On Quasilinear Algebra of Linear Interval Equations and Interval Cramer’s Rule

Abstract

Determining the solution set of a system of linear interval equations is often a difficult task. Establishing a general theory that includes the classical theory of systems of linear equations as a special case opens the door to extensive and challenging research. In this study, we aim to develop results concerning the solution sets of such systems by employing the concept of quasilinear spaces. First, we define the determinant of an interval matrix as an interval and its rank as a pair of natural numbers. Then, we introduce the concept of a quasi-inverse for interval matrices and derive several results based on this notion. Using these results, we prove a theorem, which we call the interval Cramer’s rule, concerning the solutions of certain linear interval equation systems. In addition, with respect to the existence of solutions for this type of equation, we present a theorem related to the rank of the interval matrix that models the system.

1. Introduction

Problems involving limited uncertainty are commonly referred to as problems with inexact data and arise naturally in many areas of science and engineering. In numerous applications, such problems can be formulated as systems of linear interval equations. Interval matrices therefore play a fundamental role in the analysis and development of solution methods for linear systems affected by uncertainty.

As an illustrative example, consider neural network models. If the activation functions are unbounded, the existence of an equilibrium point cannot, in general, be guaranteed. In [1], the authors investigate the existence of a unique equilibrium point for neural networks, a property that is essential for the global robust asymptotic stability of the model. Such neural network models constitute a special class of nonlinear differential equations. Meaningful estimates of system behavior can be obtained only when the entries of the parameter matrices are known to lie within prescribed closed intervals. In this setting, interval matrices provide an appropriate mathematical framework for modeling uncertainty.

More generally, interval matrices are indispensable in the solution of linear systems with inexact or incomplete data. When uncertain or missing parameters can be confined to known intervals, the corresponding linear system can be modeled as a linear interval equation system.

To illustrate this approach, we consider Example 7.5 from [2]. The mesh equations of an electric circuit are given by

$$\left(\begin{array}{cc}{R}_1+R_2& -R_2\\ -R_2& R_2+R_3\end{array}\right)\left(\begin{array}{c}{I}_1\\ I_2\end{array}\right)=\left(\begin{array}{c}{V}_1\\ -V_2\end{array}\right).$$

Here, $V_1=10,$$V_2=5,$ and $R_1=R_2=R_3=1000\pm 10\%.$ The quantities $R_i$, $I_i$, and $V_i$ denote resistances, currents, and voltages, respectively. The objective is to determine interval enclosures for the currents $I_1$ and $I_2$ under a $10\%$ variation in the resistance values. Let us consider the interval matrix

$$\mathcal{A}=\left(\begin{array}{cc}\left[1800,2200\right]& \left[-1100,-900\right]\\ \left[-1100,-900\right]& \left[1800,2200\right]\end{array}\right)$$

and (interval) vectors

$$x=\left(\begin{array}{c}{I}_1\\ I_2\end{array}\right),\mathbf{b}=\left(\begin{array}{c}{V}_1\\ -V_2\end{array}\right).$$

Then, $\mathcal{A}x=\mathbf{b}$ is the linear-interval equation that will serve the solution of the above linear interval system. In this paper, we will try to solve this problem using the quasi-inverse concept and interval Cramer’s rule and we will see that our results are close to those obtained in Example 7.5.

In general, it is highly difficult to determine whether a solution of a system of linear-interval equations exists. The main reason for this is that the set of interval vectors that we will use to find the solution of such equations is not a vector space. So, it is hard to obtain a solution method in the exact same manner as in classical linear algebra. However, fortunately, the set of all interval vectors has an algebraic structure called quasilinear space which is a generalization of linear spaces. For this reason, we have to develop more general or newer concepts than classical linear algebra concepts. Further, they must be consistent with classical linear algebra concepts. The concept was first introduced by S. M. Aseev in [3]. However, some necessary concepts such as quasispan, quasilinear dependence–independence and basis were not given in this work. These definitions and the definition of the dimension of a quasilinear space are given in references [4,5]. However, we realized that we needed to change some of the nomenclature in the definitions we gave in these works. For example, in a quasilinear space, we called an element that has no inverse with respect to addition as a singular element. In linear spaces, since every element has an additive inverse, there is no such thing as an element being singular in these spaces. However, when dealing with matrices and interval matrices, the meaning of these elements being singular is entirely different. Since interval matrices act as operators between quasilinear spaces, the definitions of singularity for the elements of the space and for interval matrices will become intertwined. Therefore, modifying some definitions from our previous work, in this study, we will call an element in a quasilinear space a “foam” if it has no additive inverse, and a “stone” if it has an additive inverse.

