A fuzzy set [3] is a class of objects with a continuum of the grade of membership. Let X be a space of points. A fuzzy set A in X is characterized by a membership function which associates with each points in X a real number μ A ( x ) in the interval [ 0 , 1 ] with the value of μ A at x representing the grade of membership of x in A. Thus the nearer to the value of μ A to unity, the higher the grade of membership of x in A.

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 Figure 1.

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 Figure 2.

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 Figure 3.

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 Figure 4.

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 Figure 5.

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 Figure 6.

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 Figure 7.

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 Figure 8.

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 Figure 9.

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 Figure 10.

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 Figure 11.

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 Figure 12.