 L ( x ) = α = ( − 2 + 12 x ) ( 3 + 4 x ) , for 1 6 ≤ x ≤ 5 8

and

x = 8 − 3 α 6 + 2 α ⇒ μ 3 U ( x ) = α = ( 8 − 6 x ) ( 3 + 2 x ) , for 5 8 ≤ x ≤ 4 3 .

Thus the membership function of x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) is

μ x ¯ 3 ( x ) = { ( − 2 + 12 x ) ( 3 + 4 x ) , for 1 6 ≤ x ≤ 5 8 ( 8 − 6 x ) ( 3 + 2 x ) , for 5 8 ≤ x ≤ 4 3 0 , otherwise (47)

and its graph is shown in

Finally the graph of the extension principle solution

X ¯ e = [ x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) ]

is shown in

Consider the system of fuzzy linear equation in matrix form

[ ( 1 / 2 / 3 ) 0 0 0 ( 3 / 4 / 5 ) 0 0 0 ( 6 / 8 / 12 ) ] [ x ¯ 1 x ¯ 2 x ¯ 3 ] = [ ( − 1 / 1 / 2 ) ( 1 / 2 / 3 ) ( 2 / 5 / 8 ) ]

We solve this fuzzy matrix equation using α-cut and interval arithmetic.

Here, a ¯ 11 = ( 1 / 2 / 3 ) , a ¯ 12 = 0 , a ¯ 13 = 0 , a ¯ 21 = 0 , a ¯ 22 = ( 3 / 4 / 5 ) , a ¯ 23 = 0 , a ¯ 31 = 0 , a ¯ 32 = 0 and a ¯ 33 = ( 6 / 8 / 12 ) . Also we have, b ¯ 1 = ( − 1 / 1 / 2 ) , b ¯ 2 = ( 1 / 2 / 3 ) and b ¯ 3 = ( 2 / 5 / 8 ) .

Then the α-cuts are:

a ¯ 11 [ α ] = [ a 11 L ( α ) , a 11 U ( α ) ] = [ 1 + α , 3 − α ] ,

a ¯ 22 [ α ] = [ a 22 L ( α ) , a 22 U ( α ) ] = [ 3 + α , 5 − α ] ,

a ¯ 33 [ α ] = [ a 33 L ( α ) , a 33 U ( α ) ] = [ 6 + 2 α , 12 − 4 α ] ,

b ¯ 1 [ α ] = [ b 1 L ( α ) , b 1 U ( α ) ] = [ − 1 + 2 α , 2 − α ] ,

b ¯ 2 [ α ] = [ b 2 L ( α ) , a 2 U ( α ) ] = [ 1 + α , 3 − α ] ,

b ¯ 3 [ α ] = [ b 3 L ( α ) , a 3 U ( α ) ] = [ 2 + 3 α , 8 − 3 α ] , ∀ α ∈ [ 0 , 1 ] .

Now the crisp solutions are

x 1 = b 1 ( a 22 a 33 − a 23 a 32 ) − a 12 ( a 33 b 2 − a 23 b 3 ) + a 13 ( a 32 b 2 − a 22 b 3 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) = a 22 a 33 b 1 a 11 a 22 a 33 = b 1 a 11 (48)

x 2 = a 11 ( a 33 b 2 − a 23 b 3 ) − b 1 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 b 3 − a 31 b 2 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) = a 11 a 33 b 2 a 11 a 22 a 33 = b 2 a 22 (49)

x 3 = a 11 ( a 22 b 3 − a 32 b 2 ) − a 12 ( a 21 b 3 − a 31 b 2 ) + b 1 ( a 21 a 32 − a 31 a 22 ) a 11 ( a 22 a 33 − a 32 a 23 ) − a 12 ( a 21 a 33 − a 31 a 23 ) + a 13 ( a 21 a 32 − a 31 a 22 ) = a 11 a 22 b 3 a 11 a 22 a 33 = b 3 a 33 (50)

We replace a 11 , a 22 , a 33 , b 1 , b 2 , b 3 by a ¯ 11 , a ¯ 22 , a ¯ 33 , b ¯ 1 , b ¯ 2 , b ¯ 3 respectively in Equations (48)-(50).