Another important contribution to the algebraic structure of interval vectors, and more generally of sets, can be found in the work of S. Markov [6]. The study of parametric linear interval systems and the solution sets of parametric interval matrix equation systems belongs to the same class of problems. For further related results, the reader is referred to [7,8,9,10]. Further new definitions of the rank of an interval matrix have been studied in several different ways in the literature [11,12]. In contrast to these existing approaches, we begin by observing that an interval matrix naturally induces a quasilinear operator between quasilinear spaces. Consequently, it is necessary to first define the rank of a quasilinear operator. Prior to the introduction of the concept of a quasilinear space by Aseev [3], it was already known that interval matrices define transformations between interval vectors. However, once it was established that the set of interval vectors possesses the algebraic structure of a quasilinear space in the sense of Aseev, it follows naturally that interval matrices act as quasilinear operators between such spaces. As in linear algebra, it is necessary to introduce a notion of rank for quasilinear operators and, consequently, for interval matrices. Otherwise, inconsistencies arise with the classical definition of rank for ordinary matrices, which constitute a special case of interval matrices. In classical linear algebra, the rank of a linear operator is defined as the dimension of its image space (range). In the quasilinear setting, however, the range of a quasilinear operator must be treated differently. While the range of a linear operator is always a subspace, the image of a quasilinear operator need not be a subspace. This phenomenon is illustrated by a simple example in Remark 3 below. Nevertheless, the quasispan of the image set always forms a subspace. Therefore, in order to define the rank of a quasilinear operator, we consider the dimension of the subspace quasispanned by the range of the operator. Since the range of a linear operator is itself a subspace, the subspace spanned by its range coincides with the range, and thus no additional operation is required in the classical definition of rank. Accordingly, we define the rank of a quasilinear operator as the dimension of the space quasispanned by its range. When the quasilinear operator is linear, this definition coincides exactly with the classical notion of rank. The definition of the rank of quasilinear operators relies essentially on the concepts of quasispan and dimension of a quasilinear space, which were introduced in our earlier works [4,5]. In particular, the concept of dimension in quasilinear spaces must be defined in a manner compatible with the classical notion of dimension in linear spaces since every linear space is a quasilinear space under the equality relation. For other properties and applications related to quasilinear algebra, see [13,14,15,16]. In these works, the dimension of a quasilinear space was defined as an ordered pair of natural numbers. Based on this definition, we define the rank of a quasilinear operator and hence of an interval matrix as a pair of natural numbers.

In a similar manner to linear algebra, we will define the row and column rank of an interval matrix as the dimensions of the quasilinear spaces quasispanned by its row and column vectors, respectively. We will also present an example demonstrating that, for a non-degenerate interval matrix (that is, one which is not a classical matrix), the row rank and column rank may differ. If these two ranks coincide, we refer to their common value as the rank of the interval matrix. From this perspective, the notion of rank introduced here differs substantially from other existing definitions of the rank of an interval matrix. In this regard, Aseev’s definition of a quasilinear space proves to be more suitable than that of Markov, as it better captures the structural properties required in this context. Moreover, according to Aseev’s quasilinear operator conjecture, interval matrices can be viewed as quasilinear operators, a correspondence that is not present in Markov’s framework. Since such a correspondence is fundamental in linear algebra, this constitutes an additional advantage of Aseev’s approach.

Building on this framework, we introduce notions such as the determinant and the quasi-inverse of an interval matrix. These concepts enable us to derive an enclosure for the solution set of certain systems of linear interval equations. We refer to this result as the interval Cramer’s rule. When a solution exists, this rule provides a straightforward method for constructing an enclosure containing the solution set. The main challenge lies in obtaining an enclosure that is sufficiently narrow.

