Methods for solving LR-bipolar fuzzy linear systems

Abstract

In this paper, we propose a technique to solve LR-bipolar fuzzy linear system(BFLS), LR-complex bipolar fuzzy linear (CBFL) system with real coefficients and LR-complex bipolar fuzzy linear (CBFL) system with complex coefficients of equations. Initially, we solve the LR-BFLS of equations using a pair of positive$(*)$ and negative$(\bullet )$ of two $n \times n$ LR-real linear systems by using mean values and left-right spread systems. We also provide the necessary and sufficient conditions for the solution of LR-BFLS of equations. We illustrate the method by using some numerical examples of symmetric and asymmetric LR-BFLS equations and obtain the strong and weak solutions to the systems. Further, we solve the LR-CBFL system of equations with real coefficients and LR-CBFL system of equations with complex coefficients by pair of positive$(*)$ and negative$(\bullet )$ two $n \times n$ real and complex LR-bipolar fuzzy linear systems by using mean values and left-right spread systems. Finally, we show the usage of technique to solve the current flow circuit which is represented by LR-CBFL system with complex coefficients and obtain the unknown current in term of LR-bipolar fuzzy complex number.

1 Introduction

The concepts of fuzzy sets were introduced by Zadeh (1965, 1971, 1975). Fuzzy set theory is a useful mathematical model for computing the uncertainty and vagueness. Dubois and Prade (1978) discussed the basic arithmetic operations of fuzzy numbers. Zhang (1998) introduced the idea of the term Yin and Yang which were based on double-sided or bipolar judgmental thinking on a positive and negative side. Zhang (1994) introduced the concept of bipolar fuzzy set in 1994 and bipolar fuzzy number in 1996 which was the extension of fuzzy set. Akram (2011) introduced the concept of bipolar fuzzy graph in 2011. Alghamdi et al. (2018) considered the multi-criteria decision-making techniques in bipolar fuzzy environment. The term simultaneous linear equations play vital role in different kinds of fields including mathematics, physics, networking, circuit analysis, economic model and attribute decision making. Fuzzy linear system (FLS) is a powerful tool to measure the fuzziness and uncertainty. In many places, a bipolar fuzzy linear system (BFLS) is used for linear optimization of system when system is represented in bipolar form. Complex fuzzy linear system is specially used in circuit analysis to measure unknown current in complex form. First the solution of FLS was examined by Friedman et al. (1998), and they used the embedding method to solve the system. They used the technique in which $n \times n$ FLS of equations whose coefficient matrix is a real number matrix and right side column vector is fuzzy number vector containing parameter r is replaced by $2n \times 2n$ real linear system. Abbasbandy and Alavi (2005) also used another method to solve $n \times n$ FLS of equations for which coefficient matrix is a real number matrix and right side column vector is fuzzy number vector containing parameter r. They used a procedure in which $n \times n$ FLS of equations is replaced by two $n \times n$ real linear systems. Allahviranloo (2019) proposed different methods to solve uncertain linear system of equations. Some numerical techniques, to solve FLS of equations, for example, Jacobi, Gauss-Seidel, SOR iterative and steepest descent method, are given in Allahviranloo (2005a, 2004, 2005b) and Abbasbandy and Jafarian (2006). Some other numerical techniques are proposed by Akram et al. (2019b, 2019c, 2019d) to solve BFLS equations, fully bipolar fuzzy linear system FBFLS of equations and linear system of equations in m-polar fuzzy environment. In many cases, general fuzzy linear system having unknown parameter in terms of parametric form of fuzzy number containing parameter r and to solve this system the system was extended into real linear system. It is not an easy task to solve the extended fuzzy linear system containing parameter r, $0 \le r \le 1$, which makes their calculation inconvenient in some sense. To make the multiplication easy, Dubois and Prade (1978) introduced the concept of LR fuzzy number. Dehghan et al. (2007) proposed a method to solve the fully fuzzy linear system $\widetilde{A}\widetilde{x}=\widetilde{b}$ whose coefficients matrix and right-side vector are LR fuzzy numbers. Allahviranloo et al. (2013) also discussed a method to solve LR fuzzy linear systems based on a 1-level expansion. The concept of fuzzy complex set and fuzzy complex number was introduced by Buckley (1989). Rahgooy et al. (2009) studied the $n \times n$ complex fuzzy linear systems with complex coefficients and proposed a method to solve complex fuzzy linear system; it was applied on problem of circuit analysis (CA) to find the current in terms of complex fuzzy number. Jahantigh et al. (2010) studied the $n \times n$ complex fuzzy linear system and discussed the general technique to solve the system. Behera and Chakraverty (2012) also proposed a numerical method to solve FLS of equations and CFLS system with complex coefficients by using fuzzy center and width. Ghanbari et al. (2020) discussed the graph of generalized fuzzy complex number and find algebraic and general solutions of rectangular fuzzy complex linear system.

In this article, we present a technique to solve $n \times n$ LR-BFLS of equations, LR-CBFL system with real coefficients and LR-CBFL system with complex coefficients for which coefficient matrix is a real and complex number and right-hand side column vector is LR-bipolar fuzzy number (BFN) and LR-bipolar fuzzy complex number (BFCN), respectively. Initially, we propose a method to solve LR-BFLS of equations we replace $n \times n$ LR-BFLS of equations by pair of positive$(*)$ and negative$(\bullet )$ of $n \times n$ LR-real linear system and then solve the system by using mean value and left-right spread system which are free from parameters r and s. Some numerical examples are also discuss to describe the efficiency of the technique. In the next section, we discuss the solution of two different systems: one is LR-CBFL system with real coefficients and the other is LR-CBFL system with complex coefficients. First, we propose a method to solve LR-CBFL system with real coefficients we replace $n \times n$ LR-CBFL system with real coefficients by pair of real and imaginary parts of $n \times n$ LR-BFLS of equation and then solve this system by using mean value and left-right spread system. Further, we extent the technique of Guo and Zhang (2016) to solve the LR-CBFL system with complex coefficients in which coefficient matrix is complex number and right side column vector is a LR-BFCN. In this method, we can replace $n \times n$ LR-CBFL system with complex coefficients by pair of positive$(*)$ and negative$(\bullet )$ of two $2n \times 2n$ mean value system and $4n \times 4n$ left-right spread system. We utilize the current flow circuit system given in Rahgooy et al. (2009) which is represented by LR-CBFL system with complex coefficients to find unknown current in the circuit in terms of LR-BFCN. In many real life problems, we deal with uncertainty of two kinds, one for the positivity and other for the negativity. Bipolar fuzzy numbers and more generally bipolar fuzzy system of linear equations are used to handle such problems. In present paper, LR-bipolar fuzzy systems of linear equations and bipolar fuzzy complex system are studied. The study of LR-bipolar fuzzy systems deal the problems that have uncertainty of two faces: certain property and its counter property. In addition, parametric BFN and BFCN, containing parameters r and s are replaced by LR-BFN and LR-BFCN for efficient mathematical calculations. For other notions and applications, readers are referred to Akram et al. (2021, 2020a, 2020b), Koam et al. (2020), Saqib et al. (2020a, 2020b), Akram et al. (2019a), Allahviranloo et al. (2008, 2014), Amirfakhrian (2012), Amirfakhrian et al. (2018), Ezzati (2011), Fariborzi Araghi and Fallahzadeh (2013), Moloudzadeh et al. (2013), Zheng and Wang (2006) and Rao et al. (2020a, 2020b).

2 Preliminaries

In this section, we describe some basic definitions like LR-bipolar fuzzy number (BFN), LR-bipolar fuzzy linear system, LR-bipolar fuzzy complex number (BFCN) and LR-complex bipolar fuzzy linear (CBFL) system of equations.

Definition 1

Abbasbandy and Alavi (2005) A fuzzy number is a fuzzy set $\nu : \mathfrak {R}\rightarrow [0,1]$ which satisfies:

• 1. $\nu$ is upper semi continuous.

• 2. $\nu (x)=0$ outside some interval [a,d].

• 3. There are real numbers $b,c : a \le b \le c \le d$ for which

• (a) $\nu (x)$ is monotonic non-decreasing on [a,b],

• (b) $\nu$ is monotonic non-increasing on [c,d],

• (c) $\nu (x) = 1$, $b \le x \le c.$

The membership function $\nu (x)$ can be represented as:

$$\begin{aligned} \nu (x)= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} 0, &{} \mathrm{if}~ x \le a,\\ h(x), &{} \mathrm{if}~ a \le x \le b,\\ 1, &{} \mathrm{if}~ b \le x \le c,\\ g(x), &{} \mathrm{if}~ c \le x \le d,\\ 0, &{} \mathrm{if}~ x \ge d, \end{array} \end{array}\right. }\end{aligned}$$

where h(x) is non-decreasing function, called left membership function, and g(x) is non-increasing function, called the right membership function.

The set of all fuzzy numbers on $\mathfrak {R}$ is denoted by $E^{1}$.

Definition 2

Guo and Zhang (2016) A fuzzy number $\widetilde{H}$ is said to be a LR fuzzy number if there are real numbers $h,~ \mu > 0$ and $\nu > 0$ such that

$$\begin{aligned} \sigma _{\widetilde{H}}(x)= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} L\left( \frac{h-x}{\mu }\right) , &{} \mathrm{if}~ x \le h,~~ \mu> 0,\\ R\left( \frac{x-h}{\nu }\right) , &{} \mathrm{if}~ x \ge h,~~ \nu > 0, \end{array} \end{array}\right. }\end{aligned}$$

where h, $\mu$ and $\nu$ are called the mean value and left and right spreads of $\widetilde{H}$, respectively.

The LR fuzzy number can also be represented as

$$\begin{aligned}\widetilde{H}=(h,\mu ,\nu )_{LR}.\end{aligned}$$

A real number a can be denoted by $a=(a,0,0)_{LR}$.

Let L and R both be decreasing functions from positive real numbers $\mathfrak {R}^{+}$ to the interval [0,1] such that the following conditions:

• 1. $L(x)< 1$, for $x > 0$,

• 2. $L(x)> 0$, for $x < 1$,

• 3. $L(0)=1$, $L(1)=0$.

If the two functions L and R are in the form

$$\begin{aligned}T(x)= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} 1-x, &{} \mathrm{if}~ 0 \le x \le 1,\\ 0, &{} \mathrm{Otherwise}, \end{array} \end{array}\right. }\end{aligned}$$

then the LR fuzzy number is called triangular fuzzy number. $\widetilde{H}=(h,\mu ,\nu )_{LR}$ is called symmetric fuzzy number if and only if $\mu =\nu$.

Clearly, two LR fuzzy numbers $\widetilde{H}=(h,\mu ,\nu )_{LR}$ and $\widetilde{G}=(g,\acute{\mu },\acute{\nu })_{LR}$ are said to be equal if and only if $h=g$, $\mu =\acute{\mu }$ and $\nu =\acute{\nu }$. Also, $\widetilde{H}=(h,\mu ,\nu )_{LR}$ is called positive (negative) if and only if $h-\mu > 0(h+\nu <0)$.

Definition 3

Guo and Zhang (2016) For arbitrary two LR fuzzy numbers $\widetilde{H}=(h,\mu ,\nu )_{LR}$ and $\widetilde{G}=(g,\alpha ,\beta )_{LR}$ and c is arbitrary scalar number, we define addition, subtraction, product and scalar multiplication by c as follows:

• (i) Addition:

  $$\begin{aligned} \widetilde{H} \oplus \widetilde{G}= & {} (h,\mu ,\nu )_{LR} \oplus (g,\alpha ,\beta )_{LR},\\= & {} (h+g,\mu + \alpha , \nu + \beta )_{LR},\end{aligned}$$

• (ii) Subtraction:

  $$\begin{aligned} \widetilde{H} - \widetilde{G}= & {} (h,\mu ,\nu )_{LR} - (g,\alpha ,\beta )_{LR},\\= & {} (h-g,\mu - \beta , \nu - \alpha )_{LR},\end{aligned}$$

• (iii) Multiplication:

  $If \widetilde{H} > 0$ and $\widetilde{G}> 0,$ then

  $$\begin{aligned} \widetilde{H} \otimes \widetilde{G}= & {} (h,\mu ,\nu )_{LR} \otimes (g,\alpha ,\beta )_{LR},\\\cong & {} (hg,h\alpha +g\mu ,h\beta +g\nu )_{LR}, \end{aligned}$$

• (iv) Scalar multiplication:

  $$\begin{aligned} c \otimes \widetilde{H}&= c \otimes (h,\mu ,\nu )_{LR},\\&{\cong } {\left\{ \begin{array}{ll} (ch,c\mu ,c\nu )_{LR},&{} ~~ c\ge 0,\\ (ch,-c\nu ,-c\mu )_{LR},&{}~~c < 0. \end{array}\right. } \end{aligned}$$

Definition 4

Akram and Arshad (2019) A bipolar fuzzy number (BFN)

$$\begin{aligned} \xi = \prec M , N \succ = \prec [m_{1},m_{2},m_{3},m_{4}],[n_{1},n_{2},n_{3},n_{4}]\succ \end{aligned}$$

is a bipolar fuzzy set of the mapping $\sigma : R \longrightarrow [0,1]\times [-1,0]$, with satisfaction degree $\sigma _{M}$ and dissatisfaction degree $\sigma _{N}$ such that:

$$\begin{aligned} \sigma _{M}(\kappa )= & {} {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} \sigma _{M}^{L}(\kappa ), &{} \mathrm{if}~ \kappa ~ \in ~ [m_{1},m_{2}],\\ 1, &{} \mathrm{if}~ \kappa ~ \in ~ [m_{2},m_{3}],\\ \sigma _{M}^{R}(\kappa ), &{} \mathrm{if}~ \kappa ~ \in ~ [m_{3},m_{4}],\\ 0, &{} \mathrm{otherwise}, \end{array} \end{array}\right. }\\ \sigma _{N}(\kappa )= & {} {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} \sigma _{N}^{L}(\kappa ), &{} \mathrm{if}~ \kappa ~ \in ~ [n_{1},n_{2}],\\ -1, &{} \mathrm{if}~ \kappa ~ \in ~ [n_{2},n_{3}],\\ \sigma _{N}^{R}(\kappa ), &{} \mathrm{if}~ \kappa ~ \in ~ [n_{3},n_{4}],\\ 0, &{} \mathrm{otherwise}, \end{array} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \sigma _{M}^{L}(\kappa ):[m_{1},m_{2}]\rightarrow [0,1],~~\sigma _{M}^{R}(\kappa ):[m_{3},m_{4}]\rightarrow [0,1],\\ \sigma _{N}^{L}(\kappa ):[n_{1},n_{2}]\rightarrow [-1,0],~~\sigma _{N}^{R}(\kappa ):[n_{3},n_{4}]\rightarrow [-1,0], \end{aligned}$$

where $\sigma _{M}^{L}(\kappa )$ and $\sigma _{N}^{L}(\kappa )$ represent the left membership functions for $\sigma _{M}(\kappa )$ and $\sigma _{N}(\kappa )$, respectively. Similarly, $\sigma _{M}^{R}(\kappa )$ and $\sigma _{N}^{R}(\kappa )$ represent right membership functions for $\sigma _{M}(\kappa )$ and $\sigma _{N}(\kappa )$, respectively.

A BFN in a parametric form is given as $h=\prec [\underline{h}^{*}(r),\overline{h}^{*}(r)],[\underline{h}^{\bullet }(s),\overline{h}^{\bullet }(s)]\succ$ of the mappings $\underline{h}^{*}(r),\overline{h}^{*}(r),\underline{h}^{\bullet }(s)$ and $\overline{h}^{\bullet }(s)$ where $0 \le r \le 1,$ $-1 \le s \le 0$. A bipolar fuzzy number in parametric form always contains some parameters named r and s which create some difficulties to solve bipolar fuzzy systems. For the alleviation of this problem, we have introduced new class of bipolar fuzzy number which is called LR-bipolar fuzzy number in this form parameters r and s are eliminated.

Definition 5

A bipolar fuzzy number $\widetilde{H}$ is said to be a LR-bipolar fuzzy number of the form $\widetilde{H}=\prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ$, where $h^{*},h^{\bullet }$ are real numbers and $\mu _{l}^{*},\mu _{l}^{\bullet }> 0$ and $\nu _{r}^{*},\nu _{r}^{\bullet } > 0$ such that

$$\begin{aligned}\sigma _{\widetilde{H}^{*}}(\kappa )= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} L\left( \frac{h^{*}-\kappa }{\mu _{l}^{*}}\right) , &{} \mathrm{if}~ \kappa \le h^{*},~~ \mu _{l}^{*}> 0,\\ R\left( \frac{\kappa -h^{*}}{\nu _{r}^{*}}\right) , &{} \mathrm{if}~ \kappa \ge h^{*},~~ \nu _{r}^{*} > 0, \end{array} \end{array}\right. }\end{aligned}$$

where $h^{*}$, $\mu _{l}^{*}$ and $\nu _{r}^{*}$ are called the mean value and left and right spreads of positive side of $\widetilde{H}$, respectively. Also L and R both are decreasing functions from positive real numbers $\mathfrak {R}^{+}$ to the interval [0,1] such that the following conditions hold:

• (1) $L(\kappa )< 1$, for $\kappa > 0$,

• (2) $L(\kappa )> 0$, for $\kappa < 1$,

• (3) $L(0)=1$, $L(1)=0$.

and,

$$\begin{aligned} \sigma _{\widetilde{H}^{\bullet }}(\kappa )= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} L\left( \frac{h^{\bullet }-\kappa }{\mu _{l}^{\bullet }}\right) , &{} \mathrm{if}~ \kappa \le h^{\bullet },~~ \mu _{l}^{\bullet }> 0,\\ R\left( \frac{\kappa -h^{\bullet }}{\nu _{r}^{\bullet }}\right) , &{} \mathrm{if}~ \kappa \ge h^{\bullet },~~ \nu _{r}^{\bullet } > 0, \end{array} \end{array}\right. }\end{aligned}$$

where $h^{\bullet }$, $\mu _{l}^{\bullet }$ and $\nu _{r}^{\bullet }$ are called the mean value and left and right spreads of negative side of $\widetilde{H}$, respectively. Also, L and R both are decreasing functions from positive real numbers $\mathfrak {R}^{+}$ to the interval [$-1,0$] such that the following conditions hold:

• (1) $L(\kappa )> -1$, for $\kappa < 0$,

• (2) $L(\kappa ) < 0$, for $\kappa > -1$,

• (3) $L(0)=-1$, $L(1)=0$.

If the two functions L and R for the positive$(*)$ part are in the form,

$$\begin{aligned} T(\kappa )= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} 1-\kappa , &{} \mathrm{if}~ 0 \le \kappa \le 1,\\ 0, &{} \mathrm{Otherwise}. \end{array} \end{array}\right. }\end{aligned}$$

for the negative$(\bullet )$ part

$$\begin{aligned} T(\kappa )= {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} -1-\kappa , &{} \mathrm{if}~ -1 \le \kappa \le 0,\\ 0, &{} \mathrm{Otherwise}, \end{array} \end{array}\right. }\end{aligned}$$

then the LR-bipolar fuzzy number (BFN) is called triangular LR-bipolar fuzzy number.

$\widetilde{H}=\prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ$ is called symmetric LR-bipolar fuzzy number if and only if $\mu _{l}^{*}=\nu _{r}^{*}$ and $\mu _{l}^{\bullet }=\nu _{r}^{\bullet }$.

For example, $\widetilde{H} =\prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ =\prec [7,2,3]_{LR},[5,3,2]_{LR}\succ$ is LR-bipolar triangular fuzzy number and $\widetilde{G} =\prec [g^{*},\alpha _{l}^{*},\beta _{r}^{*}]_{LR},[g^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet }]_{LR}\succ =\prec [5,2,2]_{LR},[1,2,2]_{LR}\succ$ is a symmetric LR-bipolar triangular fuzzy number.