Then,

x ¯ 1 [ α ] = b ¯ 1 [ α ] a ¯ 11 [ α ] = [ b 1 L ( α ) , b 1 U ( α ) ] [ a 11 L ( α ) , a 11 U ( α ) ] = [ b 1 L ( α ) a 11 U ( α ) , b 1 U ( α ) a 11 L ( α ) ] = [ − 1 + 2 α 3 − α , 2 − α 1 + α ]

Similarly,

x ¯ 2 [ α ] = b ¯ 2 [ α ] a ¯ 22 [ α ] = [ b 2 L ( α ) , b 2 U ( α ) ] [ a 22 L ( α ) , a 22 U ( α ) ] = [ b 2 L ( α ) a 22 U ( α ) , b 2 U ( α ) a 22 L ( α ) ] = [ 1 + α 5 − α , 3 − α 3 + α ] .

Also,

x ¯ 3 [ α ] = b ¯ 3 [ α ] a ¯ 33 [ α ] = [ b 3 L ( α ) , b 3 U ( α ) ] [ a 33 L ( α ) , a 33 U ( α ) ] = [ b 3 L ( α ) a 33 U ( α ) , b 3 U ( α ) a 33 L ( α ) ] = [ 2 + 3 α 12 − 4 α , 8 − 3 α 6 + 2 α ] .

Here,

x 1 L ( α ) = − 1 + 2 α 3 − α , x 2 L ( α ) = 1 + α 5 − α , x 3 L ( α ) = 2 + 3 α 12 − 4 α , x 1 U ( α ) = 2 − α 1 + α , x 2 U ( α ) = 3 − α 3 + α , x 3 U ( α ) = 8 − 3 α 6 + 2 α .

We find that,

∂ ∂ α ( x 1 L ( α ) ) = 5 ( 3 − α ) 2 > 0 ; ∂ ∂ α ( x 2 L ( α ) ) = 6 ( 5 − α ) 2 > 0 ;

∂ ∂ α ( x 3 L ( α ) ) = 44 ( 12 − 4 α ) 2 > 0 ; ∂ ∂ α ( x 1 U ( α ) ) = − 3 ( 1 + α ) 2 < 0 ;

∂ ∂ α ( x 2 U ( α ) ) = − 6 ( 3 + α ) 2 < 0 ; ∂ ∂ α ( x 3 U ( α ) ) = − 34 ( 6 + 2 α ) 2 < 0 .

That is, x 1 L ( α ) , x 2 L ( α ) and x 3 L ( α ) are increasing functions of α ∈ [ 0 , 1 ] and x 1 U ( α ) , x 2 U ( α ) and x 3 U ( α ) are decreasing functions of α ∈ [ 0 , 1 ] . Also x 1 L ( 1 ) = x 1 U ( 1 ) , x 2 L ( 1 ) = x 2 U ( 1 ) and x 3 L ( 1 ) = x 3 U ( 1 ) .

Hence,

x ¯ 1 [ α ] = [ − 1 + 2 α 3 − α , 2 − α 1 + α ] , x ¯ 2 [ α ] = [ 1 + α 5 − α , 3 − α 3 + α ] and x ¯ 3 [ α ] = [ 2 + 3 α 12 − 4 α , 8 − 3 α 6 + 2 α ]

define the α-cuts of three fuzzy numbers respectively.

Now the support of x ¯ 1 is x ¯ 1 [ 0 ] = [ − 1 + 2 × 0 3 − 0 , 2 − 0 1 + 0 ] = [ − 1 3 , 2 ] and modal of x ¯ 1 is x ¯ 1 [ 1 ] = [ − 1 + 2 × 1 3 − 1 , 2 − 1 1 + 1 ] = [ 1 2 , 1 2 ] = 1 2 ;

The support of x ¯ 2 is x ¯ 2 [ 0 ] = [ 1 + 0 5 − 0 , 3 − 0 3 + 0 ] = [ 1 5 , 1 ] and modal of x ¯ 2 is x ¯ 2 [ 1 ] = [ 1 + 1 5 − 1 , 3 − 1 3 + 1 ] = [ 1 2 , 1 2 ] = 1 2 ; and

The support of x ¯ 3 is x ¯ 3 [ 0 ] = [ 2 + 3.0 12 − 4.0 , 8 − 3.0 6 + 2.0 ] = [ 1 6 , 4 3 ] and modal of x ¯ 3 is x ¯ 3 [ 1 ] = [ 2 + 3 × 1 12 − 4 × 1 , 8 − 3 × 1 6 + 2 × 1 ] = [ 5 8 , 5 8 ] = 5 8 .

Therefore we can say that, the α-cut and interval arithmetic solution X ¯ I exists and its components are continuous triangular shaped fuzzy numbers

x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) , x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) and x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) .