In the important reference [2], several authors derived reasonable enclosures for the solutions of certain simple linear interval equations using different techniques. For instance, in Example 7.5 of [2], a satisfactory enclosure for the solution set is obtained. Applying the interval Cramer’s rule developed in the present work to the same problem, we obtain an enclosure that is comparably narrow. This suggests that our method is effective in many cases.

Basic studies on the solution of systems of linear interval equations given by a square matrix are given in references [17,18,19]. However, earlier studies on the subject were given by Farkas [20] and Oettli [21]. Later, in [22,23], important contributions were made to the solution of linear systems of interval equations.

Some important system types for which we can use the results obtained in this study are known as parametric or fuzzy linear systems. These types of systems are systems in which each component of each parameter varies within a certain closed interval in multiple parameters. Therefore, they exhibit behavior similar to linear interval equation systems. In science and engineering, we model many scientific problems involving limited uncertainty using these types of systems and attempt to develop solution methods. In recent years, there have been many studies that have been modeled using these types of systems (see [24,25,26]). Furthermore, to see the most recent studies regarding interval analysis and linear interval equations, please refer to [27,28,29,30,31].

In this work, for the sake of originality, we first try to define the concept of rank of a quasilinear operator and so of an interval matrix as a pair of natural numbers. Furthermore, we introduce the notion of quasi-inverse and determinant of an interval matrix and obtain some results based on these concepts. Moreover, in light of these, we defined the concept of the adjoint of an interval matrix. After maturing these concepts, we aim to prove a theorem that we call interval Cramer’s rule regarding the solution of some linear interval equation systems. Of course, we also examine the consistency of our method (interval Cramer’s rule) by solving some linear interval equations that had been solved using other certain methods. In addition, regarding the existence of solutions to this type of systems, we give a theorem related to the rank of an interval matrix that models the equation.

2. Interval Vectors and Matrices

An n-dimensional interval vector $\mathbf{x}=\left(\left[\underline{x_1},\overline{x_1}\right],\dots ,\left[\underline{x_n},\overline{x_n}\right]\right)$ is a set in ${\mathbb{R}}^n$ such that each component $x_i=\left[\underline{x_i},\overline{x_i}\right]$ is a closed real interval for $i=1,2,...,n.$ In some citations, the equivalent notation can be written as

$$\begin{array}{ccc}\hfill \mathbf{x}& =& \left[\underline{x},\overline{x}\right]\hfill \\ & =& \{x=\left(x_i\right):\underline{x}\le x\le \overline{x},i.e.,\underline{x_i}\le x_i\le \overline{x_i},i=1,2,...,n\}.\hfill \end{array}$$

We think that the first notation is more suitable for our work. We denote the set of all n-dimensional interval vectors using ${\mathbb{I}}_{\mathbb{R}}^n$, and the set of all closed intervals ${\mathbb{I}}_{\mathbb{R}}$ is just ${\mathbb{I}}_{\mathbb{R}}^1$ in this notation. In actuality, saying ${\mathbb{I}}_{\mathbb{R}}^n$ is n-dimensional is somewhat of a misnomer since ${\mathbb{I}}_{\mathbb{R}}^n$ is not a vector space. It is just a word-of-mouth concept. In order to properly understand the concept of the dimension of ${\mathbb{I}}_{\mathbb{R}}^n$, we need to construct the concept of dimension for quasilinear spaces. We have attempted to accomplish this aim in some earlier works.

The scalar product for any real scalar $\lambda$ with an interval vector is defined by

$$\begin{array}{ccc}\hfill \lambda .\mathbf{x}& =& \left(\lambda \left[\underline{x_i},\overline{x_i}\right]\right)\mathrm{where}\hfill \\ \hfill \lambda \left[\underline{x_i},\overline{x_i}\right]& =& \left\{\begin{array}{cc}\left[\lambda \underline{x_i},\lambda \overline{x_i}\right],\hfill & \lambda \ge 0\hfill \\ \left[\lambda \overline{x_i},\lambda \underline{x_i}\right],\hfill & \lambda <0\hfill \end{array}\right..\hfill \end{array}$$

Additionally, the sum of two interval vectors is the coordinate sum of the intervals. To transform the set of all interval vectors into a quasilinear space structure, a partial order relation is also required. This relation is defined as follows: For any $\mathbf{x},\mathbf{y}\in {\mathbb{I}}_{\mathbb{R}}^n$

$$\mathbf{x}⪯\mathbf{y}\iff \left[\underline{x_i},\overline{x_i}\right]\subseteq \left[\underline{y_i},\overline{y_i}\right],\mathrm{for}\mathrm{each}1\le i\le n.$$

Furthermore, since interval matrices define a quasilinear operator between quasilinear spaces, we must first define the concept of a quasilinear space.