Definition 6

For arbitrary two LR-bipolar fuzzy numbers $\widetilde{H}=\prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ$ and $\widetilde{G}=\prec [g^{*},\alpha _{l}^{*},\beta _{r}^{*}]_{LR},[g^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet }]_{LR}\succ$ and c is arbitrary scalar number, we define addition, subtraction, product and scalar multiplication by c as follows:

• (i) Addition:

  $$\begin{aligned} \widetilde{H} \oplus \widetilde{G}= & {} \prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ \\&\oplus \prec [g^{*},\alpha _{l}^{*},\beta _{r}^{*}]_{LR},[g^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet }]_{LR}\succ ,\\= & {} \prec [h^{*}+g^{*},\mu _{l}^{*}+\alpha _{l}^{*},\nu _{r}^{*}+\beta _{r}^{*}]_{LR},\\&[h^{\bullet }+g^{\bullet },\mu _{l}^{\bullet }+\alpha _{l}^{\bullet },\nu _{r}^{\bullet }+\beta _{r}^{\bullet }]_{LR}\succ , \end{aligned}$$

• (ii) Subtraction:

  $$\begin{aligned} \widetilde{H} - \widetilde{G}= & {} \prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ \\&- \prec [g^{*},\alpha _{l}^{*},\beta _{r}^{*}]_{LR},[g^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet }]_{LR}\succ ,\\= & {} \prec [h^{*}-g^{*},\mu _{l}^{*}-\beta _{r}^{*},\nu _{r}^{*}-\alpha _{l}^{*}]_{LR},\\&[h^{\bullet }-g^{\bullet },\mu _{l}^{\bullet }-\beta _{r}^{\bullet },\nu _{r}^{\bullet }-\alpha _{l}^{\bullet }]_{LR}\succ ,\end{aligned}$$

• (iii) Multiplication:

  $If \widetilde{H} > 0$ and $\widetilde{G}> 0,$ then

  $$\begin{aligned} \widetilde{H} \otimes \widetilde{G}= & {} \prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ \\&\otimes \prec [g^{*},\alpha _{l}^{*},\beta _{r}^{*}]_{LR},[g^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet }]_{LR}\succ ,\\\cong & {} \prec [h^{*}g^{*},h^{*}\alpha _{l}^{*}+g^{*}\mu _{l}^{*},h^{*}\beta _{r}^{*}+g^{*}\nu _{r}^{*}]_{LR},\\&[h^{\bullet }g^{\bullet },h^{\bullet }\alpha _{l}^{\bullet }+g^{\bullet }\mu _{l}^{\bullet },h^{\bullet }\beta _{r}^{\bullet }+g^{\bullet }\nu _{r}^{\bullet }]_{LR}\succ , \end{aligned}$$

• (iv) Scalar multiplication:

  $$\begin{aligned} c \otimes \widetilde{H}= & {} c \otimes \prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ ,\\\cong & {} {\left\{ \begin{array}{ll} \prec [ch^{*},c\mu _{l}^{*},c\nu _{r}^{*}]_{LR},[ch^{\bullet },c\mu _{l}^{\bullet },c\nu _{r}^{\bullet }]_{LR}\succ , &{}~~ c\ge 0,\\ \prec [ch^{*},-c\nu _{r}^{*},-c\mu _{l}^{*}]_{LR},[ch^{\bullet },-c\nu _{r}^{\bullet },-c\mu _{l}^{\bullet }]_{LR}\succ ,&{}~~c < 0. \end{array}\right. } \end{aligned}$$

Definition 7

Jha (2018) Let $\zeta$ be a complex set. Then, the fuzzy subset $\widetilde{Z}$ is called fuzzy complex set of $\zeta$ is defined by the membership function $\mu _{\widetilde{z}}: \zeta \rightarrow [0,1]$.

Definition 8

Buckley (1989) A fuzzy complex number(FCN) $\widetilde{H}$ is a fuzzy complex set with its membership function $\pounds (h|\widetilde{H})$ is a mapping $\pounds : C \rightarrow [0,1]$, which may be represented as $\widetilde{H}= u + iv$, where $u=(\underline{u}(r),\overline{u}(r))$ and $v=(\underline{v}(r),\overline{v}(r))$, for all $0 \le r \le 1$.

Definition 9

Let $\zeta$ be a complex set. Then, the bipolar fuzzy subset $\widetilde{Z}$ is called bipolar fuzzy complex set of $\zeta$ is defined by the membership function $\mu _{\widetilde{z}}: \zeta \rightarrow [-1,0]\times [0,1]$.

Definition 10

A bipolar fuzzy complex number (BFCN) $\widetilde{H}$ is bipolar fuzzy complex set with its membership function $\pounds (h|\widetilde{H})$ is a mapping $\pounds : C \rightarrow [-1,0]\times [0,1]$, which may be represented as $\widetilde{H}= u + iv$, where $u=\prec [\underline{u}^{*}(r),\overline{u}^{*}(r)],[\underline{u}^{\bullet }(s),\overline{u}^{\bullet }(s)]\succ$ and

$$\begin{aligned} v=\prec [\underline{v}^{*}(r),\overline{v}^{*}(r)],[\underline{v}^{\bullet }(s),\overline{v}^{\bullet }(s)]\succ , \end{aligned}$$

for all $0 \le r \le 1$ and $-1 \le s \le 0$.

Definition 11

For any two arbitrary BFCN $\widetilde{h}_{1}= u_{1}+iv_{1}$ and $\widetilde{h}_{2}=u_{2}+iv_{2}$ (where $u_{1}, u_{2}, v_{1}$ and $v_{2}$ are BFN), the bipolar fuzzy complex arithmetic for addition and multiplication is given as:

• (i) $\widetilde{h}_{1} + \widetilde{h}_{2}= (u_{1}+u_{2}) + i(v_{1}+v_{2})$,

• (ii) $\widetilde{h}_{1} \times \widetilde{h}_{2}= \{(u_{1} \times u_{2})-(v_{1} \times v_{2})\} + i \{(u_{1} \times v_{2})+(v_{1} \times u_{2})\}.$

Definition 12

An LR-BFCN should be represented as $\widetilde{h}=\widetilde{u}+i\widetilde{v}$, where $\widetilde{u}=\prec [u^{*},u_{l}^{*},u_{r}^{*}]_{LR},[u^{\bullet },u_{l}^{\bullet },u_{r}^{\bullet }]_{LR}\succ$ and $\widetilde{v}=\prec [v^{*},v_{l}^{*},v_{r}^{*}]_{LR},[v^{\bullet },v_{l}^{\bullet },v_{r}^{\bullet }]_{LR}\succ$. Further, $\widetilde{h}$ can be written as $\prec [u^{*},u_{l}^{*},u_{r}^{*}]_{LR},[u^{\bullet },u_{l}^{\bullet },u_{r}^{\bullet }]_{LR}\succ$+i $\prec [v^{*},v_{l}^{*},v_{r}^{*}]_{LR},[v^{\bullet },v_{l}^{\bullet },v_{r}^{\bullet }]_{LR}\succ$.

Definition 13

For any two arbitrary LR-BFCN $\widetilde{h}_{1}= \widetilde{u}_{1}+i\widetilde{v}_{1}$ and $\widetilde{h}_{2}=\widetilde{u}_{2}+i\widetilde{v}_{2}$ (where $\widetilde{u}_{1}, \widetilde{u}_{2}, \widetilde{v}_{1}$ and $\widetilde{v}_{2}$ are LR-BFCN), the LR bipolar fuzzy complex arithmetic for addition and multiplication is given as:

• (i) $\widetilde{h}_{1} + \widetilde{h}_{2}= (\widetilde{u}_{1}+\widetilde{u}_{2}) + i(\widetilde{v}_{1}+\widetilde{v}_{2})$,

• (ii) $c\widetilde{h}= c\widetilde{u}+ic\widetilde{v},$   $c~ \epsilon ~ \mathfrak {R},$

• (iii) $\widetilde{h}_{1} \times \widetilde{h}_{2}= \{(\widetilde{u}_{1} \times \widetilde{u}_{2})-(\widetilde{v}_{1} \times \widetilde{v}_{2})\} + i \{(\widetilde{u}_{1} \times \widetilde{v}_{2})+(\widetilde{v}_{1} \times \widetilde{u}_{2})\}.$

Definition 14

Consider the $n \times n$ Linear system of equations is given by:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}h_{1} + l_{12}h_{2} + \cdots + l_{1n}h_{n}&{}= z_{1},\\ l_{21}h_{1} + l_{22}h_{2} + \cdots + l_{2n}h_{n}&{}= z_{2},\\ .\\ .\\ .\\ l_{n1}h_{1} + l_{n2}h_{2} + \cdots + l_{nn}h_{n}&{}= z_{n},\\ \end{array}\right. } \end{aligned}$$

(1)

where the coefficient elements $(l_{pq}), 1 \le p,q \le n$ is a real matrix of order $n \times n$ and each $z_{q}, 1 \le q \le n$ is a LR form of bipolar fuzzy number and each unknown $h_{p}, 1 \le p \le n$ is also LR form of bipolar fuzzy number, is known as the LR-bipolar fuzzy linear system of equations (LR-BFLS).

The LR-BFLS of equations given in (1) can be written in matrix form as:

$$\begin{aligned} L\widetilde{H}=\widetilde{Z}, \end{aligned}$$

(2)

we can rewrite the above-mentioned LR-BFLS of Eqs. (2) as follows:

$$\begin{aligned}&L\prec [h^{*},\mu _{l}^{*},\nu _{r}^{*}]_{LR},[h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet }]_{LR}\succ \nonumber \\&\quad =\prec [z^{*},\alpha _{l}^{*},\beta _{r}^{*}]_{LR},[z^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet }]_{LR}\succ . \end{aligned}$$

(3)

We separate the positive$(*)$ and negative$(\bullet )$ parts of Eq. (3) to obtain two LR-real linear systems:

$$\begin{aligned}&L(h^{*},\mu _{l}^{*},\nu _{r}^{*})_{LR}=(z^{*},\alpha _{l}^{*},\beta _{r}^{*})_{LR},\nonumber \\&L(h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet })_{LR}=(z^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet })_{LR}. \end{aligned}$$

(4)

Definition 15

A LR-BFNs vector $H= (\widetilde{h}_{1},\widetilde{h}_{2},\ldots .\widetilde{h}_{n})^{T}$ is the form of $H_{q}=\prec [h_{q}^{*},\mu _{lq}^{*},\nu _{rq}^{*}]_{LR},[h_{q}^{\bullet },\mu _{lq}^{\bullet },\nu _{rq}^{\bullet }]_{LR}\succ$ where $1 \le q \le n$ is called the solution of LR-BFLS of Eq. (1) if:

$$\begin{aligned} {\left\{ \begin{array}{ll} \sum \nolimits _{q=1}^{n}l_{pq}h_{q}^{*} = z_{p}^{*},\\ \sum \nolimits _{q=1}^{n}l_{pq}\mu _{lq}^{*} = \alpha _{lp}^{*} ,\\ \sum \nolimits _{q=1}^{n}l_{pq}\nu _{rq}^{*} = \beta _{rp}^{*}, \\ \sum \nolimits _{q=1}^{n}l_{pq}h_{q}^{\bullet } = z_{p}^{\bullet },\\ \sum \nolimits _{q=1}^{n}l_{pq}\mu _{lq}^{\bullet } = \alpha _{lp}^{\bullet } ,\\ \sum \nolimits _{q=1}^{n}l_{pq}\nu _{rq}^{\bullet } = \beta _{rp}^{\bullet }. \\ \end{array}\right. } \end{aligned}$$

(5)

Definition 16

For any LR-BFLS $L\widetilde{H}=\widetilde{Z}$, we define a matrix A obtained from non-negative entries of L and defined a matrix B obtained from absolute of the negative entries of L. Then, $L=A-B$ and we define $L^{+}=A+B.$

Definition 17

Abbasbandy and Alavi (2005) A square matrix is known as permutation matrix in which each row and column contain exactly one unit element and all other entries in each row and column are zero.

Definition 18

Abbasbandy and Alavi (2005) A matrix L is known as absolutely permutation matrix if absolute of the entries of matrix L, i.e., $(L^{+})$ is a permutation matrix.

Remark 1

$L^{^-1}$ is an absolutely permutation matrix if L is an absolutely permutation matrix and $LL^{T}= I$.

Theorem 1

Minc (1988) The inverse of non-negative matrix L is non-negative if L is a permutation matrix.

3 Solution procedure for n $\times$ n LR-BFLS of equations

To solve the LR-BFLS of equations in (1), first we separate the positive$(*)$ and negative$(\bullet )$ parts of system of equations to get two $n \times n$ LR-real linear systems given as

$$\begin{aligned} L(h^{*},\mu _{l}^{*},\nu _{r}^{*})_{LR}= & {} (z^{*},\alpha _{l}^{*},\beta _{r}^{*})_{LR},\end{aligned}$$

(6)

$$\begin{aligned} L(h^{\bullet },\mu _{l}^{\bullet },\nu _{r}^{\bullet })_{LR}= & {} (z^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet })_{LR}. \end{aligned}$$

(7)

To solve positive part of LR-BFLS of equations $L(h^{*},\mu _{l}^{*}, \nu _{r}^{*})_{LR}=(z^{*},\alpha _{l}^{*},\beta _{r}^{*})_{LR}$, we resolve this system into two real linear systems, one is $n \times n$ mean value system and other is $2n \times 2n$ left-right spread system.

The $n \times n$ mean value system for the positive part of (1) is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}h_{1}^{*} + l_{12}h_{2}^{*} + \cdots + l_{1n}h_{n}^{*}&{}= z_{1}^{*},\\ l_{21}h_{1}^{*} + l_{22}h_{2}^{*} + \cdots + l_{2n}h_{n}^{*}&{}= z_{2}^{*},\\ .\\ .\\ .\\ l_{n1}h_{1}^{*} + l_{n2}h_{2}^{*} + \cdots + l_{nn}h_{n}^{*}&{}= z_{n}^{*}.\\ \end{array}\right. } \end{aligned}$$

(8)

The matrix notation of system (8) is given as

$$\begin{aligned}LH_{q}^{*}=Z_{p}^{*},\end{aligned}$$

where $1 \le p,q \le n$.

To solve left-right spread system of Eq. (1), we must solve the $2n \times 2n$ real linear system and where the right side column is function vector $(\alpha _{l1}^{*},\alpha _{l2}^{*},\ldots ,\alpha _{ln}^{*},\beta _{r1}^{*},\beta _{r2}^{*},\ldots ,\beta _{rn}^{*})^{T}$. Now rearrange the left-right spread system, so the unknown parameters are $\mu _{lq}^{*},\nu _{rq}^{*}$, $1 \le q \le n$ and column of right side vector are

$$\begin{aligned} (\alpha _{l1}^{*},\alpha _{l2}^{*},\ldots ,\alpha _{ln}^{*},\beta _{r1}^{*},\beta _{r2}^{*},\ldots ,\beta _{rn}^{*})^{T}. \end{aligned}$$

We get first $2n \times 2n$ left-right spread system of positive$(*)$ part of LR-BFLS of equations,

${\left\{ \begin{array}{ll} t_{11}\mu _{l1}^{*}+t_{12}\mu _{l2}^{*}+ \cdots + t_{1n}\mu _{ln}^{*}+t_{1,n+1}(-\nu _{r1}^{*})+ t_{1,n+2}(-\nu _{r2}^{*})\\ \quad + \cdots +t_{1,2n}(-\nu _{rn}^{*}) = \alpha _{l1}^{*},\\ .\\ .\\ .\\ t_{n1}\mu _{l1}^{*}+t_{n2}\mu _{l2}^{*}+ \cdots + t_{nn}\mu _{ln}^{*}+t_{n,n+1}(-\nu _{r1}^{*})+ t_{n,n+2}(-\nu _{r2}^{*})\\ \quad + \cdots +t_{n,2n}(-\nu _{rn}^{*}) = \alpha _{ln}^{*},\\ t_{n+1,1}\mu _{l1}^{*}+t_{n+1,2}\mu _{l2}^{*}+ \cdots + t_{n+1,n}\mu _{ln}^{*}+t_{n+1,n+1}(-\nu _{r1}^{*})\\ \quad + t_{n+1,n+2}(-\nu _{r2}^{*})+ \cdots +t_{n+1,2n}(-\nu _{rn}^{*}) = \beta _{r1}^{*},\\ .\\ .\\ .\\ t_{2n,1}\mu _{l1}^{*}+t_{2n,2}\mu _{l2}^{*}+ \cdots + t_{2n,n}\mu _{ln}^{*}+t_{2n,n+1}(-\nu _{r1}^{*})+ t_{2n,n+2}\\ \quad (-\nu _{r2}^{*}) + \cdots +t_{2n,2n}(-\nu _{rn}^{*}) = \beta _{rn}^{*},\\ \end{array}\right. }$

where $l_{pq}$ are determined as follows:

$$\begin{aligned} l_{pq}&\ge 0 \Longrightarrow t_{pq}= l_{pq}, t_{p+n,q+n} = l_{pq},\nonumber \\ l_{pq}&< 0 \Longrightarrow t_{p,q+n}= -l_{pq}, t_{p+n,q}= -l_{pq}, \end{aligned}$$

(9)

where $1 \le p,q \le 2n$ and any $t_{pq}$ which is not determined from Eq. (9) is zero.

Using the matrix notation, we obtain

$$\begin{aligned} M(\mu _{l}^{*},\nu _{r}^{*})=(\alpha _{l}^{*},\beta _{r}^{*}), \end{aligned}$$

(10)

where $M=l_{pq}$, $1 \le p,q \le 2n$

$$\begin{aligned}&(\mu _{l1}^{*},\mu _{l2}^{*},\ldots ,\mu _{ln}^{*},\nu _{r1}^{*},\nu _{r2}^{*},\ldots ,\nu _{rn}^{*})^{T} and~\\&\quad (\alpha _{l1}^{*},\alpha _{l2}^{*},\ldots ,\alpha _{ln}^{*},\beta _{r1}^{*},\beta _{r2}^{*},\ldots ,\beta _{rn}^{*})^{T}.\end{aligned}$$

We can also write the matrix form of Eq. (10) as:

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right) \left( \begin{array}{c} \mu _{l}^{*} \\ \nu _{r}^{*} \\ \end{array} \right) = \left( \begin{array}{c} \alpha _{l}^{*}\\ \beta _{r}^{*} \\ \end{array} \right) . \end{aligned}$$

(11)

In the similar way we can take the negative$(\bullet )$ part of the LR-BFLS of equations given in (1).

To solve the negative part of LR-BFLS of equation $L(h^{\bullet },\mu _{l}^{\bullet }, \nu _{r}^{\bullet })_{LR}=(z^{\bullet },\alpha _{l}^{\bullet },\beta _{r}^{\bullet })_{LR}$, we resolve this system into two real linear systems, one is $n \times n$ mean value system and other is $2n \times 2n$ left-right spread system for negative part.

The $n \times n$ mean value system for the negative part of (1) is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}h_{1}^{\bullet } + l_{12}h_{2}^{\bullet } + \cdots + l_{1n}h_{n}^{\bullet }&{}= z_{1}^{\bullet },\\ l_{21}h_{1}^{\bullet } + l_{22}h_{2}^{\bullet } + \cdots + l_{2n}h_{n}^{\bullet }&{}= z_{2}^{\bullet },\\ .\\ .\\ .\\ l_{n1}h_{1}^{\bullet } + l_{n2}h_{2}^{\bullet } + \cdots + l_{nn}h_{n}^{\bullet }&{}= z_{n}^{\bullet }.\\ \end{array}\right. } \end{aligned}$$

(12)

The matrix notation of system (12) is given as

$$\begin{aligned}LH_{q}^{\bullet }=Z_{p}^{\bullet },\end{aligned}$$

where $1 \le p,q \le n$.