The membership function of the triangularly shaped number x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) is x = ( − 1 + 2 α ) ( 3 − α ) ⇒ μ 1 L ( x ) = α = ( 1 + 3 x ) ( 2 + x ) , for − 1 3 ≤ x ≤ 1 2 , and

x = 2 − α 1 + α ⇒ μ 1 U ( x ) = α = ( 2 − x ) ( 1 + x ) , for 1 2 ≤ x ≤ 2 .

Thus the membership function of x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) is

μ x ¯ 1 ( x ) = { ( 1 + 3 x ) ( 2 + x ) , for − 1 3 ≤ x ≤ 1 2 ( 2 − x ) ( 1 + x ) , for 1 2 ≤ x ≤ 2 0 , otherwise (51)

and its graph is shown in

The membership function of the triangular shaped number x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) is

x = ( 1 + α ) ( 5 − α ) ⇒ μ 2 L ( x ) = α = ( − 1 + 5 x ) ( 1 + x ) , for 1 5 ≤ x ≤ 1 2 , and

x = ( 3 − α ) ( 3 + α ) ⇒ μ 2 U ( x ) = α = ( 3 − 3 x ) ( 1 + x ) , for 1 2 ≤ x ≤ 1 .

Thus the membership function of x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) is

μ x ¯ 2 ( x ) = { ( − 1 + 5 x ) ( 1 + x ) , for 1 5 ≤ x ≤ 1 2 ( 3 − 3 x ) ( 1 + x ) , for 1 2 ≤ x ≤ 1 0 , otherwise (52)

and its graph is shown in

And the membership function of the triangularly shaped number x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) is x = 2 + 3 α 12 − 4 α ⇒ μ 3 L ( x ) = α = ( − 2 + 12 x ) ( 3 + 4 x ) , for 1 6 ≤ x ≤ 5 8 , and x = 8 − 3 α 6 + 2 α ⇒ μ 3 U ( x ) = α = ( 8 − 6 x ) ( 3 + 2 x ) , for 5 8 ≤ x ≤ 4 3 .

Thus the membership function of x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) is

μ x ¯ 3 ( x ) = { ( − 2 + 12 x ) ( 3 + 4 x ) , for 1 6 ≤ x ≤ 5 8 ( 8 − 6 x ) ( 3 + 2 x ) , for 5 8 ≤ x ≤ 4 3 0 , otherwise (53)

and its graph is shown in

Finally, the graph of the α-cut and interval arithmetic solution

X ¯ I = [ x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) ]

is shown in

Now we compare the classical solution X ¯ c , extension principle solution X ¯ e , and α-cut and interval arithmetic solution X ¯ I for the system

From the above discussion, we get the solutions as follows

X ¯ c = [ x ¯ 1 ≈ ( − 1 / 1 2 / 2 3 ) x ¯ 2 ≈ ( 1 3 / 1 2 / 3 5 ) x ¯ 3 ≈ ( 1 3 / 5 8 / 2 3 ) ] , X ¯ e = [ x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) ] and X ¯ I = [ x ¯ 1 ≈ ( − 1 3 / 1 2 / 2 ) x ¯ 2 ≈ ( 1 5 / 1 2 / 1 ) x ¯ 3 ≈ ( 1 6 / 5 8 / 4 3 ) ] .

Here we see that, extension principle solution X ¯ e and α-cut and interval arithmetic solution X ¯ I are equal. That is, X ¯ e = X ¯ I . Hence we can say, X ¯ c ≤ X ¯ e ≤ X ¯ I . The comparison among the classical solution X ¯ c , extension principle solution X ¯ e , and α-cut and interval arithmetic solution X ¯ I is shown in

The system of fuzzy linear equations undoubtedly plays a vital role in presently applied mathematics. Here our intention was to establish some models of solving that system and we presented three different methods of with their applications. We came to know by the above discussion that among the three models extension principle solution X ¯ e , and α-cut and interval arithmetic solution X ¯ I give the same results. In the graphical representation we, find that the extension principle solution X ¯ e , and α-cut and interval arithmetic solution X ¯ I meet at the same point but the classical solution X ¯ c is deviated a bit from the other two. Actually the solving techniques and their comparison are the ultimate findings of our work.

The authors declare no conflicts of interest regarding the publication of this paper.

Islam, S., Saiduzzaman, Md., Islam, Md.S. and Sultana, A. (2019) Comparison of Classical Method, Extension Principle and α-Cuts and Interval Arithmetic Method in Solving System of Fuzzy Linear Equations. American Journal of Computational Mathematics, 9, 1-24. https://doi.org/10.4236/ajcm.2019.91001