A set X is called a quasilinear space [3] on the field $\mathbb{K}$ of real or complex numbers if X is first a partially ordered set by a relation “⪯”, an algebraic sum operation $(+)$, and a scalar product $\left(\right)$ is defined on X in such a way that $\left(X,+\right)$ is an abelian ordered monoid with the zero $\theta \in X$; furthermore, the following conditions are hold for all $x,y,z,v\in X$ and for all $\alpha ,\beta \in \mathbb{K}$:

$$\alpha (x+y)=\alpha x+\alpha y,$$

(1)

$$1x=x,$$

(2)

$$0x=\theta ,$$

(3)

$$(\alpha +\beta )x⪯\alpha x+\beta x,$$

(4)

$$x+z⪯y+v\mathrm{if}x⪯y\mathrm{and}z⪯v,$$

(5)

$$\alpha x⪯\alpha y\mathrm{if}x⪯y.$$

(6)

These conditions are those given by Aseev in cite [3] to form the definition.

Any element x of a quasilinear space (briefly QLS) is again called a “vector”, just as in the linear spaces. Any linear space is a QLS with the partial order relation “=”, but not conversely. In a QLS X, the zero is a minimal element, i.e., $x⪯\theta$ implies $x=\theta$. An element $x^{\prime }$ is called additive inverse of $x\in X$ if $x+x^{\prime }=x^{\prime }+x=\theta$. The inverse is unique whenever it exists. An element x possessing the (additive) inverse is called a stone; otherwise, it is called a foam. We proved in [13] that each stone is a minimal element in a QLS.

Lemma 1

([3]). Suppose that each element x in QLS X has an inverse element $x^{\prime }\in X$. Then, the partial order in X is determined by equality, the distributivity conditions hold; consequently, X is a linear space.

In any linear space, the equality is the only way to define a partial order such that QLS conditions hold [3].

It will be assumed in what follows that $-x=(-1)x.$ Note that $x^{\prime }$ may not exist, but if it exists then $x^{\prime }=-x.$ For example, the interval $\left[1,2\right]$ is a foam in ${\mathbb{I}}_{\mathbb{R}},$ a nonlinear QLS since the additive inverse of the element $\left[1,2\right]$ does not exist. However $-\left[1,2\right]=\left[-2,-1\right]\in {\mathbb{I}}_{\mathbb{R}}.$ All degenerate intervals are stones and all non-degenerate intervals are foams in ${\mathbb{I}}_{\mathbb{R}}$. Let us give an easy characterization of stones. An element x is a stone in any QLS if and only if $x^{\prime }=-x$, or equivalently, $x-x=\theta$. We should note that in a linear QLS, briefly in a linear space, each element is a stone. Hence, the notions of stone and foam in linear spaces are redundant. An element x in a QLS X is said to be balanced whenever $-x=x,$ and $X_b$ denotes the set of all such elements in X.

Suppose X is a QLS and $Y\subseteq X$. Then Y is said to be a subspace of X whenever Y is a QLS with the same partial order and with the same algebraic operations on X. In [3] the concept of a subspace for a QLS was not defined. After detailed investigations we saw that the characterization of the definition must be the same as in linear subspaces: Y is a subspace of X if and only if for every $x,y\in Y$ and $\alpha ,\beta \in \mathbb{R},$ $\alpha x+\beta y\in Y$ [13]. There exist three important subspaces of any QLS X: The space $X_s$, which is the class of all stones; $X_f$, the class of all foams with zero; and $X_b$, the class of all balanced elements. We call $X_s$ and $X_f$ the stone and foam subspaces of X, respectively. $X_b$ is known as the balanced subspace of X. Note that the quasilinear space $X_s$ is a linear space but $X_f$ and $X_b$ are not. That is why we denote the linear part of X using $X_s$. Further, $X_s\cap X_f=\left\{\theta \right\}.$