To solve negative part of left-right spread system of Eq. (1), we must solve the $2n \times 2n$ real linear system and where the right side column is function vector $(\alpha _{l1}^{\bullet },\alpha _{l2}^{\bullet },\ldots ,\alpha _{ln}^{\bullet },\beta _{r1}^{\bullet }, \beta _{r2}^{\bullet },\ldots ,\beta _{rn}^{\bullet })^{T}$. Now rearrange the left-right spread system, so the unknown parameters are $\mu _{lq}^{\bullet },\nu _{rq}^{\bullet }$, $1 \le q \le n$ and column of right side vector are

$$\begin{aligned}(\alpha _{l1}^{\bullet },\alpha _{l2}^{\bullet },\ldots ,\alpha _{ln}^{\bullet },\beta _{r1}^{\bullet },\beta _{r2}^{\bullet },\ldots ,\beta _{rn}^{\bullet })^{T}.\end{aligned}$$

We get first $2n \times 2n$ left-right spread system of negative$(\bullet )$ part of LR-BFLS of equations,

${\left\{ \begin{array}{ll} t_{11}\mu _{l1}^{\bullet }+t_{12}\mu _{l2}^{\bullet }+ \cdots + t_{1n}\mu _{ln}^{\bullet }+t_{1,n+1}(-\nu _{r1}^{\bullet })+ t_{1,n+2}(-\nu _{r2}^{\bullet })\\ \quad + \cdots +t_{1,2n}(-\nu _{rn}^{\bullet }) = \alpha _{l1}^{\bullet },\\ .\\ .\\ .\\ t_{n1}\mu _{l1}^{\bullet }+t_{n2}\mu _{l2}^{\bullet }+ \cdots + t_{nn}\mu _{ln}^{\bullet }+t_{n,n+1}(-\nu _{r1}^{\bullet })+ t_{n,n+2}(-\nu _{r2}^{\bullet })\\ \quad + \cdots +t_{n,2n}(-\nu _{rn}^{\bullet }) = \alpha _{ln}^{\bullet },\\ t_{n+1,1}\mu _{l1}^{\bullet }+t_{n+1,2}\mu _{l2}^{\bullet }+ \cdots + t_{n+1,n}\mu _{ln}^{\bullet }+t_{n+1,n+1}(-\nu _{r1}^{\bullet })\\ \quad + t_{n+1,n+2}(-\nu _{r2}^{\bullet })+ \cdots +t_{n+1,2n}(-\nu _{rn}^{\bullet }) = \beta _{r1}^{\bullet },\\ .\\ .\\ .\\ t_{2n,1}\mu _{l1}^{\bullet }+t_{2n,2}\mu _{l2}^{\bullet }+ \cdots + t_{2n,n}\mu _{ln}^{\bullet }+t_{2n,n+1}(-\nu _{r1}^{\bullet })+ t_{2n,n+2}\\ \quad (-\nu _{r2}^{\bullet })+ \cdots +t_{2n,2n}(-\nu _{rn}^{\bullet }) = \beta _{rn}^{\bullet },\\ \end{array}\right. }$

where $l_{pq}$ are determined as follows:

$$\begin{aligned} l_{pq}&\ge 0 \Longrightarrow t_{pq}= l_{pq}, t_{p+n,q+n} = l_{pq},\nonumber \\ l_{pq}&< 0 \Longrightarrow t_{p,q+n}= -l_{pq}, t_{p+n,q}= -l_{pq}, \end{aligned}$$

(13)

where $1 \le p,q \le 2n$ and any $t_{pq}$ which is not determined from Eq. (13) is zero.

Using the matrix notation, we obtain

$$\begin{aligned} M(\mu _{l}^{\bullet },\nu _{r}^{\bullet })=(\alpha _{l}^{\bullet },\beta _{r}^{\bullet }), \end{aligned}$$

(14)

where $M=l_{pq}$, $1 \le p,q \le 2n$

$$\begin{aligned}&(\mu _{l1}^{\bullet },\mu _{l2}^{\bullet },\ldots ,\mu _{ln}^{\bullet },\nu _{r1}^{\bullet },\nu _{r2}^{\bullet },\ldots ,\nu _{rn}^{\bullet })^{T} and~\\&\quad (\alpha _{l1}^{\bullet },\alpha _{l2}^{\bullet },\ldots ,\alpha _{ln}^{\bullet },\beta _{r1}^{\bullet },\beta _{r2}^{\bullet },\ldots ,\beta _{rn}^{\bullet })^{T}.\end{aligned}$$

We can also write the matrix form of Eq. (14) as:

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right) \left( \begin{array}{c} \mu _{l}^{\bullet } \\ \nu _{r}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} \alpha _{l}^{\bullet }\\ \beta _{r}^{\bullet } \\ \end{array} \right) . \end{aligned}$$

(15)

To find the solution of LR-BFLS of Eq. (1), we can find the solution of Eqs. (8), (11), (12) and (15) by using matrix inverse method.

Theorem 2

For any arbitrary LR bipolar fuzzy vector Z the unique solution $\widetilde{H}$ is a LR bipolar fuzzy vector of LR-BFLS of equations in (1) if and only if the matrix M= $\left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right)$ and matrix L is non-singular, the necessary and sufficient condition that the matrices $(A + B)^{-1}$ and $(A - B)^{-1}$ exist.

Example 1

Consider the $2 \times 2$ non-symmetric LR-BFLS of equations

$$\begin{aligned} \widetilde{h}_{1} - \widetilde{h}_{2}= & {} \prec [1,1,1]_{LR},[-2,2,2]_{LR}\succ ,\\ \widetilde{h}_{1} + 3\widetilde{h}_{2}= & {} \prec [5,1,2]_{LR},[4,2,3]_{LR}\succ . \end{aligned}$$

To solve the above LR-BFLS equation first we put $\widetilde{h}_{1}=$ $\prec [h_{1}^{*}, \mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ$ and $\widetilde{h}_{2}= \prec [h_{2}^{*},$ $\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ$ in original system and get,

$$\begin{aligned}&\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad - \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [1,1,1]_{LR},[-2,2,2]_{LR}\succ ,\\&\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad + 3\prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [5,1,2]_{LR},[4,2,3]_{LR}\succ . \end{aligned}$$

To solve the LR-BFLS of equation in (1), first we take the positive$(*)$ part of the system (1) given as:

$$\begin{aligned}&(h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*})_{LR} - (h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*})_{LR}= (1,1,1)_{LR},\\&(h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*})_{LR} + 3 (h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*})_{LR}= (5,1,2)_{LR}.\end{aligned}$$

To solve this positive part of LR-BFLS of equations we can express this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for positive part is given as:

$$\begin{aligned}&h_{1}^{*} - h_{2}^{*}= 1,\\&h_{1}^{*} + 3 h_{2}^{*}= 5,\end{aligned}$$

by solving above equations, we have $h_{1}^{*}=2$ and $h_{2}^{*}=1$. Now the extended $4 \times 4$ form of positive part of left-right spread is given as:

$$\begin{aligned} 1(\mu _{l1}^{*}) + 0(\mu _{l2}^{*}) + 0(\nu _{r1}^{*}) + 1(\nu _{r2}^{*})&= 1,\\ 1(\mu _{l1}^{*}) + 3(\mu _{l2}^{*}) + 0(\nu _{r1}^{*}) + 0(\nu _{r2}^{*})&= 1,\\ 0(\mu _{l1}^{*}) + 1(\mu _{l2}^{*}) + 1(\nu _{r1}^{*}) + 0(\nu _{r2}^{*})&= 1,\\ 0(\mu _{l1}^{*}) + 0(\mu _{l2}^{*}) + 1(\nu _{r1}^{*}) + 3(\nu _{r2}^{*})&= 2.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 3 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 3 \\ \end{array} \right) \left( \begin{array}{c} \mu _{l1}^{*} \\ \mu _{l2}^{*} \\ \nu _{r1}^{*} \\ \nu _{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 1\\ 1 \\ 1 \\ 2 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned} M(\mu _{l}^{*},\nu _{r}^{*})=(\alpha _{l}^{*},\beta _{r}^{*}), \end{aligned}$$

and the solution of above-mentioned system is,

$\left( \begin{array}{c} \mu _{l1}^{*} \\ \mu _{l2}^{*} \\ \nu _{r1}^{*} \\ \nu _{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.1250 &{} -0.1250 &{} 0.3750 &{} -0.3750 \\ -0.3750 &{} 0.3750 &{} -0.1250 &{} 0.1250 \\ 0.3750 &{} -0.3750 &{} 1.1250 &{} -0.1250 \\ -0.1250 &{} 0.1250 &{} -0.3750 &{} 0.3750 \\ \end{array} \right)$

$\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 2 \\ \end{array} \right)$

$$\begin{aligned} \mu _{l1}^{*}= & {} 0.6250, ~~~~\mu _{l2}^{*}=0.1250,\\ \nu _{r1}^{*}= & {} 0.8750, ~~~~\nu _{r2}^{*}=0.3750. \end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of the system (1) given as:

$$\begin{aligned}&(h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet })_{LR} - (h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet })_{LR}= (-2,2,2)_{LR},\\&\quad (h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet })_{LR} + 3 (h_{2}^{*},\mu _{l2}^{\bullet },\nu _{r2}^{\bullet })_{LR}= (4,2,3)_{LR}.\end{aligned}$$

To solve this negative part of LR-BFLS of equations we can express this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for negative part is given as:

$$\begin{aligned}&h_{1}^{\bullet } - h_{2}^{\bullet }= -2,\\&h_{1}^{\bullet } + 3 h_{2}^{\bullet }= 4,\end{aligned}$$

by solving above equations, we have $h_{1}^{\bullet }=\frac{-1}{2}$ and $h_{2}^{\bullet }=\frac{3}{2}$. Now the extended $4 \times 4$ form of negative part of left-right spread is given as:

$$\begin{aligned} 1(\mu _{l1}^{\bullet }) + 0(\mu _{l2}^{\bullet }) + 0(\nu _{r1}^{\bullet }) + 1(\nu _{r2}^{\bullet })&= 2,\\ 1(\mu _{l1}^{\bullet }) + 3(\mu _{l2}^{\bullet }) + 0(\nu _{r1}^{\bullet }) + 0(\nu _{r2}^{\bullet })&= 2,\\ 0(\mu _{l1}^{\bullet }) + 1(\mu _{l2}^{\bullet }) + 1(\nu _{r1}^{\bullet }) + 0(\nu _{r2}^{\bullet })&= 2,\\ 0(\mu _{l1}^{\bullet }) + 0(\mu _{l2}^{\bullet }) + 1(\nu _{r1}^{\bullet }) + 3(\nu _{r2}^{\bullet })&= 3.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 3 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 3 \\ \end{array} \right) \left( \begin{array}{c} \mu _{l1}^{\bullet } \\ \mu _{l2}^{\bullet } \\ \nu _{r1}^{\bullet } \\ \nu _{r2}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} 2\\ 2 \\ 2 \\ 3 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned} M(\mu _{l}^{\bullet },\nu _{r}^{\bullet })=(\alpha _{l}^{\bullet },\beta _{r}^{\bullet }),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} \mu _{l1}^{\bullet } \\ \mu _{l2}^{\bullet } \\ \nu _{r1}^{\bullet } \\ \nu _{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.1250 &{} -0.1250 &{} 0.3750 &{} -0.3750 \\ -0.3750 &{} 0.3750 &{} -0.1250 &{} 0.1250 \\ 0.3750 &{} -0.3750 &{} 1.1250 &{} -0.1250 \\ -0.1250 &{} 0.1250 &{} -0.3750 &{} 0.3750 \\ \end{array} \right)$

$\left( \begin{array}{c} 2 \\ 2 \\ 2 \\ 3 \\ \end{array} \right)$

$$\begin{aligned} \mu _{l1}^{\bullet }= & {} 1.6250, ~~~~\mu _{l2}^{\bullet }=0.1250,\\ \nu _{r1}^{\bullet }= & {} 1.8750, ~~~~\nu _{r2}^{\bullet }=0.3750. \end{aligned}$$

Hence, the solution of LR-BFLS of equation in (1) is given as:

$$\begin{aligned}&\widetilde{h}_{1}=\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad =\prec [2,0.625,0.875]_{LR},\left[ \frac{-1}{2},1.6250,1.8750\right] _{LR}\succ ,\\&\widetilde{h}_{2}=\prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [1,0.1250,0.3750]_{LR},\left[ \frac{3}{2},0.1250,0.3750\right] _{LR}\succ .\end{aligned}$$

Example 2

Consider the $2 \times 2$ symmetric LR-BFLS of equations

$$\begin{aligned} \widetilde{h}_{1} - \widetilde{h}_{2}= & {} \prec [1,1,1]_{LR},[2,3,3]_{LR}\succ ,\\ \widetilde{h}_{1} + 3\widetilde{h}_{2}= & {} \prec [6,2,2]_{LR},[-2,7,7]_{LR}\succ .\end{aligned}$$

To solve the above LR-BFLS of equations first we put $\widetilde{h}_{1}=\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ$ and $\widetilde{h}_{2}= \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ$ in original system and get,

$$\begin{aligned}&\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad - \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [1,1,1]_{LR},[2,3,3]_{LR}\succ ,\\&\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad + 3\prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [6,2,2]_{LR},[-2,7,7]_{LR}\succ . \end{aligned}$$

To solve the LR-BFLS of equations in (2), first we take the positive$(*)$ part of the system (2) given as:

$$\begin{aligned}&(h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*})_{LR} - (h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*})_{LR}= (1,1,1)_{LR},\\&(h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*})_{LR} + 3 (h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*})_{LR}= (6,2,2)_{LR}.\end{aligned}$$

To solve this positive part of LR-BFLS of equation we can restructure this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for positive part is given as:

$$\begin{aligned} h_{1}^{*} - h_{2}^{*}= & {} 1,\\ h_{1}^{*} + 3 h_{2}^{*}= & {} 6,\end{aligned}$$

by solving above equations, we have $h_{1}^{*}=\frac{9}{4}$ and $h_{2}^{*}=\frac{5}{4}$. Now the extended $4 \times 4$ form of positive part of left-right spread is given as:

$$\begin{aligned} 1(\mu _{l1}^{*}) + 0(\mu _{l2}^{*}) + 0(\nu _{r1}^{*}) + 1(\nu _{r2}^{*})&= 1,\\ 1(\mu _{l1}^{*}) + 3(\mu _{l2}^{*}) + 0(\nu _{r1}^{*}) + 0(\nu _{r2}^{*})&= 2,\\ 0(\mu _{l1}^{*}) + 1(\mu _{l2}^{*}) + 1(\nu _{r1}^{*}) + 0(\nu _{r2}^{*})&= 1,\\ 0(\mu _{l1}^{*}) + 0(\mu _{l2}^{*}) + 1(\nu _{r1}^{*}) + 3(\nu _{r2}^{*})&= 2.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 3 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 3 \\ \end{array} \right) \left( \begin{array}{c} \mu _{l1}^{*} \\ \mu _{l2}^{*} \\ \nu _{r1}^{*} \\ \nu _{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 1\\ 2 \\ 1 \\ 2 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned}M(\mu _{l}^{*},\nu _{r}^{*})=(\alpha _{l}^{*},\beta _{r}^{*}),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} \mu _{l1}^{*} \\ \mu _{l2}^{*} \\ \nu _{r1}^{*} \\ \nu _{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.1250 &{} -0.1250 &{} 0.3750 &{} -0.3750 \\ -0.3750 &{} 0.3750 &{} -0.1250 &{} 0.1250 \\ 0.3750 &{} -0.3750 &{} 1.1250 &{} -0.1250 \\ -0.1250 &{} 0.1250 &{} -0.3750 &{} 0.3750 \\ \end{array} \right)$

$\left( \begin{array}{c} 1 \\ 2 \\ 1 \\ 2 \\ \end{array} \right)$

$$\begin{aligned}\mu _{l1}^{*}= 0.5, ~~~~\mu _{l2}^{*}=0.5,\\ \nu _{r1}^{*}= 0.5, ~~~~\nu _{r2}^{*}=0.5.\end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of the system (2) given as:

$$\begin{aligned}(h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet })_{LR} - (h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet })_{LR}= (2,3,3)_{LR},\\ (h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet })_{LR} + 3 (h_{2}^{*},\mu _{l2}^{\bullet },\nu _{r2}^{\bullet })_{LR}= (-2,7,7)_{LR}.\end{aligned}$$

To solve this negative part of LR-BFLS of equations given in (2) we can restructure this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for negative part is given as:

$$\begin{aligned}&h_{1}^{\bullet } - h_{2}^{\bullet }= 2,\\&h_{1}^{\bullet } + 3 h_{2}^{\bullet }= -2,\end{aligned}$$

by solving above equations, we have $h_{1}^{\bullet }=1$ and $h_{2}^{\bullet }=-1$. Now the extended $4 \times 4$ form of negative part of left-right spread is given as:

$$\begin{aligned} 1(\mu _{l1}^{\bullet }) + 0(\mu _{l2}^{\bullet }) + 0(\nu _{r1}^{\bullet }) + 1(\nu _{r2}^{\bullet })&= 3,\\ 1(\mu _{l1}^{\bullet }) + 3(\mu _{l2}^{\bullet }) + 0(\nu _{r1}^{\bullet }) + 0(\nu _{r2}^{\bullet })&= 7,\\ 0(\mu _{l1}^{\bullet }) + 1(\mu _{l2}^{\bullet }) + 1(\nu _{r1}^{\bullet }) + 0(\nu _{r2}^{\bullet })&= 3,\\ 0(\mu _{l1}^{\bullet }) + 0(\mu _{l2}^{\bullet }) + 1(\nu _{r1}^{\bullet }) + 3(\nu _{r2}^{\bullet })&= 7.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 3 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 3 \\ \end{array} \right) \left( \begin{array}{c} \mu _{l1}^{\bullet } \\ \mu _{l2}^{\bullet } \\ \nu _{r1}^{\bullet } \\ \nu _{r2}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} 3\\ 7 \\ 3 \\ 7 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned}M(\mu _{l}^{\bullet },\nu _{r}^{\bullet })=(\alpha _{l}^{\bullet },\beta _{r}^{\bullet }),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} \mu _{l1}^{\bullet } \\ \mu _{l2}^{\bullet } \\ \nu _{r1}^{\bullet } \\ \nu _{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.1250 &{} -0.1250 &{} 0.3750 &{} -0.3750 \\ -0.3750 &{} 0.3750 &{} -0.1250 &{} 0.1250 \\ 0.3750 &{} -0.3750 &{} 1.1250 &{} -0.1250 \\ -0.1250 &{} 0.1250 &{} -0.3750 &{} 0.3750 \\ \end{array} \right)$

$\left( \begin{array}{c} 3 \\ 7 \\ 3 \\ 7 \\ \end{array} \right)$

$$\begin{aligned} \mu _{l1}^{\bullet }= & {} 1, ~~~~\mu _{l2}^{\bullet }=2,\\ \nu _{r1}^{\bullet }= & {} 1, ~~~~\nu _{r2}^{\bullet }=2.\end{aligned}$$

Hence, the solution of LR-BFLS of equation in (2) is symmetric given as:

$$\begin{aligned} \widetilde{h}_{1}= & {} \prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\= & {} \prec \left[ \frac{9}{4},0.5,0.5\right] _{LR},[1,1,1]_{LR}\succ ,\\ \widetilde{h}_{2}= & {} \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\= & {} \prec \left[ \frac{5}{4},0.5,0.5\right] _{LR},[-1,2,2]_{LR}\succ .\end{aligned}$$

3.1 Weak LR-Bipolar Fuzzy Solution

We restrict our discussion to LR-bipolar fuzzy triangular number, i.e., if $\widetilde{h}_{p}=\prec [h_{p}^{*},\mu _{lp}^{*},\nu _{rp}^{*}]_{LR},[h_{p}^{\bullet },\mu _{lp}^{\bullet },\nu _{rp}^{\bullet }]_{LR}\succ$ is the unique solution of (1) such that $\mu _{lp}^{*} > 0$, $\nu _{rp}^{*} > 0$, $\mu _{lp}^{\bullet } > 0$ and $\nu _{rp}^{\bullet } > 0$ then $\widetilde{h}_{p}=\prec [h_{p}^{*},\mu _{lp}^{*},\nu _{rp}^{*}]_{LR}, [h_{p}^{\bullet },\mu _{lp}^{\bullet }, \nu _{rp}^{\bullet }]_{LR}\succ$ is called the strong LR-bipolar fuzzy solution of (1). Meanwhile, if one of them from $\mu _{lp}^{*}, \nu _{rp}^{*}, \mu _{lp}^{\bullet }, \nu _{rp}^{\bullet }$ be negative, then is said to be a weak LR-bipolar fuzzy solution of system (1) so the strong solution is given as:

$$\begin{aligned}\widetilde{h}_{p}^{*}= & {} {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} (h_{p}^{*},\mu _{lp}^{*},\nu _{rp}^{*})_{LR}, &{} \mu _{lp}^{*}> 0,~~ \nu _{rp}^{*}> 0, \\ (h_{p}^{*},0,\max \{-\mu _{lp}^{*},\nu _{rp}^{*}\})_{LR}, &{} \mu _{lp}^{*}< 0,~~ \nu _{rp}^{*}> 0, \\ (h_{p}^{*},\max \{\mu _{lp}^{*},-\nu _{rp}^{*}\},0)_{LR}, &{} \mu _{lp}^{*}> 0,~~ \nu _{rp}^{*}< 0, \\ (h_{p}^{*},-\nu _{rp}^{*},-\mu _{lp}^{*})_{LR}, &{} \mu _{lp}^{*}< 0,~~ \nu _{rp}^{*}< 0, \end{array} \end{array}\right. }\\ \widetilde{h}_{p}^{\bullet }= & {} {\left\{ \begin{array}{ll} \begin{array}{l@{\quad }l} (h_{p}^{\bullet },\mu _{lp}^{\bullet },\nu _{rp}^{\bullet })_{LR}, &{} \mu _{lp}^{\bullet }> 0,~~ \nu _{rp}^{\bullet }> 0, \\ (h_{p}^{\bullet },0,\max \{-\mu _{lp}^{\bullet },\nu _{rp}^{\bullet }\})_{LR}, &{} \mu _{lp}^{\bullet }< 0,~~ \nu _{rp}^{\bullet }> 0, \\ (h_{p}^{\bullet },\max \{\mu _{lp}^{\bullet },-\nu _{rp}^{\bullet }\},0)_{LR}, &{} \mu _{lp}^{\bullet } > 0,~~ \nu _{rp}^{\bullet }< 0, \\ (h_{p}^{\bullet },-\nu _{rp}^{\bullet },-\mu _{lp}^{\bullet })_{LR}, &{} \mu _{lp}^{\bullet }< 0,~~ \nu _{rp}^{\bullet } < 0, \end{array} \end{array}\right. }\end{aligned}$$

where $p=1,2,3,\ldots ,n$.