An $m\times n$ interval matrix $\mathcal{A}=\left[\underline{A},\overline{A}\right]$ is defined as the set $\left\{A\in M^{m\times n}:\underline{A}\le A\le \overline{A}\right\}$ of all real-term $m\times n$ matrices A such that $\underline{A}=\left(\underline{a_{ij}}\right)$ and $\overline{A}=\left(\overline{a_{ij}}\right)$ are fixed $m\times n$ matrices and are lower and upper bounds of $\mathcal{A}$, respectively. Writing interval matrices with their rows and columns explicitly shown will make our next results more understandable. Hence, let us use the notation

$$\mathcal{A}=\left(\begin{array}{cccc}{\mathcal{A}}_{11}& {\mathcal{A}}_{12}& \dots & {\mathcal{A}}_{1n}\\ {\mathcal{A}}_{21}& {\mathcal{A}}_{22}& \dots & {\mathcal{A}}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ {\mathcal{A}}_{m1}& {\mathcal{A}}_{m2}& \dots & {\mathcal{A}}_{mn}\end{array}\right)=\left(\begin{array}{ccc}\left[\underline{a_{11}},\overline{a_{11}}\right]& \dots & \left[\underline{a_{1n}},\overline{a_{1n}}\right]\\ \left[\underline{a_{21}},\overline{a_{21}}\right]& \dots & \left[\underline{a_{2n}},\overline{a_{2n}}\right]\\ ⋮& \ddots & ⋮\\ \left[\underline{a_{m1}},\overline{a_{m1}}\right]& \dots & \left[\underline{a_{mn}},\overline{a_{mn}}\right]\end{array}\right)$$

from now on, where ${\mathcal{A}}_{ij}=\left[\underline{a_{ij}},\overline{a_{ij}}\right]$. Let us denote the family of all $m\times n-$ interval matrices as ${\mathcal{IM}}^{m\times n}$. Thus, from the former notation ${\mathcal{IM}}^{m\times n}$ is just

$${\mathcal{IM}}^{m\times n}=\left\{\mathcal{A}:\mathcal{A}=\left[\underline{A},\overline{A}\right]\mathrm{where}\underline{A},\overline{A}\mathrm{are}\mathrm{bounds}\right\}.$$

As a special case, ${\mathcal{IM}}^n$ denotes the family ${\mathcal{IM}}^{n\times n}.$ If $\overline{a_{ij}}=\underline{a_{ij}},$ for each $i,j$ then $\mathcal{A}$ is called degenerate and any degenerate interval matrix is a singleton including only one classical real-term matrix $A.$ In this case, we can write $\mathcal{A}=\left\{A\right\},$ or sometimes $\mathcal{A}=A$. For two elements $\mathcal{A}$ and $\mathcal{B}$ of ${\mathcal{IM}}^{m\times n},$ addition operation is defined by

$$\begin{array}{ccc}\hfill \mathcal{A}\oplus \mathcal{B}& =& \left\{A+B:\underline{A}\le A\le \overline{A}\mathrm{and}\underline{B}\le B\le \overline{B}\right\}\hfill \\ & =& \left\{{\mathcal{A}}_{ij}+{\mathcal{B}}_{ij}:{\mathcal{A}}_{ij}\in \mathcal{A}\mathrm{and}{\mathcal{B}}_{ij}\in \mathcal{B}\right\}\hfill \end{array}$$