Example 3

Consider the $2 \times 2$ symmetric LR-BFLS of equations

$$\begin{aligned} \widetilde{h}_{1} - \widetilde{h}_{2}= & {} \prec [-3,2,2]_{LR},[-5,1,1]_{LR}\succ ,\\ \widetilde{h}_{1} - 2\widetilde{h}_{2}= & {} \prec [7,5,5]_{LR},[12,5,5]_{LR}\succ . \end{aligned}$$

To solve the above LR-BFLS of equations first we put $\widetilde{h}_{1}=\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ$ and $\widetilde{h}_{2}= \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ$ in original system and get,

$$\begin{aligned}&\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad - \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [-3,2,2]_{LR},[-5,1,1]_{LR}\succ ,\\&\prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\&\quad - 2\prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [7,5,5]_{LR},[12,5,5]_{LR}\succ . \end{aligned}$$

To solve the LR-BFLS of equation in (3), first we take the positive$(*)$ part of the system (3) given as:

$$\begin{aligned}&(h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*})_{LR} - (h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*})_{LR}= (-3,2,2)_{LR},\\&(h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*})_{LR} - 2 (h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*})_{LR}= (7,5,5)_{LR}.\end{aligned}$$

To solve this positive part of LR-BFLS of equations we can divide this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for positive part is given as:

$$\begin{aligned}h_{1}^{*} - h_{2}^{*}= -3,\\ h_{1}^{*} - 2h_{2}^{*}= 7,\end{aligned}$$

by solving above mentioned equations, we have $h_{1}^{*}=-13$ and $h_{2}^{*}=-10$. Now the extended $4 \times 4$ form of positive part of left-right spread is given as:

$$\begin{aligned} 1(\mu _{l1}^{*}) + 0(\mu _{l2}^{*}) + 0(\nu _{r1}^{*}) + 1(\nu _{r2}^{*})&= 2,\\ 1(\mu _{l1}^{*}) + 0(\mu _{l2}^{*}) + 0(\nu _{r1}^{*}) + 2(\nu _{r2}^{*})&= 5,\\ 0(\mu _{l1}^{*}) + 1(\mu _{l2}^{*}) + 1(\nu _{r1}^{*}) + 0(\nu _{r2}^{*})&= 2,\\ 0(\mu _{l1}^{*}) + 2(\mu _{l2}^{*}) + 1(\nu _{r1}^{*}) + 0(\nu _{r2}^{*})&= 5.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 0 &{} 0 &{} 2 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 2 &{} 1 &{} 0 \\ \end{array} \right) \left( \begin{array}{c} \mu _{l1}^{*} \\ \mu _{l2}^{*} \\ \nu _{r1}^{*} \\ \nu _{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 2\\ 5 \\ 2 \\ 5 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned}M(\mu _{l}^{*},\nu _{r}^{*})=(\alpha _{l}^{*},\beta _{r}^{*}),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} \mu _{l1}^{*} \\ \mu _{l2}^{*} \\ \nu _{r1}^{*} \\ \nu _{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 2 &{} -1 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 &{} 1 \\ 0 &{} 0 &{} 2 &{} -1 \\ -1 &{} 1 &{} 0 &{} 0 \\ \end{array} \right) \left( \begin{array}{c} 2 \\ 5 \\ 2 \\ 5 \\ \end{array} \right)$

$$\begin{aligned}\mu _{l1}^{*}= -1, ~~~~\mu _{l2}^{*}=3,\\ \nu _{r1}^{*}= -1, ~~~~\nu _{r2}^{*}=3.\end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of the system (3) given as:

$$\begin{aligned}&(h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet })_{LR} - (h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet })_{LR}= (-5,1,1)_{LR},\\&(h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet })_{LR} - 2 (h_{2}^{*},\mu _{l2}^{\bullet },\nu _{r2}^{\bullet })_{LR}= (12,5,5)_{LR}.\end{aligned}$$

To solve this negative part we can divide this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for negative part is given as:

$$\begin{aligned}&h_{1}^{\bullet } - h_{2}^{\bullet }= -5,\\&h_{1}^{\bullet } - 2 h_{2}^{\bullet }= 12,\end{aligned}$$

by solving above equations, we have $h_{1}^{\bullet }=-22$ and $h_{2}^{\bullet }=-17$. Now the extended $4 \times 4$ form of negative part of left-right spread is given as:

$$\begin{aligned} 1(\mu _{l1}^{\bullet }) + 0(\mu _{l2}^{\bullet }) + 0(\nu _{r1}^{\bullet }) + 1(\nu _{r2}^{\bullet })&= 1,\\ 1(\mu _{l1}^{\bullet }) + 0(\mu _{l2}^{\bullet }) + 0(\nu _{r1}^{\bullet }) + 2(\nu _{r2}^{\bullet })&= 5,\\ 0(\mu _{l1}^{\bullet }) + 1(\mu _{l2}^{\bullet }) + 1(\nu _{r1}^{\bullet }) + 0(\nu _{r2}^{\bullet })&= 1,\\ 0(\mu _{l1}^{\bullet }) + 2(\mu _{l2}^{\bullet }) + 1(\nu _{r1}^{\bullet }) + 0(\nu _{r2}^{\bullet })&= 5.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 0 &{} 0 &{} 2 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 2 &{} 1 &{} 0 \\ \end{array} \right) \left( \begin{array}{c} \mu _{l1}^{\bullet } \\ \mu _{l2}^{\bullet } \\ \nu _{r1}^{\bullet } \\ \nu _{r2}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} 1\\ 5 \\ 1 \\ 5 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned}M(\mu _{l}^{\bullet },\nu _{r}^{\bullet })=(\alpha _{l}^{\bullet },\beta _{r}^{\bullet }),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} \mu _{l1}^{\bullet } \\ \mu _{l2}^{\bullet } \\ \nu _{r1}^{\bullet } \\ \nu _{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 2 &{} -1 &{} 0 &{} 0 \\ 0 &{} 0 &{} -1 &{} 1 \\ 0 &{} 0 &{} 2 &{} -1 \\ -1 &{} 1 &{} 0 &{} 0 \\ \end{array} \right) \left( \begin{array}{c} 1 \\ 5 \\ 1 \\ 5 \\ \end{array} \right)$

$$\begin{aligned}\mu _{l1}^{\bullet }= -3, ~~~~\mu _{l2}^{\bullet }=4,\\ \nu _{r1}^{\bullet }= -3, ~~~~\nu _{r2}^{\bullet }=4.\end{aligned}$$

Hence, the solution of LR-BFLS of equations in (3) is symmetric given as:

$$\begin{aligned} \widetilde{h}_{1}= & {} \prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\= & {} \prec [-13,-1,-1]_{LR},[-22,-3,-3]_{LR}\succ ,\\ \widetilde{h}_{2}= & {} \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\= & {} \prec [-10,3,3]_{LR},[-17,4,4]_{LR}\succ .\end{aligned}$$

It is clear to see the LR-bipolar fuzzy solution that $\mu _{l1}^{*}<0$, $\nu _{r1}^{*}<0$, $\mu _{l1}^{\bullet }<0$ and $\nu _{r1}^{\bullet }<0$, so that LR-bipolar fuzzy solution is a weak solution and now solution is given as:

$$\begin{aligned} \widetilde{h}_{1}= & {} \prec [h_{1}^{*},\mu _{l1}^{*},\nu _{r1}^{*}]_{LR},[h_{1}^{\bullet },\mu _{l1}^{\bullet },\nu _{r1}^{\bullet }]_{LR}\succ \\ {}= & {} \prec [-13,1,1]_{LR},[-22,3,3]_{LR}\succ ,\\ \widetilde{h}_{2}= & {} \prec [h_{2}^{*},\mu _{l2}^{*},\nu _{r2}^{*}]_{LR},[h_{2}^{\bullet },\mu _{l2}^{\bullet },\nu _{r2}^{\bullet }]_{LR}\succ \\ {}= & {} \prec [-10,3,3]_{LR},[-17,4,4]_{LR}\succ .\end{aligned}$$

4 Solution to LR complex bipolar fuzzy linear system

In this section, we develop two different methods to solve LR-CBFL system with real coefficients and LR-CBFL system with complex coefficients in which unknown parameter, right side vector are LR-BFCN and coefficients of unknown are real and complex number, respectively. In the first method, we suggest a technique to solve LR-CBFL system with real coefficients. First we compare real and imaginary parts of equation to get two LR-BFLS of equations and then solve these LR-BFLS of equations by using mean value and left-right spread linear systems. Guo and Zhang (2016) suggested a method to solve $n \times n$ LR-complex fuzzy linear system with complex coefficients in which coefficient matrix is a complex number matrix and unknown and right side vector are LR-complex fuzzy number. In the next method we extent the Xiaobin and Zhang technique to solve LR-CBFL system with complex coefficients in bipolar fuzzy environment. To solve the system we can replace $n \times n$ LR-CBFL system with complex coefficients by pair of positive$(*)$ and negative$(\bullet )$ of two $2n \times 2n$ mean value system and $4n \times 4n$ left-right spread system. We use this technique to solve the example of circuit flow and find the current which is calculated in terms of LR-BFCN.

Definition 19

Consider the $n \times n$ system of linear equations,

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}h_{1} + l_{12}h_{2} + \cdots + l_{1n}h_{n}&{}= z_{1},\\ l_{21}h_{1} + l_{22}h_{2} + \cdots + l_{2n}h_{n}&{}= z_{2},\\ .\\ .\\ .\\ l_{n1}h_{1} + l_{n2}h_{2} + \cdots + l_{nn}h_{n}&{}= z_{n},\\ \end{array}\right. } \end{aligned}$$

(16)

is the LR-complex bipolar fuzzy linear (CBFL) system of equation with real coefficients, if all the coefficient elements $l_{pq}$, $1 \le p,q \le n$ is a real number matrix of order $n \times n$ and each $Z_{q}$, $1 \le q \le n$ is a LR-BFCN, and each unknown parameter $h_{p}$, $1 \le p \le n$ is also LR-BFCN.

The CBFL system in (16) can be written in matrix form as:

$$\begin{aligned} L\widetilde{H}=\widetilde{Z}, \end{aligned}$$

(17)

where $\widetilde{H}$ and $\widetilde{Z}$ are LR-BFCN.

We can rewrite the above LR-CBFL system(17) as follows:

$$\begin{aligned} L(\widetilde{U}+i\widetilde{V}) = \widetilde{E}+i\widetilde{F}, \end{aligned}$$

(18)

where each $\widetilde{U}$, $\widetilde{V}$, $\widetilde{E}$ and $\widetilde{F}$ are LR-BFN.

If we compare the real and imaginary part of the given equation, then we obtain two LR-bipolar fuzzy linear systems given as:

$$\begin{aligned} L\widetilde{U}=\widetilde{E},~~~~~L\widetilde{V}=\widetilde{F}. \end{aligned}$$

(19)

Definition 20

A LR-BFCNs vector $H= (\widetilde{h}_{1},\widetilde{h}_{2},\ldots .\widetilde{h}_{n})^{T}$ is the form of $H_{q}=\prec [u_{q}^{*},u_{lq}^{*},u_{rq}^{*}]_{LR},[u_{q}^{\bullet },u_{lq}^{\bullet },u_{rq}^{\bullet }]_{LR}\succ + i \prec [v_{q}^{*},v_{lq}^{*},v_{rq}^{*}]_{LR},[v_{q}^{\bullet },v_{lq}^{\bullet },v_{rq}^{\bullet }]_{LR}\succ =\widetilde{u}_{q}+i\widetilde{v}_{q}$, where $1 \le q \le n$ is called the LR-bipolar fuzzy complex solution of LR-CBFL System of equations in (18) and (19) if,

$$\begin{aligned} \widetilde{U}^{\prime }=(\widetilde{u}_{1},\widetilde{u}_{2},\ldots ,\widetilde{u}_{n}),~~~~~~ \widetilde{V}^{\prime }=(\widetilde{v}_{1},\widetilde{v}_{2},\ldots ,\widetilde{v}_{n}), \end{aligned}$$

are the LR-bipolar fuzzy solution of LR-bipolar fuzzy linear system in (19), respectively, if

$$\begin{aligned} L\widetilde{U}^{\prime }=\widetilde{E}, ~~~~L\widetilde{V}^{\prime }=\widetilde{F}, \end{aligned}$$

(20)

and this solution vector also satisfied the system (16) given as:

$$\begin{aligned}&{\left\{ \begin{array}{ll} \sum \nolimits _{q=1}^{n}l_{pq}u_{q}^{*} &{}= e_{p}^{*},\\ \sum \nolimits _{q=1}^{n}l_{pq}u_{lq}^{*} &{}= e_{lp}^{*} ,\\ \sum \nolimits _{q=1}^{n}l_{pq}u_{rq}^{*} &{}= e_{rp}^{*}, \\ \sum \nolimits _{q=1}^{n}l_{pq}u_{q}^{\bullet } &{}= e_{p}^{\bullet },\\ \sum \nolimits _{q=1}^{n}l_{pq}u_{lq}^{\bullet } &{}= e_{lp}^{\bullet } ,\\ \sum \nolimits _{q=1}^{n}l_{pq}u_{rq}^{\bullet } &{}= e_{rp}^{\bullet }. \\ \end{array}\right. } \end{aligned}$$

(21)

$$\begin{aligned}&{\left\{ \begin{array}{ll} \sum \nolimits _{q=1}^{n}l_{pq}v_{q}^{*} &{}= f_{p}^{*},\\ \sum \nolimits _{q=1}^{n}l_{pq}v_{lq}^{*} &{}= f_{lp}^{*} ,\\ \sum \nolimits _{q=1}^{n}l_{pq}v_{rq}^{*} &{}= f_{rp}^{*}, \\ \sum \nolimits _{q=1}^{n}l_{pq}v_{q}^{\bullet } &{}= f_{p}^{\bullet },\\ \sum \nolimits _{q=1}^{n}l_{pq}v_{lq}^{\bullet } &{}= f_{lp}^{\bullet } ,\\ \sum \nolimits _{q=1}^{n}l_{pq}v_{rq}^{\bullet } &{}= f_{rp}^{\bullet }. \\ \end{array}\right. } \end{aligned}$$

(22)

4.1 Solution procedure to n $\times$ n LR-CBFL system with real coefficients

To solve the LR-CBFL system of equation with real coefficients in (16), first we should compare the real and imaginary part of system of equations to get two $n \times n$ LR-bipolar fuzzy linear (BFL) systems $L\widetilde{U}=\widetilde{E}$ and $L\widetilde{V}=\widetilde{F}$.

To solve real part of LR-CBFL system of equation, i.e., $L\widetilde{U}=\widetilde{E}$, first we should separate the positive$(*)$ and negative$(\bullet )$ part of system of equations to get two $n \times n$ LR-real linear systems given as

$$\begin{aligned} L(u^{*},u_{l}^{*},u_{r}^{*})_{LR}= & {} (e^{*},e_{l}^{*},e_{r}^{*})_{LR}, \end{aligned}$$

(23)

$$\begin{aligned} L(u^{\bullet },u_{l}^{\bullet },u_{r}^{\bullet })_{LR}= & {} (e^{\bullet },e_{l}^{\bullet },e_{r}^{\bullet })_{LR}. \end{aligned}$$

(24)

To solve LR-real linear system $L(u^{*},u_{l}^{*},u_{r}^{*})_{LR}=(e^{*},e_{l}^{*}, e_{r}^{*})_{LR}$, we resolve this system into two systems, one is $n \times n$ mean value system and other is $2n \times 2n$ left-right spread system.

The $n \times n$ mean value system for the positive part of $L\widetilde{U}=\widetilde{E}$ is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}u_{1}^{*} + l_{12}u_{2}^{*} + \cdots + l_{1n}u_{n}^{*}&{}= e_{1}^{*},\\ l_{21}u_{1}^{*} + l_{22}u_{2}^{*} + \cdots + l_{2n}u_{n}^{*}&{}= e_{2}^{*},\\ .\\ .\\ .\\ l_{n1}u_{1}^{*} + l_{n2}u_{2}^{*} + \cdots + l_{nn}u_{n}^{*}&{}= e_{n}^{*}.\\ \end{array}\right. } \end{aligned}$$

(25)

The matrix notation of system (25) is given as

$$\begin{aligned}LU_{q}^{*}=E_{p}^{*},\end{aligned}$$

where $1 \le p,q \le n$.

To solve left-right spread system of equation $L\widetilde{U}=\widetilde{E}$, we must solve the $2n \times 2n$ real linear system and where the right side column is function vector $(e_{l1}^{*},e_{l2}^{*},\ldots ,e_{ln}^{*},e_{r1}^{*},e_{r2}^{*},\ldots , e_{rn}^{*})^{T}$. Now rearrange the left-right spread system, so the unknown parameters are $u_{lq}^{*},u_{rq}^{*}$, $1 \le q \le n$ and column of right side vector are

$$\begin{aligned}(e_{l1}^{*},e_{l2}^{*},\ldots ,e_{ln}^{*},e_{r1}^{*},e_{r2}^{*},\ldots ,e_{rn}^{*})^{T}.\end{aligned}$$

We get first $2n \times 2n$ left-right spread system of positive$(*)$ part of LR-BFLS of equations $L\widetilde{U}=\widetilde{E}$ given as:

${\left\{ \begin{array}{ll} t_{11}u_{l1}^{*}+t_{12}u_{l2}^{*}+ \cdots + t_{1n}u_{ln}^{*}+t_{1,n+1}(-u_{r1}^{*})+ t_{1,n+2}\\ \quad (-u_{r2}^{*})+ \cdots +t_{1,2n}(-u_{rn}^{*}) = e_{l1}^{*},\\ .\\ .\\ .\\ t_{n1}u_{l1}^{*}+t_{n2}u_{l2}^{*}+ \cdots + t_{nn}u_{ln}^{*}+t_{n,n+1}(-u_{r1}^{*})+ t_{n,n+2}\\ \quad (-u_{r2}^{*})+ \cdots +t_{n,2n}(-u_{rn}^{*}) = e_{ln}^{*},\\ t_{n+1,1}u_{l1}^{*}+t_{n+1,2}u_{l2}^{*}+ \cdots + t_{n+1,n}u_{ln}^{*}+t_{n+1,n+1}(-u_{r1}^{*})\\ \quad + t_{n+1,n+2}(-u_{r2}^{*})+ \cdots +t_{n+1,2n}(-u_{rn}^{*}) = e_{r1}^{*},\\ .\\ .\\ .\\ t_{2n,1}u_{l1}^{*}+t_{2n,2}u_{l2}^{*}+ \cdots + t_{2n,n}u_{ln}^{*}+t_{2n,n+1}(-u_{r1}^{*})\\ \quad + t_{2n,n+2}(-u_{r2}^{*})+ \cdots +t_{2n,2n}(-u_{rn}^{*}) = e_{rn}^{*},\\ \end{array}\right. }$

where $l_{pq}$ are determined as follows:

$$\begin{aligned} l_{pq}\ge & {} 0 \Longrightarrow t_{pq}= l_{pq}, t_{p+n,q+n} = l_{pq},\nonumber \\ l_{pq}< & {} 0 \Longrightarrow t_{p,q+n}= -l_{pq}, t_{p+n,q}= -l_{pq}, \end{aligned}$$

(26)

where $1 \le p,q \le 2n$ and any $t_{pq}$ which is not determined from Eq. (26) is zero.

Using the matrix notation, we obtain

$$\begin{aligned} M(u_{l}^{*},u_{r}^{*})=(e_{l}^{*},e_{r}^{*}), \end{aligned}$$

(27)

where $M=l_{pq}$, $1 \le p,q \le 2n$

$$\begin{aligned}&(u_{l1}^{*},u_{l2}^{*},\ldots ,u_{ln}^{*},u_{r1}^{*},u_{r2}^{*},\ldots ,u_{rn}^{*})^{T} and~\\&\quad (e_{l1}^{*},e_{l2}^{*},\ldots ,e_{ln}^{*},e_{r1}^{*},e_{r2}^{*},\ldots ,e_{rn}^{*})^{T}.\end{aligned}$$

We can also write the matrix form of Eq. (27) as:

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right) \left( \begin{array}{c} u_{l}^{*} \\ u_{r}^{*} \\ \end{array} \right) = \left( \begin{array}{c} e_{l}^{*}\\ e_{r}^{*} \\ \end{array} \right) . \end{aligned}$$

(28)

In the similar way we can take the negative$(\bullet )$ part of the LR-BFLS of equation given $L\widetilde{U}=\widetilde{E}$.

To solve LR-real linear system $L(u^{\bullet },u_{l}^{\bullet },u_{r}^{\bullet })_{LR}=(e^{\bullet },e_{l}^{\bullet }, e_{r}^{\bullet })_{LR}$, we divide this system into two systems, one is $n \times n$ mean value system and other is $2n \times 2n$ left-right spread system for negative part of $L\widetilde{U}=\widetilde{E}$.

The $n \times n$ mean value system for the negative part of $L\widetilde{U}=\widetilde{E}$ is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}u_{1}^{\bullet } + l_{12}u_{2}^{\bullet } + \cdots + l_{1n}u_{n}^{\bullet }&{}= e_{1}^{\bullet },\\ l_{21}u_{1}^{\bullet } + l_{22}u_{2}^{\bullet } + \cdots + l_{2n}u_{n}^{\bullet }&{}= e_{2}^{\bullet },\\ .\\ .\\ .\\ l_{n1}u_{1}^{\bullet } + l_{n2}u_{2}^{\bullet } + \cdots + l_{nn}u_{n}^{\bullet }&{}= e_{n}^{\bullet }.\\ \end{array}\right. } \end{aligned}$$

(29)

The matrix notation of system (29) is given as

$$\begin{aligned} LU_{q}^{\bullet }=E_{p}^{\bullet },\end{aligned}$$

where $1 \le p,q \le n$.

To solve negative part of left-right spread system of equation $L\widetilde{U}=\widetilde{E}$, we must solve the $2n \times 2n$ crisp linear system and where the right side column is function vector $(e_{l1}^{\bullet },e_{l2}^{\bullet },\ldots ,e_{ln}^{\bullet },e_{r1}^{\bullet },e_{r2}^{\bullet },\ldots ,e_{rn}^{\bullet })^{T}$. Now rearrange the left-right spread system, so the unknown parameters are $u_{lq}^{\bullet },u_{rq}^{\bullet }$, $1 \le q \le n$ and column of right side vector are

$$\begin{aligned} (e_{l1}^{\bullet },e_{l2}^{\bullet },\ldots ,e_{ln}^{\bullet },e_{r1}^{\bullet },e_{r2}^{\bullet },\ldots ,e_{rn}^{\bullet })^{T}.\end{aligned}$$

We get first $2n \times 2n$ left-right spread system of negative$(\bullet )$ part of $L\widetilde{U}=\widetilde{E}$ of equation,

${\left\{ \begin{array}{ll} t_{11}u_{l1}^{\bullet }+t_{12}u_{l2}^{\bullet }+ \cdots + t_{1n}u_{ln}^{\bullet }+t_{1,n+1}(-u_{r1}^{\bullet })+ t_{1,n+2}\\ \quad (-u_{r2}^{\bullet })+ \cdots +t_{1,2n}(-u_{rn}^{\bullet }) = e_{l1}^{\bullet },\\ .\\ .\\ .\\ t_{n1}u_{l1}^{\bullet }+t_{n2}u_{l2}^{\bullet }+ \cdots + t_{nn}u_{ln}^{\bullet }+t_{n,n+1}(-u_{r1}^{\bullet })+ t_{n,n+2}\\ \quad (-u_{r2}^{\bullet })+ \cdots +t_{n,2n}(-u_{rn}^{\bullet }) = e_{ln}^{\bullet },\\ t_{n+1,1}u_{l1}^{\bullet }+t_{n+1,2}u_{l2}^{\bullet }+ \cdots + t_{n+1,n}u_{ln}^{\bullet }+t_{n+1,n+1}(-u_{r1}^{\bullet })\\ \quad + t_{n+1,n+2} (-u_{r2}^{\bullet })+ \cdots +t_{n+1,2n}(-u_{rn}^{\bullet }) = e_{r1}^{\bullet },\\ .\\ .\\ .\\ t_{2n,1}u_{l1}^{\bullet }+t_{2n,2}u_{l2}^{\bullet }+ \cdots + t_{2n,n}u_{ln}^{\bullet }+t_{2n,n+1}(-u_{r1}^{\bullet })\\ \quad + t_{2n,n+2}(-u_{r2}^{\bullet })+ \cdots +t_{2n,2n}(-u_{rn}^{\bullet }) = e_{rn}^{\bullet },\\ \end{array}\right. }$

where $l_{pq}$ are determined as follows:

$$\begin{aligned} l_{pq}&\ge 0 \Longrightarrow t_{pq}= l_{pq}, t_{p+n,q+n} = l_{pq},\nonumber \\ l_{pq}&< 0 \Longrightarrow t_{p,q+n}= -l_{pq}, t_{p+n,q}= -l_{pq}, \end{aligned}$$

(30)

where $1 \le p,q \le 2n$ and any $t_{pq}$ which is not determined from Eq. (30) is zero.

Using the matrix notation, we obtain

$$\begin{aligned} M(u_{l}^{\bullet },u_{r}^{\bullet })=(e_{l}^{\bullet },e_{r}^{\bullet }), \end{aligned}$$

(31)

where $M=l_{pq}$, $1 \le p,q \le 2n$

$$\begin{aligned}&(u_{l1}^{\bullet },u_{l2}^{\bullet },\ldots ,u_{ln}^{\bullet },u_{r1}^{\bullet },u_{r2}^{\bullet },\ldots ,u_{rn}^{\bullet })^{T} and~\\&\quad (e_{l1}^{\bullet },e_{l2}^{\bullet },\ldots ,e_{ln}^{\bullet },e_{r1}^{\bullet },e_{r2}^{\bullet },\ldots ,e_{rn}^{\bullet })^{T}.\end{aligned}$$

We can also write the matrix form of Eq. (31) as:

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right) \left( \begin{array}{c} u_{l}^{\bullet } \\ u_{r}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} e_{l}^{\bullet }\\ e_{r}^{\bullet } \\ \end{array} \right) . \end{aligned}$$

(32)

To find the solution of real part of LR-CBFL system with real coefficients in (16), we can find the solution of Eqs. (25), (28), (29) and (32) by using matrix inverse method.

Similarly, we can solve imaginary part of LR-CBFL system of equation, i.e., $L\widetilde{V}=\widetilde{F}$, first we separate the positive$(*)$ and negative$(\bullet )$ part of system of equations to get two $n \times n$ LR-real linear systems given as

$$\begin{aligned} L(v^{*},v_{l}^{*},v_{r}^{*})_{LR}= & {} (f^{*},f_{l}^{*},f_{r}^{*})_{LR}, \end{aligned}$$

(33)

$$\begin{aligned} L(v^{\bullet },v_{l}^{\bullet },v_{r}^{\bullet })_{LR}= & {} (f^{\bullet },f_{l}^{\bullet },f_{r}^{\bullet })_{LR}. \end{aligned}$$

(34)

To solve LR-real linear system $L(v^{*},v_{l}^{*},v_{r}^{*})_{LR}=(f^{*},f_{l}^{*}, f_{r}^{*})_{LR}$, we restructure this system into two systems, one is $n \times n$ mean value system and other is $2n \times 2n$ left-right spread system.

The $n \times n$ mean value system for the positive part of $L\widetilde{V}=\widetilde{F}$ is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}v_{1}^{*} + l_{12}v_{2}^{*} + \cdots + l_{1n}v_{n}^{*}&{}= f_{1}^{*},\\ l_{21}v_{1}^{*} + l_{22}v_{2}^{*} + \cdots + l_{2n}v_{n}^{*}&{}= f_{2}^{*},\\ .\\ .\\ .\\ l_{n1}v_{1}^{*} + l_{n2}v_{2}^{*} + \cdots + l_{nn}v_{n}^{*}&{}= f_{n}^{*}.\\ \end{array}\right. } \end{aligned}$$

(35)

The matrix notation of system (35) is given as

$$\begin{aligned} LV_{q}^{*}=F_{p}^{*},\end{aligned}$$

where $1 \le p,q \le n$.

To solve left-right spread system of equation $L\widetilde{V}=\widetilde{F}$, we must solve the $2n \times 2n$ real linear system and where the right side column is function vector $(f_{l1}^{*},f_{l2}^{*},\ldots ,f_{ln}^{*},f_{r1}^{*},f_{r2}^{*}, \ldots , f_{rn}^{*})^{T}$. Now rearrange the left-right spread system, so the unknown parameters are $v_{lq}^{*},v_{rq}^{*}$, $1 \le q \le n$ and column of right side vector are

$$\begin{aligned}(f_{l1}^{*},f_{l2}^{*},\ldots ,f_{ln}^{*},f_{r1}^{*},f_{r2}^{*},\ldots ,f_{rn}^{*})^{T}.\end{aligned}$$

We get first $2n \times 2n$ left-right spread system of positive$(*)$ part of LR-BFLS of equation $L\widetilde{V}=\widetilde{F}$ given as:

${\left\{ \begin{array}{ll} t_{11}v_{l1}^{*}+t_{12}v_{l2}^{*}+ \cdots + t_{1n}v_{ln}^{*}+t_{1,n+1}(-v_{r1}^{*})\\ \quad + t_{1,n+2}(-v_{r2}^{*})+ \cdots +t_{1,2n}(-v_{rn}^{*}) = f_{l1}^{*},\\ .\\ .\\ .\\ t_{n1}v_{l1}^{*}+t_{n2}v_{l2}^{*}+ \cdots + t_{nn}v_{ln}^{*}+t_{n,n+1}(-v_{r1}^{*})+ t_{n,n+2}\\ \quad (-v_{r2}^{*})+ \cdots +t_{n,2n}(-v_{rn}^{*}) = f_{ln}^{*},\\ t_{n+1,1}v_{l1}^{*}+t_{n+1,2}v_{l2}^{*}+ \cdots + t_{n+1,n}v_{ln}^{*}+t_{n+1,n+1}(-v_{r1}^{*})\\ \quad + t_{n+1,n+2}(-v_{r2}^{*})+ \cdots +t_{n+1,2n}(-v_{rn}^{*}) = f_{r1}^{*},\\ .\\ .\\ .\\ t_{2n,1}v_{l1}^{*}+t_{2n,2}v_{l2}^{*}+ \cdots + t_{2n,n}v_{ln}^{*}+t_{2n,n+1}(-v_{r1}^{*})\\ \quad + t_{2n,n+2}(-v_{r2}^{*})+ \cdots +t_{2n,2n}(-v_{rn}^{*}) = f_{rn}^{*},\\ \end{array}\right. }$

where $l_{pq}$ are determined as follows:

$$\begin{aligned} l_{pq}&\ge 0 \Longrightarrow t_{pq}= l_{pq}, t_{p+n,q+n} = l_{pq},\nonumber \\ l_{pq}&< 0 \Longrightarrow t_{p,q+n}= -l_{pq}, t_{p+n,q}= -l_{pq}, \end{aligned}$$

(36)

where $1 \le p,q \le 2n$ and any $t_{pq}$ which is not determined from Eqs. (36) is zero.

Using the matrix notation, we obtain

$$\begin{aligned} M(v_{l}^{*},v_{r}^{*})=(f_{l}^{*},f_{r}^{*}), \end{aligned}$$

(37)

where $M=l_{pq}$, $1 \le p,q \le 2n$

$$\begin{aligned}&(v_{l1}^{*},v_{l2}^{*},\ldots ,v_{ln}^{*},v_{r1}^{*},v_{r2}^{*},\ldots ,v_{rn}^{*})^{T} and~\\&\quad (f_{l1}^{*},f_{l2}^{*},\ldots ,f_{ln}^{*},f_{r1}^{*},f_{r2}^{*},\ldots ,f_{rn}^{*})^{T}.\end{aligned}$$

We can also write the matrix form of Eq. (37) as:

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right) \left( \begin{array}{c} v_{l}^{*} \\ v_{r}^{*} \\ \end{array} \right) = \left( \begin{array}{c} f_{l}^{*}\\ f_{r}^{*} \\ \end{array} \right) . \end{aligned}$$

(38)

In the similar way we can take the negative$(\bullet )$ part of the LR-BFLS of equation given $L\widetilde{V}=\widetilde{F}$.

To solve LR-real linear system $L(v^{\bullet },v_{l}^{\bullet },v_{r}^{\bullet })_{LR}=(f^{\bullet },f_{l}^{\bullet }, f_{r}^{\bullet })_{LR}$, we divide this system into two systems, one is $n \times n$ mean value system and other is $2n \times 2n$ left-right spread system for negative part of $L\widetilde{V}=\widetilde{F}$.

The $n \times n$ mean value system for the negative part of $L\widetilde{V}=\widetilde{F}$ is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}v_{1}^{\bullet } + l_{12}v_{2}^{\bullet } + \cdots + l_{1n}v_{n}^{\bullet }&{}= f_{1}^{\bullet },\\ l_{21}v_{1}^{\bullet } + l_{22}v_{2}^{\bullet } + \cdots + l_{2n}v_{n}^{\bullet }&{}= f_{2}^{\bullet },\\ .\\ .\\ .\\ l_{n1}v_{1}^{\bullet } + l_{n2}v_{2}^{\bullet } + \cdots + l_{nn}v_{n}^{\bullet }&{}= f_{n}^{\bullet }.\\ \end{array}\right. } \end{aligned}$$

(39)

The matrix notation of system (39) is given as

$$\begin{aligned}LV_{q}^{\bullet }=F_{p}^{\bullet },\end{aligned}$$

where $1 \le p,q \le n$.

To solve negative part of left-right spread system of equation $L\widetilde{V}=\widetilde{F}$, we must solve the $2n \times 2n$ real linear system and where the right side column is function vector $(f_{l1}^{\bullet },f_{l2}^{\bullet },\ldots ,f_{ln}^{\bullet },f_{r1}^{\bullet },f_{r2}^{\bullet },\ldots ,f_{rn}^{\bullet })^{T}$. Now rearrange the left-right spread system, so the unknown parameters are $v_{lq}^{\bullet },v_{rq}^{\bullet }$, $1 \le q \le n$ and column of right side vector are

$$\begin{aligned}(f_{l1}^{\bullet },f_{l2}^{\bullet },\ldots ,f_{ln}^{\bullet },f_{r1}^{\bullet },f_{r2}^{\bullet },\ldots ,f_{rn}^{\bullet })^{T}. \end{aligned}$$

We get first $2n \times 2n$ left-right spread system of negative$(\bullet )$ part of $L\widetilde{V}=\widetilde{F}$ of equation,

${\left\{ \begin{array}{ll} t_{11}v_{l1}^{\bullet }+t_{12}v_{l2}^{\bullet }+ \cdots + t_{1n}v_{ln}^{\bullet }+t_{1,n+1}\\ \quad (-v_{r1}^{\bullet })+ t_{1,n+2}(-v_{r2}^{\bullet })+ \cdots +t_{1,2n}(-v_{rn}^{\bullet }) = f_{l1}^{\bullet },\\ .\\ .\\ .\\ t_{n1}v_{l1}^{\bullet }+t_{n2}v_{l2}^{\bullet }+ \cdots + t_{nn}v_{ln}^{\bullet }+t_{n,n+1}(-v_{r1}^{\bullet })+ t_{n,n+2}\\ \quad (-v_{r2}^{\bullet })+ \cdots +t_{n,2n}(-v_{rn}^{\bullet }) = f_{ln}^{\bullet },\\ t_{n+1,1}v_{l1}^{\bullet }+t_{n+1,2}v_{l2}^{\bullet }+ \cdots + t_{n+1,n}v_{ln}^{\bullet } +t_{n+1,n+1}(-v_{r1}^{\bullet })\\ \quad + t_{n+1,n+2}(-v_{r2}^{\bullet })+ \cdots +t_{n+1,2n}(-v_{rn}^{\bullet }) = f_{r1}^{\bullet },\\ .\\ .\\ .\\ t_{2n,1}v_{l1}^{\bullet }+t_{2n,2}v_{l2}^{\bullet }+ \cdots + t_{2n,n}v_{ln}^{\bullet }+t_{2n,n+1}(-v_{r1}^{\bullet })\\ \quad + t_{2n,n+2}(-v_{r2}^{\bullet })+ \cdots +t_{2n,2n}(-v_{rn}^{\bullet }) = f_{rn}^{\bullet },\\ \end{array}\right. }$

where $l_{pq}$ are determined as follows:

$$\begin{aligned} l_{pq}&\ge 0 \Longrightarrow t_{pq}= l_{pq}, t_{p+n,q+n} = l_{pq},\nonumber \\ l_{pq}&< 0 \Longrightarrow t_{p,q+n}= -l_{pq}, t_{p+n,q}= -l_{pq}, \end{aligned}$$

(40)

where $1 \le p,q \le 2n$ and any $t_{pq}$ which is not determined from Eq. (40) is zero.

Using the matrix notation, we obtain

$$\begin{aligned} M(v_{l}^{\bullet },v_{r}^{\bullet })=(f_{l}^{\bullet },f_{r}^{\bullet }), \end{aligned}$$

(41)

where $M=l_{pq}$, $1 \le p,q \le 2n$

$$\begin{aligned}&(v_{l1}^{\bullet },v_{l2}^{\bullet },\ldots ,v_{ln}^{\bullet },v_{r1}^{\bullet },v_{r2}^{\bullet },\ldots ,v_{rn}^{\bullet })^{T} and~\\&\quad (f_{l1}^{\bullet },f_{l2}^{\bullet },\ldots ,f_{ln}^{\bullet },f_{r1}^{\bullet },f_{r2}^{\bullet },\ldots ,f_{rn}^{\bullet })^{T}.\end{aligned}$$

We can also write the matrix form of Eq. (41) as:

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right) \left( \begin{array}{c} v_{l}^{\bullet } \\ v_{r}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} f_{l}^{\bullet }\\ f_{r}^{\bullet } \\ \end{array} \right) . \end{aligned}$$

(42)

To find the solution of imaginary part of LR-CBFL system with real coefficients in (16), we can find the solution of equations (35), (38), (39) and (42) by using matrix inverse method.

Theorem 3

For any arbitrary LR-bipolar fuzzy complex vector $\widetilde{Z}$ the unique solution $\widetilde{H}$ is a LR-bipolar fuzzy complex vector of LR-CBFL system of equations with real coefficients in (16) if and only if the matrix M= $\left( \begin{array}{r@{\quad }r} A &{} -B \\ -B &{} A \\ \end{array} \right)$ is non-singular, the necessary and sufficient condition that the matrices $(A + B)^{-1}$ and $(A - B)^{-1}$ exist.