From this operation, ${\mathcal{IM}}^{m\times n}$ is an abelian monoid with the identity interval matrix zero, which is a degenerate (classical) $m\times n$ zero matrix. Obviously, $({\mathcal{IM}}^{m\times n},\oplus )$ is not a group since some elements have no additive inverses. Let us call a degenerate (classical) interval matrix $\mathcal{A}$ a stone since it has an additive inverse and call it a foam since it has no additive inverse. Just as in classical matrices, we use the term inverse only for the multiplicative inverse in interval matrices. Although the additive inverse always exists in classical matrices, this is not the case in non-degenerate (pure) interval matrices; this is why we introduced the concepts of stone and foam. It is easy to prove that any interval matrix is degenerate if and only if it is a stone. Let us denote the class of all degenerate elements (stones) using ${\mathcal{IM}}_s^{m\times n}$ and denote the class of all foams in ${\mathcal{IM}}^{m\times n}$ using ${\mathcal{IM}}_f^{m\times n}$. The function $f:{\mathcal{IM}}_s^{m\times n}\to M^{m\times n},$ $f\left(\left\{A\right\}\right)=A$ is a bijection, and hence we can see that the set $M^{m\times n}$ of all classical real $m\times n$ matrices is equivalent to ${\mathcal{IM}}_s^{m\times n}$.

For two elements $\mathcal{A}$ and $\mathcal{B}$ of ${\mathcal{IM}}^{m\times n},$ the relation

$$\mathcal{A}\subseteq \mathcal{B}if f{\mathcal{A}}_{ij}\subseteq {\mathcal{B}}_{ij}\mathrm{for}\mathrm{each}{\mathcal{A}}_{ij}\in \mathcal{A}\mathrm{and}{\mathcal{B}}_{ij}\in \mathcal{B},$$

is a partial order and hence $({\mathcal{IM}}^{m\times n},\oplus$,$\subseteq )$ is a partially ordered monoid with the compatibility condition:

$$\mathcal{A}\subseteq \mathcal{B}\mathrm{and}\mathcal{C}\subseteq \mathcal{D}\mathrm{implies}\mathcal{A}\oplus \mathcal{C}\subseteq \mathcal{B}\oplus \mathcal{D}.$$

If $\mathcal{A}$ is a stone and $\mathcal{B}$ is a foam then $\mathcal{A}\subseteq \mathcal{B}$ means $A\in \mathcal{B}.$ If $\mathcal{A}$ and $\mathcal{B}$ are both stones then they are classical matrices and $\mathcal{A}\subseteq \mathcal{B}$ means $\mathcal{A}=\mathcal{B}$. If $\mathcal{A}$ is a foam, $\mathcal{B}$ is a stone then the assumption $\mathcal{A}\subseteq \mathcal{B}$ indicates that $\mathcal{B}$ also has to be a foam. We can summarize the last case as follows: “any foam cannot be a subset of a stone”. The following proposition states this assertion, and it can easily be proved.

Proposition 1.

The zero interval matrix θ and moreover all stones are minimal elements in ordered monoid $({\mathcal{IM}}^{m\times n},\oplus$,$\subseteq ).$

For the field $\mathbb{R}$, the law $·:\mathbb{R}\times {\mathcal{IM}}^{m\times n}\to {\mathcal{IM}}^{m\times n}$ is known as the scalar product on ${\mathcal{IM}}^{m\times n}$ and has the following properties: for all elements $\mathcal{A},\mathcal{B},\mathcal{C},\mathcal{D}\in {\mathcal{IM}}^{m\times n}$ and for all $\alpha ,\beta \in \mathbb{R}$,

$$\alpha ·(\beta ·\mathcal{A})=\left(\alpha \beta \right)·\mathcal{A},$$

(7)

$$\alpha ·(\mathcal{A}\oplus \mathcal{B})=\alpha ·\mathcal{A}\oplus \alpha ·\mathcal{B},$$

(8)

$$1·\mathcal{A}=\mathcal{A},$$

(9)

$$0·\mathcal{A}=\theta ,$$

(10)

$$(\alpha +\beta )·\mathcal{A}\subseteq \alpha ·\mathcal{A}\oplus \beta ·\mathcal{A},$$

(11)

$$\mathcal{A}\oplus \mathcal{C}\subseteq \mathcal{B}\oplus \mathcal{D}\mathrm{if}\mathcal{A}\subseteq \mathcal{B}\mathrm{and}\mathcal{C}\subseteq \mathcal{D},$$

(12)

$$\alpha ·\mathcal{A}\subseteq \alpha ·\mathcal{B}\mathrm{if}\mathcal{A}\subseteq \mathcal{B}.$$

(13)

Using these properties, we construct an algebraic structure $({\mathcal{IM}}^{m\times n},\oplus$,$·,\subseteq )$. We will again write $\alpha \mathcal{A}$ for $\alpha ·\mathcal{A}=\left\{\alpha A:A\in \mathcal{A}\right\}$ in the sequel. In this respect,

$$({\mathcal{IM}}^{m\times n},\oplus ,·,\subseteq )$$

is a quasilinear space on the field $\mathbb{R}$.