Example 4

Consider the $2 \times 2$ LR-CBFL system with real coefficients

$$\begin{aligned} \widetilde{h}_{1} - \widetilde{h}_{2}= & {} \prec [-3,3,2]_{LR},[-3,3,3]_{LR}\succ \\&+i\prec [-4,4,4]_{LR},[-5,2,3]_{LR}\succ ,\\ \widetilde{h}_{1} + 2\widetilde{h}_{2}= & {} \prec [21,4,3]_{LR},[15,6,3]_{LR}\succ \\&+i\prec [32,7,4]_{LR},[25,3,4]_{LR}\succ . \end{aligned}$$

To solve this LR-CBFL system first we put $\widetilde{h}_{1}= \widetilde{u}_{1}+i\widetilde{v}_{1}$ and $\widetilde{h}_{2}= \widetilde{u}_{2}+i\widetilde{v}_{2}$ in original system and get,

$$\begin{aligned}&(\widetilde{u}_{1}+i\widetilde{v}_{1}) - (\widetilde{u}_{2}+i\widetilde{v}_{2})\\&\quad =\prec [-3,3,2]_{LR},[-3,3,3]_{LR}\succ \\&\quad +i\prec [-4,4,4]_{LR},[-5,2,3]_{LR}\succ ,\\&(\widetilde{u}_{1}+i\widetilde{v}_{1}) + 2(\widetilde{u}_{2}+i\widetilde{v}_{2})\\&\quad =\prec [21,4,3]_{LR},[15,6,3]_{LR}\succ \\&\quad +i\prec [32,7,4]_{LR},[25,3,4]_{LR}\succ , \end{aligned}$$

compare the real and imaginary parts to both side of the equations and we obtain two LR-BFLS of equations given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} \widetilde{u}_{1} - \widetilde{u}_{2}=\prec [-3,3,2]_{LR},[-3,3,3]_{LR}\succ ,\\ \widetilde{u}_{1} + 2\widetilde{u}_{2}=\prec [21,4,3]_{LR},[15,6,3]_{LR}\succ , \end{array}\right. } \end{aligned}$$

(43)

and,

$$\begin{aligned} {\left\{ \begin{array}{ll} v_{1} - v_{2}=\prec [-4,4,4]_{LR},[-5,2,3]_{LR}\succ ,\\ v_{1} + 2v_{2}=\prec [32,7,4]_{LR},[25,3,4]_{LR}\succ , \end{array}\right. } \end{aligned}$$

(44)

To solve real part of LR-CBFL system with real coefficients given in (43) first we put $\widetilde{u}_{1} =\prec [u_{1}^{*}, u_{l1}^{*}, u_{r1}^{*}]_{LR},[u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet }]_{LR}\succ$ and $\widetilde{u}_{2}= \prec [u_{2}^{*},u_{l2}^{*},u_{r2}^{*}]_{LR},[u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet }]_{LR}\succ$ in original system and get,

$$\begin{aligned}&\prec [u_{1}^{*},u_{l1}^{*},u_{r1}^{*}]_{LR},[u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet }]_{LR}\succ \\&\quad - \prec [u_{2}^{*},u_{l2}^{*},u_{r2}^{*}]_{LR},[u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [-3,3,2]_{LR},[-3,3,3]_{LR}\succ ,\\&\prec [u_{1}^{*},u_{l1}^{*},u_{r1}^{*}]_{LR},[u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet }]_{LR}\succ \\&\quad + 2\prec [u_{2}^{*},u_{l2}^{*},u_{r2}^{*}]_{LR},[u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [21,4,3]_{LR},[15,6,3]_{LR}\succ . \end{aligned}$$

To solve the LR-BFLS of equation in (43), first we take the positive$(*)$ part of the system (43) given as:

$$\begin{aligned}(u_{1}^{*},u_{l1}^{*},u_{r1}^{*})_{LR} - (u_{2}^{*},u_{l2}^{*},u_{r2}^{*})_{LR}= (-3,2,2)_{LR},\\ (u_{1}^{*},u_{l1}^{*},u_{r1}^{*})_{LR} + 2 (u_{2}^{*},u_{l2}^{*},u_{r2}^{*})_{LR}= (21,4,3)_{LR}. \end{aligned}$$

To solve this positive part we can divide this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for positive part of (43) is given as:

$$\begin{aligned} u_{1}^{*} - u_{2}^{*}= -3,\\ u_{1}^{*} + 2 u_{2}^{*}= 21, \end{aligned}$$

by solving above equations, we have $u_{1}^{*}=5$ and $u_{2}^{*}=8$. Now the extended $4 \times 4$ form of positive part of left-right spread is given as:

$$\begin{aligned} 1(u_{l1}^{*}) + 0(u_{l2}^{*}) + 0(u_{r1}^{*}) + 1(u_{r2}^{*})&= 3,\\ 1(u_{l1}^{*}) + 2(u_{l2}^{*}) + 0(u_{r1}^{*}) + 0(u_{r2}^{*})&= 4,\\ 0(u_{l1}^{*}) + 1(u_{l2}^{*}) + 1(u_{r1}^{*}) + 0(u_{r2}^{*})&= 2,\\ 0(u_{l1}^{*}) + 0(u_{l2}^{*}) + 1(u_{r1}^{*}) + 2(u_{r2}^{*})&= 3.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 2 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 2 \\ \end{array} \right) \left( \begin{array}{c} u_{l1}^{*} \\ u_{l2}^{*} \\ u_{r1}^{*} \\ u_{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 3\\ 4 \\ 2 \\ 3 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned} M(u_{l}^{*},u_{r}^{*})=(e_{l}^{*},e_{r}^{*}),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} u_{l1}^{*} \\ u_{l2}^{*} \\ u_{r1}^{*} \\ u_{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.3333 &{} -0.3333 &{} 0.6667 &{} -0.6667 \\ -0.6667 &{} 0.6667 &{} -0.3333 &{} 0.3333 \\ 0.6667 &{} -0.6667 &{} 1.3333 &{} -0.3333 \\ -0.3333 &{} 0.3333 &{} -0.6667 &{} 0.6667 \\ \end{array} \right)$

$\left( \begin{array}{c} 3 \\ 4 \\ 2 \\ 3 \\ \end{array} \right)$

$$\begin{aligned}u_{l1}^{*}= 2, ~~~~u_{l2}^{*}=1,\\ u_{r1}^{*}= 1, ~~~~u_{r2}^{*}=1.\end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of the system (43) given as:

$$\begin{aligned}&(u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet })_{LR} - (u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet })_{LR}= (-3,3,3)_{LR},\\&(u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet })_{LR} + 2 (u_{2}^{*},u_{l2}^{\bullet },u_{r2}^{\bullet })_{LR}= (15,6,3)_{LR}.\end{aligned}$$

To solve this negative part of LR-BFLS of equation we can restructure this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for negative part is given as:

$$\begin{aligned} u_{1}^{\bullet } - u_{2}^{\bullet }= & {} -3,\\ u_{1}^{\bullet } + 2 u_{2}^{\bullet }= & {} 15,\end{aligned}$$

by solving above equations, we have $u_{1}^{\bullet }=3$ and $u_{2}^{\bullet }=6$. Now the extended $4 \times 4$ form of negative part of left-right spread is given as:

$$\begin{aligned} 1(u_{l1}^{\bullet }) + 0(u_{l2}^{\bullet }) + 0(u_{r1}^{\bullet }) + 1(u_{r2}^{\bullet })&= 3,\\ 1(u_{l1}^{\bullet }) + 2(u_{l2}^{\bullet }) + 0(u_{r1}^{\bullet }) + 0(u_{r2}^{\bullet })&= 6,\\ 0(u_{l1}^{\bullet }) + 1(u_{l2}^{\bullet }) + 1(u_{r1}^{\bullet }) + 0(u_{r2}^{\bullet })&= 3,\\ 0(u_{l1}^{\bullet }) + 0(u_{l2}^{\bullet }) + 1(u_{r1}^{\bullet }) + 2(u_{r2}^{\bullet })&= 3. \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 2 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 2 \\ \end{array} \right) \left( \begin{array}{c} u_{l1}^{\bullet } \\ u_{l2}^{\bullet } \\ u_{r1}^{\bullet } \\ u_{r2}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} 3\\ 6 \\ 3 \\ 3 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned} M(u_{l}^{\bullet },u_{r}^{\bullet })=(e_{l}^{\bullet },e_{r}^{\bullet }),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} u_{l1}^{\bullet } \\ u_{l2}^{\bullet } \\ u_{r1}^{\bullet } \\ u_{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.3333 &{} -0.3333 &{} 0.6667 &{} -0.6667 \\ -0.6667 &{} 0.6667 &{} -0.3333 &{} 0.3333 \\ 0.6667 &{} -0.6667 &{} 1.3333 &{} -0.3333 \\ -0.3333 &{} 0.3333 &{} -0.6667 &{} 0.6667 \\ \end{array} \right)$

$\left( \begin{array}{c} 3 \\ 6 \\ 3 \\ 3 \\ \end{array} \right)$

$$\begin{aligned} u_{l1}^{\bullet }= & {} 2, ~~~~u_{l2}^{\bullet }=2,\\ u_{r1}^{\bullet }= & {} 1, ~~~~u_{r2}^{\bullet }=1. \end{aligned}$$

Hence, the solution for real part of LR-CBFL system of equation with real coefficients in (4) is given as:

$$\begin{aligned} \widetilde{U}_{1}= & {} \prec [u_{1}^{*},u_{l1}^{*},u_{r1}^{*}]_{LR},[u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet }]_{LR}\succ \\&\quad =\prec [5,2,1]_{LR},[3,2,1]_{LR}\succ ,\\ \widetilde{U}_{2}= & {} \prec [u_{2}^{*},u_{l2}^{*},u_{r2}^{*}]_{LR},[u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [8,1,1]_{LR},[6,2,1]_{LR}\succ . \end{aligned}$$

Similarly, we can solve imaginary part of LR-CBFL system with real coefficients given in (44) first we put $\widetilde{v}_{1}=\prec [v_{1}^{*},v_{l1}^{*},v_{r1}^{*}]_{LR},[v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ$ and $\widetilde{v}_{2}= \prec [v_{2}^{*},v_{l2}^{*},v_{r2}^{*}]_{LR},[v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet }]_{LR}\succ$ in original system and get,

$$\begin{aligned}&\prec [v_{1}^{*},v_{l1}^{*},v_{r1}^{*}]_{LR},[v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ \\&\quad - \prec [v_{2}^{*},v_{l2}^{*},v_{r2}^{*}]_{LR},[v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [-4,4,4]_{LR},[-5,2,3]_{LR}\succ ,\\&\prec [v_{1}^{*},v_{l1}^{*},v_{r1}^{*}]_{LR},[v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ \\&\quad + 2\prec [v_{2}^{*},v_{l2}^{*},v_{r2}^{*}]_{LR},[v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet }]_{LR}\succ \\&\quad =\prec [32,7,4]_{LR},[25,3,4]_{LR}\succ . \end{aligned}$$

To solve the LR-BFLS equation in (44), first we take the positive$(*)$ part of the system (44) given as:

$$\begin{aligned}&(v_{1}^{*},v_{l1}^{*},v_{r1}^{*})_{LR} - (v_{2}^{*},v_{l2}^{*},v_{r2}^{*})_{LR}= (-4,4,4)_{LR},\\&(v_{1}^{*},v_{l1}^{*},v_{r1}^{*})_{LR} + 2 (v_{2}^{*},v_{l2}^{*},v_{r2}^{*})_{LR}= (32,7,4)_{LR}. \end{aligned}$$

To solve this positive part we can divide this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for positive part of (44) is given as:

$$\begin{aligned} v_{1}^{*} - v_{2}^{*}= & {} -4,\\ v_{1}^{*} + 2 v_{2}^{*}= & {} 32,\end{aligned}$$

by solving above equations, we have $v_{1}^{*}=8$ and $v_{2}^{*}=12$. Now the extended $4 \times 4$ form of positive part of left-right spread is given as:

$$\begin{aligned} 1(v_{l1}^{*}) + 0(v_{l2}^{*}) + 0(v_{r1}^{*}) + 1(v_{r2}^{*})&= 4,\\ 1(v_{l1}^{*}) + 2(v_{l2}^{*}) + 0(v_{r1}^{*}) + 0(v_{r2}^{*})&= 7,\\ 0(v_{l1}^{*}) + 1(v_{l2}^{*}) + 1(v_{r1}^{*}) + 0(v_{r2}^{*})&= 4,\\ 0(v_{l1}^{*}) + 0(v_{l2}^{*}) + 1(v_{r1}^{*}) + 2(v_{r2}^{*})&= 4. \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 2 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 2 \\ \end{array} \right) \left( \begin{array}{c} v_{l1}^{*} \\ v_{l2}^{*} \\ v_{r1}^{*} \\ v_{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 4\\ 7 \\ 4 \\ 4 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned} M(v_{l}^{*},v_{r}^{*})=(f_{l}^{*},f_{r}^{*}),\end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} v_{l1}^{*} \\ v_{l2}^{*} \\ v_{r1}^{*} \\ v_{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.3333 &{} -0.3333 &{} 0.6667 &{} -0.6667 \\ -0.6667 &{} 0.6667 &{} -0.3333 &{} 0.3333 \\ 0.6667 &{} -0.6667 &{} 1.3333 &{} -0.3333 \\ -0.3333 &{} 0.3333 &{} -0.6667 &{} 0.6667 \\ \end{array} \right)$

$\left( \begin{array}{c} 4 \\ 7 \\ 4 \\ 4 \\ \end{array} \right)$

$$\begin{aligned}v_{l1}^{*}= 2, ~~~~v_{l2}^{*}=1,\\ v_{r1}^{*}= 1, ~~~~v_{r2}^{*}=1.\end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of the system (44) given as:

$$\begin{aligned}&(v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet })_{LR} - (v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet })_{LR}= (-5,2,3)_{LR},\\&(v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet })_{LR} + 2 (v_{2}^{*},v_{l2}^{\bullet },v_{r2}^{\bullet })_{LR}= (25,3,4)_{LR}.\end{aligned}$$

To solve this negative part of LR-BFLS of equation we can divide this system into two systems, one is mean value system and other is left-right spread system.

The mean value system for negative part is given as:

$$\begin{aligned}&v_{1}^{\bullet } - v_{2}^{\bullet }= -5,\\&v_{1}^{\bullet } + 2 v_{2}^{\bullet }= 25,\end{aligned}$$

by solving above equations, we have $v_{1}^{\bullet }=5$ and $v_{2}^{\bullet }=10$. Now the extended $4 \times 4$ form of negative part of left-right spread is given as:

$$\begin{aligned} 1(v_{l1}^{\bullet }) + 0(v_{l2}^{\bullet }) + 0(v_{r1}^{\bullet }) + 1(v_{r2}^{\bullet })&= 2,\\ 1(v_{l1}^{\bullet }) + 2(v_{l2}^{\bullet }) + 0(v_{r1}^{\bullet }) + 0(v_{r2}^{\bullet })&= 3,\\ 0(v_{l1}^{\bullet }) + 1(v_{l2}^{\bullet }) + 1(v_{r1}^{\bullet }) + 0(v_{r2}^{\bullet })&= 3,\\ 0(v_{l1}^{\bullet }) + 0(v_{l2}^{\bullet }) + 1(v_{r1}^{\bullet }) + 2(v_{r2}^{\bullet })&= 4.\\ \end{aligned}$$

The matrix form of the above-mentioned system is

$$\begin{aligned}\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1 &{} 0 &{} 0 &{} 1 \\ 1 &{} 2 &{} 0 &{} 0 \\ 0 &{} 1 &{} 1 &{} 0 \\ 0 &{} 0 &{} 1 &{} 2 \\ \end{array} \right) \left( \begin{array}{c} v_{l1}^{\bullet } \\ v_{l2}^{\bullet } \\ v_{r1}^{\bullet } \\ v_{r2}^{\bullet } \\ \end{array} \right) = \left( \begin{array}{c} 2\\ 3 \\ 3 \\ 4 \\ \end{array} \right) , \end{aligned}$$

this can also be written as:

$$\begin{aligned} M(v_{l}^{\bullet },v_{r}^{\bullet })=(f_{l}^{\bullet },f_{r}^{\bullet }), \end{aligned}$$

and the solution of above system is,

$\left( \begin{array}{c} v_{l1}^{\bullet } \\ v_{l2}^{\bullet } \\ v_{r1}^{\bullet } \\ v_{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 1.3333 &{} -0.3333 &{} 0.6667 &{} -0.6667 \\ -0.6667 &{} 0.6667 &{} -0.3333 &{} 0.3333 \\ 0.6667 &{} -0.6667 &{} 1.3333 &{} -0.3333 \\ -0.3333 &{} 0.3333 &{} -0.6667 &{} 0.6667 \\ \end{array} \right)$

$\left( \begin{array}{c} 2 \\ 3 \\ 3 \\ 4 \\ \end{array} \right)$

$$\begin{aligned} v_{l1}^{\bullet }= & {} 1, ~~~~v_{l2}^{\bullet }=1,\\ v_{r1}^{\bullet }= & {} 2, ~~~~v_{r2}^{\bullet }=1.\end{aligned}$$

Hence, the solution for imaginary part of LR-CBFL system of equations with real coefficients in (4) is given as:

$$\begin{aligned} \widetilde{V}_{1}= & {} \prec [v_{1}^{*},v_{l1}^{*},v_{r1}^{*}]_{LR},[v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ \\ {}= & {} \prec [8,3,2]_{LR},[5,1,2]_{LR}\succ ,\\ \widetilde{V}_{2}= & {} \prec [v_{2}^{*},v_{l2}^{*},v_{r2}^{*}]_{LR},[v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet }]_{LR}\succ \\ {}= & {} \prec [12,2,1]_{LR},[10,1,1]_{LR}\succ . \end{aligned}$$

Hence, the solution for LR-CBFL system of equation with real coefficients is given as:

$$\begin{aligned} \widetilde{H}_{1}= & {} \widetilde{u}_{1} + i \widetilde{v}_{1}=\prec [5,2,1]_{LR},[3,2,1]_{LR}\succ \\&\quad + i \prec [8,3,2]_{LR},[5,1,2]_{LR}\succ ,\\ \widetilde{H}_{2}= & {} \widetilde{u}_{2} + i \widetilde{v}_{2}=\prec [8,1,1]_{LR},[6,2,1]_{LR}\succ \\&\quad + i \prec [12,2,1]_{LR},[10,1,1]_{LR}\succ . \end{aligned}$$

4.2 Solution procedure to n $\times$ n LR-CBFL system with complex coefficients

Definition 21

Consider the $n\times n$ complex linear system of equations given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} l_{11}h_{1} + l_{12}h_{2} + \cdots + l_{1n}h_{n}&{}= z_{1},\\ l_{21}h_{1} + l_{22}h_{2} + \cdots + l_{2n}h_{n}&{}= z_{2},\\ .\\ .\\ .\\ l_{n1}h_{1} + l_{n2}h_{2} + \cdots + l_{nn}h_{n}&{}= z_{n},\\ \end{array}\right. } \end{aligned}$$

(45)

is the LR-complex bipolar fuzzy linear (CBFL) system of equations with complex coefficients, if all the coefficient elements $l_{pq}$, $1 \le p,q \le n$ is a complex number matrix of order $n \times n$ and each $Z_{q}$, $1 \le q \le n$ is a LR-BFCN, and each unknown parameter $h_{p}$, $1 \le p \le n$ is also LR-BFCN.

The matrix form of the LR-CBFL system of equation with complex coefficients in (45) given as:

$$\begin{aligned} L\widetilde{H}_{p}=\widetilde{Z}_{q}. \end{aligned}$$

(46)

A LR-bipolar fuzzy complex number vector

$$\begin{aligned} \widetilde{H}=(\widetilde{h}_{1},\widetilde{h}_{2},\ldots ,\widetilde{h}_{n})^{T}, \end{aligned}$$

(47)

where $\widetilde{h}_{q}=\widetilde{u}_{q} + i \widetilde{v}_{q}$, $1 \le q \le n$ is called a LR-bipolar fuzzy complex solution of the LR-CBFL system given in (45), if $\widetilde{H}$ satisfies (46).

Definition 22

An arbitrary LR-bipolar fuzzy complex vector should be represented as $\widetilde{h}_{q}=\widetilde{u}_{q} +i\widetilde{v}_{q}$, where $\widetilde{u}_{q}=\prec [u_{q}^{*},u_{lq}^{*},u_{rq}^{*}]_{LR},[u_{q}^{\bullet },u_{lq}^{\bullet },u_{rq}^{\bullet }]_{LR}\succ$ and $\widetilde{v}_{q}=\prec [v_{q}^{*},v_{lq}^{*},v_{rq}^{*}]_{LR},[v_{q}^{\bullet },v_{lq}^{\bullet },v_{rq}^{\bullet }]_{LR}\succ$ are two LR-bipolar fuzzy number vectors. The LR-bipolar fuzzy complex vector can be written as:

$$\begin{aligned} \widetilde{H}_{q}= & {} \prec [u_{q}^{*},u_{lq}^{*},u_{rq}^{*}]_{LR},[u_{q}^{\bullet },u_{lq}^{\bullet },u_{rq}^{\bullet }]_{LR}\succ \\&+i\prec [v_{q}^{*},v_{lq}^{*},v_{rq}^{*}]_{LR},[v_{q}^{\bullet },v_{lq}^{\bullet },v_{rq}^{\bullet }]_{LR}\succ . \end{aligned}$$

Theorem 4

The $n \times n$ LR-CBFL system of equations with complex coefficients in (45) is equivalent to pair of $2n \times 2n$ order LR-bipolar fuzzy linear system

$$\begin{aligned} M\widetilde{h}_{q}^{*}=\widetilde{z}_{q}^{*},~~~~~M\widetilde{h}_{q}^{\bullet }=\widetilde{z}_{q}^{\bullet }, \end{aligned}$$

(48)

where,

$$\begin{aligned} M&=\left( \begin{array}{r@{\quad }r} L_{1} &{} -L_{2} \\ L_{2} &{} L_{1} \\ \end{array} \right) , ~~~~~\widetilde{h}_{q}^{*}=\left( \begin{array}{c} \widetilde{u}_{q}^{*} \\ \widetilde{v}_{q}^{*} \\ \end{array} \right) , ~~~~~ \widetilde{z}_{q}^{*}=\left( \begin{array}{c} \widetilde{e}_{q}^{*} \\ \widetilde{f}_{q}^{*} \\ \end{array} \right) ,\nonumber \\ {h}_{q}^{*}&=\left( \begin{array}{c} \widetilde{u}_{q}^{*} \\ \widetilde{v}_{q}^{*} \\ \end{array} \right) , ~~~~~ \widetilde{z}_{q}^{*}=\left( \begin{array}{c} \widetilde{e}_{q}^{*} \\ \widetilde{f}_{q}^{*} \\ \end{array} \right) . \end{aligned}$$

(49)

Proof

We defined $L=L_{1}+i L_{2}$ where $L_{1}, L_{2} \epsilon \mathfrak {R}$ and $\widetilde{z}_{q}= \widetilde{e}_{q}+i\widetilde{f}_{q}$ where $\widetilde{e}_{q}$ and $\widetilde{f}_{q}$ are LR-bipolar fuzzy number vector. We also suppose the unknown vector $\widetilde{H}_{q}=\widetilde{u}_{q}+i\widetilde{v}_{q}$, where $\widetilde{u}_{q}$ and $\widetilde{v}_{q}$ are two unknown LR-bipolar fuzzy number vectors.

First we take the positive$(*)$ part of the system given in (46), i.e.,

$$\begin{aligned} L\widetilde{H}_{q}^{*}=\widetilde{z}_{q}^{*},\end{aligned}$$

the above-mentioned equation can also be written as

$$\begin{aligned} (L_{1}+i L_{2})(\widetilde{u}_{q}^{*}+i\widetilde{v}_{q}^{*})=\widetilde{e}_{q}^{*}+i\widetilde{f}_{q}^{*}.\end{aligned}$$

That is,

$$\begin{aligned} (L_{1}\widetilde{u}_{q}^{*}-L_{2}\widetilde{v}_{q}^{*})+i(L_{1}\widetilde{v}_{q}^{*}+L_{2}\widetilde{u}_{q}^{*})=\widetilde{e}_{q}^{*}+i\widetilde{f}_{q}^{*}.\end{aligned}$$

Compare real and complex coefficients of above equation we get,

$$\begin{aligned} L_{1}\widetilde{u}_{q}^{*}-L_{2}\widetilde{v}_{q}^{*}&=\widetilde{e}_{q}^{*},\nonumber \\ L_{1}\widetilde{v}_{q}^{*}+L_{2}\widetilde{u}_{q}^{*}&=\widetilde{f}_{q}^{*}. \end{aligned}$$

(50)

The matrix form of above-mentioned system is give as

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} L_{1} &{} -L_{2} \\ L_{2} &{} L_{1} \\ \end{array} \right) \left( \begin{array}{c} \widetilde{u}_{q}^{*} \\ \widetilde{v}_{q}^{*} \\ \end{array} \right) =\left( \begin{array}{c} \widetilde{e}_{q}^{*} \\ \widetilde{f}_{q}^{*} \\ \end{array} \right) , \end{aligned}$$

(51)

which is $2n \times 2n$ order LR-BFLS equations.

we can express it in matrix form as:

$$\begin{aligned} M\widetilde{h}_{q}^{*}=\widetilde{z}_{q}^{*}.\end{aligned}$$

Similarly, we can take the negative $(\bullet )$ part of the system given in (46), i.e.,

$$\begin{aligned} L\widetilde{H}_{q}^{\bullet }=\widetilde{z}_{q}^{\bullet },\end{aligned}$$

the above-mentioned equation can also be written as

$$\begin{aligned} (L_{1}+i L_{2})(\widetilde{u}_{q}^{\bullet }+i\widetilde{v}_{q}^{\bullet })=\widetilde{e}_{q}^{\bullet }+i\widetilde{f}_{q}^{\bullet }.\end{aligned}$$

That is,

$$\begin{aligned} (L_{1}\widetilde{u}_{q}^{\bullet }-L_{2}\widetilde{v}_{q}^{\bullet })+i(L_{1}\widetilde{v}_{q}^{\bullet }+L_{2}\widetilde{u}_{q}^{\bullet })=\widetilde{e}_{q}^{\bullet }+i\widetilde{f}_{q}^{\bullet }.\end{aligned}$$

Compare real and complex coefficients of above equation we get,

$$\begin{aligned} L_{1}\widetilde{u}_{q}^{\bullet }-L_{2}\widetilde{v}_{q}^{\bullet }&=\widetilde{e}_{q}^{\bullet },\nonumber \\ L_{1}\widetilde{v}_{q}^{\bullet }+L_{2}\widetilde{u}_{q}^{\bullet }&=\widetilde{f}_{q}^{\bullet }. \end{aligned}$$

(52)

The matrix form of above-mentioned system is give as

$$\begin{aligned} \left( \begin{array}{r@{\quad }r} L_{1} &{} -L_{2} \\ L_{2} &{} L_{1} \\ \end{array} \right) \left( \begin{array}{c} \widetilde{u}_{q}^{\bullet } \\ \widetilde{v}_{q}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{c} \widetilde{e}_{q}^{\bullet } \\ \widetilde{f}_{q}^{\bullet } \\ \end{array} \right) , \end{aligned}$$

(53)

which is $2n \times 2n$ order LR-BFLS of equation.

we can express it in matrix form as:

$$\begin{aligned} M\widetilde{h}_{q}^{\bullet }=\widetilde{z}_{q}^{\bullet }.\end{aligned}$$

$\square$

Theorem 5

The LR-bipolar fuzzy linear systems in (48) can be extended into the following linear system of matrix equations:

$$\begin{aligned}&(M^{+}+M^{-})h_{q}^{*}=z_{q}^{*},~~~~~(M^{+}+M^{-})h_{q}^{\bullet }=z_{q}^{\bullet },\\&\left( \begin{array}{r@{\quad }r} M^{+} &{} -M^{-} \\ -M^{-} &{} M^{+} \\ \end{array} \right) \left( \begin{array}{c} h_{lq}^{*} \\ h_{rq}^{*} \\ \end{array} \right) =\left( \begin{array}{c} z_{lq}^{*} \\ z_{rq}^{*} \\ \end{array} \right) ,\\&\left( \begin{array}{r@{\quad }r} M^{+} &{} -M^{-} \\ -M^{-} &{} M^{+} \\ \end{array} \right) \left( \begin{array}{c} h_{lq}^{\bullet } \\ h_{rq}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{c} z_{lq}^{\bullet } \\ z_{rq}^{\bullet } \\ \end{array} \right) ,\end{aligned}$$

where

$$\begin{aligned} M= & {} \left( \begin{array}{r@{\quad }r} M^{+} &{} -M^{-} \\ -M^{-} &{} M^{+} \\ \end{array} \right) = M^{+}+M^{-},\\ \widetilde{h}_{q}= & {} \prec [h_{q}^{*},h_{lq}^{*},h_{rq}^{*}]_{LR},[h_{q}^{\bullet },h_{lq}^{\bullet },h_{rq}^{\bullet }]_{LR}\succ \\= & {} \prec \left( \begin{array}{l@{\quad }l@{\quad }c} \left( \begin{array}{c} u_{q}^{*} \\ v_{q}^{*} \\ \end{array} \right) &{} \left( \begin{array}{c} u_{lq}^{*} \\ v_{lq}^{*} \\ \end{array} \right) &{} \left( \begin{array}{c} u_{rq}^{*} \\ v_{rq}^{*} \\ \end{array} \right) \\ \end{array} \right) ,\\&\quad \left( \begin{array}{l@{\quad }l@{\quad }c} \left( \begin{array}{c} u_{q}^{\bullet } \\ v_{q}^{\bullet } \\ \end{array} \right) &{} \left( \begin{array}{c} u_{lq}^{\bullet } \\ v_{lq}^{\bullet } \\ \end{array} \right) &{} \left( \begin{array}{c} u_{rq}^{\bullet } \\ v_{rq}^{\bullet } \\ \end{array} \right) \\ \end{array} \right) \succ , \\ \widetilde{z}_{q}= & {} \prec [z_{q}^{*},z_{lq}^{*},z_{rq}^{*}]_{LR},[z_{q}^{\bullet },z_{lq}^{\bullet },z_{rq}^{\bullet }]_{LR}\succ \\= & {} \prec \left( \begin{array}{l@{\quad }l@{\quad }c} \left( \begin{array}{c} e_{q}^{*} \\ f_{q}^{*} \\ \end{array} \right) &{} \left( \begin{array}{c} e_{lq}^{*} \\ f_{lq}^{*} \\ \end{array} \right) &{} \left( \begin{array}{c} e_{rq}^{*} \\ f_{rq}^{*} \\ \end{array} \right) \\ \end{array} \right) ,\\&\quad \left( \begin{array}{l@{\quad }l@{\quad }c} \left( \begin{array}{c} e_{q}^{\bullet } \\ f_{q}^{\bullet } \\ \end{array} \right) &{} \left( \begin{array}{c} e_{lq}^{\bullet } \\ f_{lq}^{\bullet } \\ \end{array} \right) &{} \left( \begin{array}{c} e_{rq}^{\bullet } \\ f_{rq}^{\bullet } \\ \end{array} \right) \\ \end{array} \right) \succ , \end{aligned}$$

where the elements $m_{pq}^{+}$ and $m_{pq}^{-}$ are determined as follow: if $m_{pq}\ge 0$, $m_{pq}^{+}=m_{pq}$ otherwise $m_{pq}^{+}=0$, $1 \le p,q \le 2n$; if $m_{pq} < 0$, $m_{pq}^{-}=m_{pq}$ otherwise $m_{pq}^{-}=0$, $1 \le p,q \le 2n$.

Proof

We denote

$$\begin{aligned} \widetilde{H}_{q}&=\widetilde{u}_{q}+i\widetilde{v}_{q}=\prec [u_{q}^{*},u_{lq}^{*},u_{rq}^{*}]_{LR},[u_{q}^{\bullet },u_{lq}^{\bullet },u_{rq}^{\bullet }]_{LR}\succ \\&\quad +i\prec [v_{q}^{*},v_{lq}^{*},v_{rq}^{*}]_{LR},[v_{q}^{\bullet },v_{lq}^{\bullet },v_{rq}^{\bullet }]_{LR}\succ ,\\ \widetilde{z}_{q}&=\widetilde{u}_{q}+i\widetilde{v}_{q}=\prec [e_{q}^{*},e_{lq}^{*},e_{rq}^{*}]_{LR},[e_{q}^{\bullet },e_{lq}^{\bullet },e_{rq}^{\bullet }]_{LR}\succ \\&\quad +i\prec [f_{q}^{*},f_{lq}^{*},f_{rq}^{*}]_{LR},[f_{q}^{\bullet },f_{lq}^{\bullet },f_{rq}^{\bullet }]_{LR}\succ , \end{aligned}$$

where $u_{q}^{*}, v_{q}^{*}$ are the mean values and $u_{lq}^{*}, v_{lq}^{*}, u_{rq}^{*}$ and $v_{rq}^{*}$ are left-right spread values of LR-bipolar fuzzy number vector $\widetilde{u}_{q},\widetilde{v}_{q}$, respectively.

First we take the positive$(*)$ part of LR-bipolar fuzzy linear system $M\widetilde{h}_{q}^{*}=\widetilde{z}_{q}^{*}$ is

$$\begin{aligned} M(h_{q}^{*},h_{lq}^{*},h_{rq}^{*})_{LR}=(z_{q}^{*},z_{lq}^{*},z_{rq}^{*})_{LR}. \end{aligned}$$

(54)

Let

$$\begin{aligned} M=\left( \begin{array}{r@{\quad }r} L_{1} &{} -L_{2} \\ L_{2} &{} L_{1} \\ \end{array} \right) =M^{+}+M^{-}, \end{aligned}$$

(55)

where $m_{pq}^{+}$ and $m_{pq}^{-}$ are determined as follows: if $m_{pq}\ge 0$, $m_{pq}^{+}=m_{pq}$ otherwise $m_{pq}^{+}=0$, $1 \le p,q \le 2n$; if $m_{pq} < 0$, $m_{pq}^{-}=m_{pq}$ other $m_{pq}^{-}=0$, $1 \le p,q \le 2n$.

Since

$$\begin{aligned} c\widetilde{h}_{q}= {\left\{ \begin{array}{ll} (ch,ch_{lq},ch_{rq})_{LR},&{} ~~ c\ge 0,\\ (ch,-ch_{rq},-ch_{lq})_{LR},&{}~~c \le 0, \end{array}\right. } \end{aligned}$$

(56)

we have

$$\begin{aligned} G\widetilde{h}_{q}^{*}= {\left\{ \begin{array}{ll} (Gh_{q}^{*},Gh_{lq}^{*},Gh_{rq}^{*})_{LR}, &{} ~~ G\ge 0,\\ (Gh_{q}^{*},-Gh_{rq}^{*},-Gh_{lq}^{*})_{LR}, &{}~~ G\le 0. \end{array}\right. } \end{aligned}$$

(57)

Equation (54) becomes

$$\begin{aligned} (M^{+}+M^{-})(h_{q}^{*},h_{lq}^{*},h_{rq}^{*})_{LR}=(z_{q}^{*},z_{lq}^{*},z_{rq}^{*})_{LR},\end{aligned}$$

which is equivalent to,

$$\begin{aligned}&M^{+}(h_{q}^{*},h_{lq}^{*},h_{rq}^{*})_{LR}+M^{-}(h_{q}^{*},h_{lq}^{*},h_{rq}^{*})_{LR}\\&\quad =(z_{q}^{*},z_{lq}^{*},z_{rq}^{*})_{LR},\\&(M^{+}h_{q}^{*},M^{+}h_{lq}^{*},M^{+}h_{rq}^{*})_{LR}+(M^{-}h_{q}^{*},-M^{-}h_{rq}^{*},-M^{-}h_{lq}^{*})_{LR}\\&\quad =(z_{q}^{*},z_{lq}^{*},z_{rq}^{*})_{LR},\\&(M^{+}h_{q}^{*}+M^{-}h_{q}^{*},M^{+}h_{lq}^{*}-M^{-}h_{rq}^{*},M^{+}h_{rq}^{*}-M^{-}h_{lq}^{*})_{LR}\\&\quad =(z_{q}^{*},z_{lq}^{*},z_{rq}^{*})_{LR}, \end{aligned}$$

by comparing we have,

$$\begin{aligned} (M^{+}+M^{-})h_{q}^{*}=z_{q}^{*},\nonumber \\ M^{+}h_{lq}^{*}-M^{-}h_{rq}^{*}=z_{lq}^{*},\nonumber \\ M^{+}h_{rq}^{*}-M^{-}h_{lq}^{*}=z_{rq}^{*}. \end{aligned}$$

(58)

Representing the system of Eq. (58) in matrix form, we have

$$\begin{aligned}&(M^{+}+M^{-})\left( \begin{array}{c} u_{q}^{*} \\ v_{q}^{*} \\ \end{array} \right) =\left( \begin{array}{c} e_{q}^{*} \\ f_{q}^{*} \\ \end{array} \right) , \\&\left( \begin{array}{r@{\quad }r} M^{+} &{} -M^{-} \\ -M^{-} &{} M^{+} \\ \end{array} \right) \left( \begin{array}{c} u_{lq}^{*} \\ v_{lq}^{*} \\ u_{rq}^{*} \\ v_{rq}^{*} \\ \end{array} \right) =\left( \begin{array}{c} e_{lq}^{*} \\ f_{lq}^{*} \\ e_{rq}^{*} \\ f_{rq}^{*} \\ \end{array} \right) . \end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of LR-bipolar fuzzy linear system $M\widetilde{h}_{q}^{\bullet }=\widetilde{z}_{q}^{\bullet }$ is

$$\begin{aligned} M(h_{q}^{\bullet },h_{lq}^{\bullet },h_{rq}^{\bullet })_{LR}=(z_{q}^{\bullet },z_{lq}^{\bullet },z_{rq}^{\bullet })_{LR}. \end{aligned}$$

(59)

Let

$$\begin{aligned} M=\left( \begin{array}{r@{\quad }r} L_{1} &{} -L_{2} \\ L_{2} &{} L_{1} \\ \end{array} \right) =M^{+}+M^{-}, \end{aligned}$$

(60)

where $m_{pq}^{+}$ and $m_{pq}^{-}$ are determined as follows: if $m_{pq}\ge 0$, $m_{pq}^{+}=m_{pq}$ otherwise $m_{pq}^{+}=0$, $1 \le p,q \le 2n$; if $m_{pq} < 0$, $m_{pq}^{-}=m_{pq}$ otherwise $m_{pq}^{-}=0$, $1 \le p,q \le 2n$.

Fig. 1

A circuit with LR-BFCN source and current

Since

$$\begin{aligned} c\widetilde{h}_{q}= {\left\{ \begin{array}{ll} (ch,ch_{lq},ch_{rq})_{LR},&{} ~~ c\ge 0,\\ (ch,-ch_{rq},-ch_{lq})_{LR},&{}~~c \le 0, \end{array}\right. } \end{aligned}$$

(61)

we have

$$\begin{aligned} G\widetilde{h}_{q}^{\bullet }= {\left\{ \begin{array}{ll} (Gh_{q}^{\bullet },Gh_{lq}^{\bullet },Gh_{rq}^{\bullet })_{LR}, &{} ~~ G\ge 0,\\ (Gh_{q}^{\bullet },-Gh_{rq}^{\bullet },-Gh_{lq}^{\bullet })_{LR}, &{}~~ G\le 0. \end{array}\right. } \end{aligned}$$

(62)

Equation (59) becomes

$$\begin{aligned} (M^{+}+M^{-})(h_{q}^{\bullet },h_{lq}^{\bullet },h_{rq}^{\bullet })_{LR}=(z_{q}^{\bullet },z_{lq}^{\bullet },z_{rq}^{\bullet })_{LR},\end{aligned}$$

which is equivalent to,

$$\begin{aligned}&M^{+}(h_{q}^{*},h_{lq}^{\bullet },h_{rq}^{\bullet })_{LR}+M^{-}(h_{q}^{\bullet },h_{lq}^{\bullet },h_{rq}^{\bullet })_{LR}\\&\quad =(z_{q}^{\bullet },z_{lq}^{\bullet },z_{rq}^{\bullet })_{LR},\\&(M^{+}h_{q}^{\bullet },M^{+}h_{lq}^{\bullet },M^{+}h_{rq}^{\bullet })_{LR}\\&\qquad +(M^{-}h_{q}^{\bullet },-M^{-}h_{rq}^{\bullet },-M^{-}h_{lq}^{\bullet })_{LR}\\&\quad =(z_{q}^{\bullet },z_{lq}^{\bullet },z_{rq}^{\bullet })_{LR},\\&(M^{+}h_{q}^{\bullet }+M^{-}h_{q}^{\bullet },M^{+}h_{lq}^{\bullet }-M^{-}h_{rq}^{\bullet },M^{+}h_{rq}^{\bullet }-M^{-}h_{lq}^{\bullet })_{LR}\\&\quad =(z_{q}^{\bullet },z_{lq}^{\bullet },z_{rq}^{\bullet })_{LR},\end{aligned}$$

by comparing we have,

$$\begin{aligned} (M^{+}+M^{-})h_{q}^{\bullet }=z_{q}^{\bullet },\nonumber \\ M^{+}h_{lq}^{\bullet }-M^{-}h_{rq}^{\bullet }=z_{lq}^{\bullet },\nonumber \\ M^{+}h_{rq}^{\bullet }-M^{-}h_{lq}^{\bullet }=z_{rq}^{\bullet }. \end{aligned}$$

(63)

Representing the system of Eq. (63) in matrix form, we have

$$\begin{aligned}&(M^{+}+M^{-})\left( \begin{array}{c} u_{q}^{\bullet } \\ v_{q}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{c} e_{q}^{\bullet } \\ f_{q}^{\bullet } \\ \end{array} \right) , \\&\left( \begin{array}{r@{\quad }r} M^{+} &{} -M^{-} \\ -M^{-} &{} M^{+} \\ \end{array} \right) \left( \begin{array}{c} u_{lq}^{\bullet } \\ v_{lq}^{\bullet } \\ u_{rq}^{\bullet } \\ v_{rq}^{\bullet } \end{array} \right) =\left( \begin{array}{c} e_{lq}^{\bullet } \\ f_{lq}^{\bullet } \\ e_{rq}^{\bullet } \\ f_{rq}^{\bullet } \\ \end{array} \right) . \end{aligned}$$

$\square$

Example 5

We consider a simple RLC circuit (Rahgooy et al. 2009) in which current and source all are in terms of LR-BFCN. A circuit with LR-BFCN source and current is shown in Figure 1. The source and current are in terms of bipolar fuzzy number, the positive part shows likelihood of source and current and negative part shows unlikelihood of source and current.