Example 1.

Let $\mathcal{A}=\left[\underline{A},\overline{A}\right]$ where $\underline{A}=$$\left(\begin{array}{cc}-1& 0\\ -1& -1\end{array}\right)$ and $\overline{A}=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$. Alternatively, using interval notation, we write $\mathcal{A}=\left(\begin{array}{cc}\left[-1,1\right]& \left[0,0\right]\\ \left[-1,1\right]& \left[-1,1\right]\end{array}\right)$. This is a balanced interval matrix, and hence it is an element of the subspace ${\mathcal{IM}}_b^2$ of ${\mathcal{IM}}^2.$ It is also a subspace of ${\mathcal{IM}}_f^2.$ Furthermore, except for zero, all balanced interval matrices are foams. For $n=1,$${\mathcal{IM}}^1$ corresponds to ${\mathbb{I}}_{\mathbb{R}},$ the quasilinear space of all closed intervals of real numbers, and ${\mathcal{IM}}_s^1$ corresponds to $\mathbb{R}$. Further, $\mathcal{B}=\left(\begin{array}{ccc}\left[1,3\right]& \left[2,2\right]& \left[1,3\right]\\ \left[-1,-1\right]& \left[4,4\right]& \left[2,3\right]\end{array}\right)$ is a foam and an element of ${\mathcal{IM}}_f^{2\times 3}$ while

$$\mathcal{C}=\left(\begin{array}{ccc}\left[1,1\right]& \left[2,2\right]& \left[1,1\right]\\ \left[-1,-1\right]& \left[4,4\right]& \left[3,3\right]\end{array}\right)=\left(\begin{array}{ccc}1& 2& 1\\ -1& 4& 3\end{array}\right)$$

is a stone and $\mathcal{C}\in {\mathcal{IM}}_s^{2\times 3}\equiv M^{2\times 3}.$

3. Dimension and Basis in the Space of Interval Vectors

To define another notation in interval analysis, any n-dimensional interval vector $\mathbf{x}=\left(\left[\overline{x_1},\underline{x_1}\right],\dots ,\left[\overline{x_n},\underline{x_n}\right]\right)$ is also written as $\mathbf{x}=\left[\overline{x},\underline{x}\right]$ where $\overline{x}$ and $\underline{x}$ are bounds of $\mathbf{x}$, and they are n-tuples the algebraic operations and the partial order defined above, ${\mathbb{I}}_{\mathbb{R}}^n$ is a quasilinear space on the field $\mathbb{R}$. In order to properly understand the concept of the dimension of ${\mathbb{I}}_{\mathbb{R}}^n$, we need to construct the concept of dimension for quasilinear spaces.

In this section, let us present some basic results obtained in our previous works [4,5,32] by slightly changing some notations. Any quasilinear combination of the set ${\left\{{\mathbf{x}}_k\right\}}_{k=1}^n$ in a QLS X is an element $\mathbf{z}\in X$ such that ${\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2+...+{\alpha }_n{\mathbf{x}}_n⪯\mathbf{z}$ for some scalars ${\alpha }_1,{\alpha }_2,...,{\alpha }_n$. But, any linear combination of the set ${\left\{{\mathbf{x}}_k\right\}}_{k=1}^n$ in X is an element $\mathbf{z}$ of X in the form ${\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2+...+{\alpha }_n{\mathbf{x}}_n=\mathbf{z},$ just as in classical linear (vector) spaces. Hence, a linear combination of the set ${\left\{{\mathbf{x}}_k\right\}}_{k=1}^n$ is an element z of X such that

$${\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2+...+{\alpha }_n{\mathbf{x}}_n⪯\mathbf{z}\mathrm{and}\mathbf{z}⪯{\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2+...+{\alpha }_n{\mathbf{x}}_n.$$