We use Kirchhoff’s second law for the above circuit and write the linear equations for first and second loop is given as:

$$\begin{aligned} {\left\{ \begin{array}{ll} (10-7.5i)I_{1} - (6-5i)I_{2}=\prec [5,1,1]_{LR},[4,1,1]_{LR}\succ \\ \quad +i\prec [0,1,1]_{LR},[-1,1,1]_{LR}\succ ,\\ -(6-5i)I_{1} + (16+3i)I_{2}=\prec [-1,,1,1]_{LR},\\ \quad [-2,1,1]_{LR}\succ +i\prec [-2,1,1]_{LR},[-3,1,1]_{LR}\succ . \end{array}\right. } \end{aligned}$$

We can put $\widetilde{I}_{1}=\widetilde{u}_{1}+i\widetilde{v}_{1} =\prec [u_{1}^{*},u_{l1}^{*}, u_{r1}^{*}]_{LR}, [u_{1}^{\bullet }, u_{l1}^{\bullet }, u_{r1}^{\bullet }]_{LR}\succ +i\prec [v_{1}^{*},v_{l1}^{*}, v_{r1}^{*}]_{LR},[v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ$, and $\widetilde{I}_{2}=\widetilde{u}_{2}+i\widetilde{v}_{2}=\prec [u_{2}^{*},u_{l2}^{*}, u_{r2}^{*}]_{LR},[u_{2}^{\bullet }, u_{l2}^{\bullet },u_{r2}^{\bullet }]_{LR}\succ +i\prec [v_{2}^{*},v_{l2}^{*}, v_{r2}^{*}]_{LR},[v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet }]_{LR}\succ$, in above-mentioned system then system become,

$(10-7.5i)\{\prec [u_{1}^{*},u_{l1}^{*},u_{r1}^{*}]_{LR},[u_{1}^{\bullet }, u_{l1}^{\bullet },u_{r1}^{\bullet }]_{LR}\succ +i\prec [v_{1}^{*},v_{l1}^{*},v_{r1}^{*}]_{LR}, [v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ \}-(6-5i)$ $\{\prec [u_{2}^{*}, u_{l2}^{*}, u_{r2}^{*}]_{LR},[u_{2}^{\bullet },u_{l2}^{\bullet }, u_{r2}^{\bullet }]_{LR}\succ + i\prec [v_{2}^{*},v_{l2}^{*}, v_{r2}^{*}]_{LR}, [v_{2}^{\bullet }, v_{l2}^{\bullet }, v_{r2}^{\bullet }]_{LR}\succ \} =\prec [5,1,1]_{LR}, [4,1,1]_{LR}\succ +i\prec [0,1,1]_{LR},[-1,1,1]_{LR}\succ ,$ $-(6-5i)\{\prec [u_{1}^{*},u_{l1}^{*},u_{r1}^{*}]_{LR},[u_{1}^{\bullet },u_{l1}^{\bullet }, u_{r1}^{\bullet }]_{LR}\succ +i\prec [v_{1}^{*},v_{l1}^{*},v_{r1}^{*}]_{LR},[v_{1}^{\bullet }, v_{l1}^{\bullet },v_{r1}^{\bullet }]_{LR}\succ \}+(16+3i)$ $\{\prec [u_{2}^{*},u_{l2}^{*}, u_{r2}^{*}]_{LR},[u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet }]_{LR}\succ +i\prec [v_{2}^{*}, v_{l2}^{*}, v_{r2}^{*}]_{LR},[v_{2}^{\bullet },v_{l2}^{\bullet }, v_{r2}^{\bullet }]_{LR}\succ \}=\prec [-1,,1,1]_{LR},[-2,1,1]_{LR}\succ +i\prec [-2,1,1]_{LR}, [-3,1,1]_{LR}\succ .$

First, we take the positive$(*)$ part of the above-mentioned system of equations,

$(10-7.5i)\{(u_{1}^{*},u_{l1}^{*},u_{r1}^{*})_{LR}+i(v_{1}^{*},v_{l1}^{*},v_{r1}^{*})_{LR}\} - (6-5i)\{(u_{2}^{*},u_{l2}^{*},u_{r2}^{*})_{LR}+i(v_{2}^{*},v_{l2}^{*},v_{r2}^{*})_{LR}\}=(5,1,1)_{LR}+i(0,1,1)_{LR},$

$-(6-5i)\{(u_{1}^{*},u_{l1}^{*},u_{r1}^{*})_{LR}+i(v_{1}^{*},v_{l1}^{*},v_{r1}^{*})_{LR}\} + (16+3i)\{(u_{2}^{*},u_{l2}^{*},u_{r2}^{*})_{LR}+i(v_{2}^{*},v_{l2}^{*},v_{r2}^{*})_{LR}\}=(-1,,1,1)_{LR}+i(-2,1,1)_{LR}.$

According to Theorem (4), the LR-CBFL system of equation with complex coefficients is converted into positive part of LR-bipolar linear system $M\widetilde{I}_{q}^{*}=\widetilde{z}_{q}^{*}$, given as

$$\begin{aligned} \left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 10 &{} -6 &{} 7.5 &{} -5 \\ -6 &{} 16 &{} -5 &{} -3 \\ -7.5 &{} 5 &{} 10 &{} -6 \\ 5 &{} 3 &{} -6 &{} 16 \\ \end{array} \right) \left( \begin{array}{c} \widetilde{u}_{1}^{*} \\ \widetilde{u}_{2}^{*} \\ \widetilde{v}_{1}^{*} \\ \widetilde{v}_{2}^{*} \\ \end{array} \right) =\left( \begin{array}{c} (5,1,1)_{LR} \\ (-1,1,1)_{LR}\\ (0,1,1)_{LR} \\ (-2,1,1)_{LR} \\ \end{array} \right) . \end{aligned}$$

Applying Theorem (5), we get two linear system one is mean value system and other is left-right spread system given as:

$$\begin{aligned}&\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 10 &{} -6 &{} 7.5 &{} -5 \\ -6 &{} 16 &{} -5 &{} -3 \\ -7.5 &{} 5 &{} 10 &{} -6 \\ 5 &{} 3 &{} -6 &{} 16 \\ \end{array} \right) \left( \begin{array}{c} u_{1}^{*} \\ u_{2}^{*} \\ v_{1}^{*} \\ v_{2}^{*} \\ \end{array} \right) =\left( \begin{array}{c} 5 \\ -1 \\ 0 \\ -2 \\ \end{array} \right) . \\&\left( \begin{array}{cccccccc} 10 &{} 0 &{} 7.5 &{} 0 &{} 0 &{} 6 &{} 0 &{} 5 \\ 0 &{} 16 &{} 0 &{} 0 &{} 6 &{} 0 &{} 5 &{} 3 \\ 0 &{} 5 &{} 10 &{} 0 &{} 7.5 &{} 0 &{} 0 &{} 6 \\ 5 &{} 3 &{} 0 &{} 16 &{} 0 &{} 0 &{} 6 &{} 0 \\ 0 &{} 6 &{} 0 &{} 5 &{} 10 &{} 0 &{} 7.5 &{} 0 \\ 6 &{} 0 &{} 5 &{} 3 &{} 0 &{} 16 &{} 0 &{} 0 \\ 7.5 &{} 0 &{} 0 &{} 6 &{} 0 &{} 5 &{} 10 &{} 0 \\ 0 &{} 0 &{} 6 &{} 0 &{} 5 &{} 3 &{} 0 &{} 16 \\ \end{array} \right) \left( \begin{array}{c} u_{l1}^{*} \\ u_{l2}^{*} \\ v_{l1}^{*} \\ v_{l2}^{*} \\ u_{r1}^{*} \\ u_{r2}^{*} \\ v_{r1}^{*} \\ v_{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right) . \end{aligned}$$

Solution for mean value system is given as

$$\begin{aligned} \left( \begin{array}{c} u_{1}^{*} \\ u_{2}^{*} \\ v_{1}^{*} \\ v_{2}^{*} \\ \end{array} \right)= & {} \left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 0.0876 &{} 0.0389 &{} -0.0328 &{} 0.0224 \\ 0.0389 &{} 0.0640 &{} 0.0224 &{} -0.0325 \\ 0.0328 &{} -0.0224 &{} 0.0876 &{} 0.0389 \\ -0.0224 &{} -0.0325 &{} 0.0389 &{} 0.0640 \\ \end{array} \right) \left( \begin{array}{c} 5 \\ -1 \\ 0 \\ -2 \\ \end{array} \right) \\= & {} \left( \begin{array}{c} 0.3542 \\ 0.0654 \\ 0.1086 \\ -0.2072 \\ \end{array} \right) . \end{aligned}$$

Solution for left-right spread system given as

$$\begin{aligned}&\left( \begin{array}{c} u_{l1}^{*} \\ u_{l2}^{*} \\ v_{l1}^{*} \\ v_{l2}^{*} \\ u_{r1}^{*} \\ u_{r2}^{*} \\ v_{r1}^{*} \\ v_{r2}^{*} \\ \end{array} \right) =\left( \begin{array}{cccccccc} 0.1694 &{} -0.0015 &{} -0.0971 &{} 0.0061 &{} 0.0819 &{} -0.0404 &{} -0.0643 &{} -0.0612 \\ -0.0015 &{} 0.0725 &{} 0.0061 &{} 0.0171 &{} -0.0404 &{} 0.0085 &{} -0.0162 &{} -0.0154 \\ -0.0643 &{} -0.0162 &{} 0.1694 &{} -0.0015 &{} -0.971 &{} 0.0061 &{} 0.0819 &{} -0.0404 \\ -0.0162 &{} -0.0154 &{} -0.0015 &{} 0.0725 &{} 0.0061 &{} 0.0171 &{} -0.0404 &{} 0.0085 \\ 0.0819 &{} -0.0404 &{} -0.0643 &{} -0.0162 &{} 0.1694 &{} -0.0015 &{} -0.0971 &{} 0.0061 \\ -0.0404 &{} 0.0085 &{} -0.0162 &{} -0.0154 &{} -0.0015 &{} 0.0725 &{} 0.0061 &{} 0.0171 \\ -0.0971 &{} 0.0061 &{} 0.0819 &{} -0.0404 &{} -0.0643 &{} -0.0162 &{} 0.1694 &{} -0.0015 \\ -0.0061 &{} 0.0171 &{} -0.0404 &{} 0.0085 &{} -0.0162 &{} -0.0154 &{} -0.0015 &{} 0.0725 \\ \end{array} \right) \left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right) , \\&\left( \begin{array}{c} u_{l1}^{*} \\ u_{l2}^{*} \\ v_{l1}^{*} \\ v_{l2}^{*} \\ u_{r1}^{*} \\ u_{r2}^{*} \\ v_{r1}^{*} \\ v_{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 0.0378 \\ 0.0307 \\ 0.0378 \\ 0.0307 \\ 0.0378 \\ 0.0307 \\ 0.0378 \\ 0.0307 \\ \end{array} \right) . \end{aligned}$$

Similarly, we can take the negative$(\bullet )$ part of the LR-CBFL system of equations with complex coefficients,

$(10-7.5i)\{(u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet })_{LR}+i(v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet })_{LR}\} - (6-5i)\{(u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet })_{LR}+i(v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet })_{LR}\}=(4,1,1)_{LR}+i(-1,1,1)_{LR},$

$-(6-5i)\{(u_{1}^{\bullet },u_{l1}^{\bullet },u_{r1}^{\bullet })_{LR}+i(v_{1}^{\bullet },v_{l1}^{\bullet },v_{r1}^{\bullet })_{LR}\} + (16+3i)\{(u_{2}^{\bullet },u_{l2}^{\bullet },u_{r2}^{\bullet })_{LR}+i(v_{2}^{\bullet },v_{l2}^{\bullet },v_{r2}^{\bullet })_{LR}\}=(-2,1,1)_{LR}+i(-3,1,1)_{LR}.$

According to Theorem (4), the LR-CBFL system of equation with complex coefficients is converted into negative part of LR-bipolar linear system $M\widetilde{I}_{q}^{\bullet }=\widetilde{z}_{q}^{\bullet }$, given as

$$\begin{aligned} \left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 10 &{} -6 &{} 7.5 &{} -5 \\ -6 &{} 16 &{} -5 &{} -3 \\ -7.5 &{} 5 &{} 10 &{} -6 \\ 5 &{} 3 &{} -6 &{} 16 \\ \end{array} \right) \left( \begin{array}{c} \widetilde{u}_{1}^{\bullet } \\ \widetilde{u}_{2}^{\bullet } \\ \widetilde{v}_{1}^{\bullet } \\ \widetilde{v}_{2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{c} (4,1,1)_{LR} \\ (-2,1,1)_{LR} \\ (-1,1,1)_{LR} \\ (-3,1,1)_{LR} \\ \end{array} \right) . \end{aligned}$$

Applying Theorem (5), we get two linear system one is mean value system and other is left-right spread system given as:

$$\begin{aligned}&\left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 10 &{} -6 &{} 7.5 &{} -5 \\ -6 &{} 16 &{} -5 &{} -3 \\ -7.5 &{} 5 &{} 10 &{} -6 \\ 5 &{} 3 &{} -6 &{} 16 \\ \end{array} \right) \left( \begin{array}{c} u_{1}^{\bullet } \\ u_{2}^{\bullet } \\ v_{1}^{\bullet } \\ v_{2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{c} 4 \\ -2 \\ -1 \\ -3 \\ \end{array} \right) . \\&\left( \begin{array}{cccccccc} 10 &{} 0 &{} 7.5 &{} 0 &{} 0 &{} 6 &{} 0 &{} 5 \\ 0 &{} 16 &{} 0 &{} 0 &{} 6 &{} 0 &{} 5 &{} 3 \\ 0 &{} 5 &{} 10 &{} 0 &{} 7.5 &{} 0 &{} 0 &{} 6 \\ 5 &{} 3 &{} 0 &{} 16 &{} 0 &{} 0 &{} 6 &{} 0 \\ 0 &{} 6 &{} 0 &{} 5 &{} 10 &{} 0 &{} 7.5 &{} 0 \\ 6 &{} 0 &{} 5 &{} 3 &{} 0 &{} 16 &{} 0 &{} 0 \\ 7.5 &{} 0 &{} 0 &{} 6 &{} 0 &{} 5 &{} 10 &{} 0 \\ 0 &{} 0 &{} 6 &{} 0 &{} 51 &{} 3 &{} 0 &{} 16 \\ \end{array} \right) \left( \begin{array}{c} u_{l1}^{\bullet } \\ u_{l2}^{\bullet } \\ v_{l1}^{\bullet } \\ v_{l2}^{\bullet } \\ u_{r1}^{\bullet } \\ u_{r2}^{\bullet } \\ v_{r1}^{\bullet } \\ v_{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right) . \end{aligned}$$

Solution for mean value system is given as

$$\begin{aligned} \left( \begin{array}{c} u_{1}^{\bullet } \\ u_{2}^{\bullet } \\ v_{1}^{\bullet } \\ v_{2}^{\bullet } \\ \end{array} \right)= & {} \left( \begin{array}{r@{\quad }r@{\quad }r@{\quad }r} 0.0876 &{} 0.0389 &{} -0.0328 &{} 0.0224 \\ 0.0389 &{} 0.0640 &{} 0.0224 &{} -0.0325 \\ 0.0328 &{} -0.0224 &{} 0.0876 &{} 0.0389 \\ -0.0224 &{} -0.0325 &{} 0.0389 &{} 0.0640 \\ \end{array} \right) \left( \begin{array}{c} 4 \\ -2 \\ -1 \\ -3 \\ \end{array} \right) \\ {}= & {} \left( \begin{array}{c} 0.2382 \\ -0.0924 \\ -0.0283 \\ -0.2552 \\ \end{array} \right) . \end{aligned}$$

Solution for left-right spread system given as

$$\begin{aligned}&\left( \begin{array}{c} u_{l1}^{\bullet } \\ u_{l2}^{\bullet } \\ v_{l1}^{\bullet } \\ v_{l2}^{\bullet } \\ u_{r1}^{\bullet } \\ u_{r2}^{\bullet } \\ v_{r1}^{\bullet } \\ v_{r2}^{\bullet } \\ \end{array} \right) =\left( \begin{array}{cccccccc} 0.1694 &{} -0.0015 &{} -0.0971 &{} 0.0061 &{} 0.0819 &{} -0.0404 &{} -0.0643 &{} -0.0612 \\ -0.0015 &{} 0.0725 &{} 0.0061 &{} 0.0171 &{} -0.0404 &{} 0.0085 &{} -0.0162 &{} -0.0154 \\ -0.0643 &{} -0.0162 &{} 0.1694 &{} -0.0015 &{} -0.971 &{} 0.0061 &{} 0.0819 &{} -0.0404 \\ -0.0162 &{} -0.0154 &{} -0.0015 &{} 0.0725 &{} 0.0061 &{} 0.0171 &{} -0.0404 &{} 0.0085 \\ 0.0819 &{} -0.0404 &{} -0.0643 &{} -0.0162 &{} 0.1694 &{} -0.0015 &{} -0.0971 &{} 0.0061 \\ -0.0404 &{} 0.0085 &{} -0.0162 &{} -0.0154 &{} -0.0015 &{} 0.0725 &{} 0.0061 &{} 0.0171 \\ -0.0971 &{} 0.0061 &{} 0.0819 &{} -0.0404 &{} -0.0643 &{} -0.0162 &{} 0.1694 &{} -0.0015 \\ -0.0061 &{} 0.0171 &{} -0.0404 &{} 0.0085 &{} -0.0162 &{} -0.0154 &{} -0.0015 &{} 0.0725 \\ \end{array} \right) \left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right) ,\\&\left( \begin{array}{c} u_{l1}^{*} \\ u_{l2}^{*} \\ v_{l1}^{*} \\ v_{l2}^{*} \\ u_{r1}^{*} \\ u_{r2}^{*} \\ v_{r1}^{*} \\ v_{r2}^{*} \\ \end{array} \right) = \left( \begin{array}{c} 0.0378 \\ 0.0307 \\ 0.0378 \\ 0.0307 \\ 0.0378 \\ 0.0307 \\ 0.0378 \\ 0.0307 \\ \end{array} \right) . \end{aligned}$$

Hence, the solution is LR-BFCN for the LR-CBFL system of equation with complex coefficients given as

$\widetilde{I}_{1}=\prec [0.3542,0.0378,0.0378]_{LR},[0.2382,0.0378,0.0378]_{LR}\succ +i\prec [0.1086,0.0378,0.0378]_{LR}$,

$[-0.0283,0.0378,0.0378]_{LR}\succ ,$

$\widetilde{I}_{2}=\prec [0.0654,0.0307,0.0307]_{LR},[-0.0924,0.0307,0.0307]_{LR}\succ +i\prec [-0.2072,0.0307,0.0307]_{LR}$,

$[-0.2552,0.0307,0.0307]_{LR}\succ ,$

The positive part of solution shows likelihood (in percentage) and negative part of solution shows unlikelihood (in percentage) of current in circuit.

5 Conclusions

In real-world problems, most of the applications in different areas are dealt with system of linear equations to find the unknown parameters. In different situations the unknown parameters of the systems are uncertain or vague. We represent the parameters in terms of BFN and BFCN which contain parameters r and s. We usually face the problems of calculation to solve these systems due to interruption of parameters r and s. We have discussed different methods to overcome the problems of parameters and solved the system in terms of LR-fuzzy numbers, LR-bipolar fuzzy numbers and LR-bipolar fuzzy complex numbers which are free from parameters r and s. We have discussed a technique to solve LR-BFLS of equation, LR-CBFL system with real coefficients and LR-CBFL system with complex coefficients. In the first section, we have discussed a technique to solve the LR-BFLS of equations in which the $n \times n$ coefficient matrix of the system is represented by real numbers and right side vector is represented by LR-BFN. To solve the LR-BFLS of equations we replace the original system into the pair of positive$(*)$ and negative$(\bullet )$ of two $n \times n$ LR-fuzzy linear system and solve these system we use mean values and left-right spread systems. In the next section, first we have discussed a technique to solve the LR-CBFL system with real coefficients in which we can replace $n \times n$ LR-CBFL system with real coefficients by pair of real and imaginary parts of $n \times n$ LR-BFLS of equation and then solve these systems by using mean value and left-right spread systems. Further, we have extended the technique (Guo and Zhang 2016) to solve the LR-CBFL system with complex coefficients in which the $n \times n$ coefficient matrix of the system is represented by regular complex numbers and right side vector is represented by LR-BFCN. To solve the system we can replace $n \times n$ LR-CBFL system with complex coefficients by pair of positive$(*)$ and negative$(\bullet )$ of two $2n \times 2n$ mean value system and $4n \times 4n$ left-right spread system. Some numerical examples have also been solved to show the effectiveness of the technique. We plan to extend our study to solve: (1) circuit analysis and models of artificial intelligence, (2) numerical solution of bipolar fuzzy initial value problems, (3) bipolar fuzzy linear programming problems, and (4) numerical solution of m-polar fuzzy initial value problems.