In a linear space, these two definitions coincide since the relation “⪯” turns out to be the relation “=”. Clearly, a linear combination of ${\left\{{\mathbf{x}}_k\right\}}_{k=1}^n,$ is also a quasilinear combination of ${\left\{{\mathbf{x}}_k\right\}}_{k=1}^n,$ but not conversely. For any nonempty subset A of a QLS $X,$ the quasispan (q-span, for short) $QspA$ of $A,$ is defined by the set of all possible quasilinear combinations of $A,$ that is,

$$\begin{array}{ccc}\hfill QspA& =& \{\mathbf{x}\in X:\sum _{k=1}^n{\alpha }_k{\mathbf{x}}_k⪯\mathbf{x},\hfill \\ \hfill \mathrm{for}{\mathbf{x}}_1,{\mathbf{x}}_2,...,{\mathbf{x}}_n& \in & A\mathrm{and}\mathrm{for}\mathrm{some}\mathrm{scalars}{\alpha }_1,{\alpha }_2,...,{\alpha }_n\}.\hfill \end{array}$$

The span of $A,$ $SpA,$ is also defined in quasilinear spaces, justas in classical linear spaces and obviously, $SpA\subseteq QspA.$ Further, $SpA=QspA$ for some linear QLS (linear space); hence, the notion of $QspA$ is redundant in linear spaces. Moreover, we say A quasispans X whenever $QspA=X.$ We know from former works that $QspA$ is a subspace of X but $SpA$ may not be a subspace of $X.$

Definition 1

([32]). (Quasilinear independence and dependence) A set

$$A=\left\{{\mathbf{x}}_1,{\mathbf{x}}_2,...,{\mathbf{x}}_n\right\}$$

in a QLS X is called quasilinear independent (briefly ql-independent) whenever the inequality

$$\theta ⪯{\lambda }_1{\mathbf{x}}_1+{\lambda }_2{\mathbf{x}}_2+...+{\lambda }_n{\mathbf{x}}_n$$

(14)

holds if and only if ${\lambda }_1={\lambda }_2=...={\lambda }_n=0$. Otherwise, A is called quasilinear dependent (briefly ql-dependent).

If we recall again that every linear space is a QLS under the equality relation, it can be seen that the notions of quasilinear independence and dependence coincide with linear independence and dependence in these spaces.

Example 2.

Consider $A=\left\{\right[1,2\left]\right\},$ a singleton in ${\mathbb{I}}_{\mathbb{R}}$. It is obvious that $\left\{0\right\}=\left[0,0\right]\subseteq \alpha .[1,2]$ if and only if $\alpha =0$ where $\left\{0\right\}$ is the zero of ${\mathbb{I}}_{\mathbb{R}}$. Therefore, A is ql-independent. However, the singleton $B=\left\{\right[-1,2\left]\right\}$ is ql-dependent since $\left[0,0\right]\subseteq \beta .[-1,2]$ for $\beta =2\ne 0$. This is an unusual case, since a non-zero singleton is obviously linearly independent in linear spaces. On the other hand, the set $\left\{[1,2],[-1,2]\right\}$ is ql-dependent. In general, the definition implies that any subset containing an element related to zero is necessarily ql-dependent in a QLS. This extends the well-known result in linear spaces that any subset containing zero must be linearly dependent.

Example 3.

In ${\mathbb{I}}_{\mathbb{R}}^2$, let ${\mathbf{v}}_1=\left(\left[-2,1\right],\left[0,0\right]\right)$ and ${\mathbf{v}}_2=\left(\left[0,0\right],\left[-2,3\right]\right)$. Then, the set $\{{\mathbf{v}}_1,{\mathbf{v}}_2\}$ is ql-dependent since

$$(\left[0,0\right],\left[0,0\right])\subseteq {\lambda }_1.{\mathbf{v}}_1+{\lambda }_2.{\mathbf{v}}_2=\left(\left[-2,1\right],\left[-2,3\right]\right)$$

for ${\lambda }_1={\lambda }_2=1$ where $(\left[0,0\right],\left[0,0\right])$ is the zero of ${\mathbb{I}}_{\mathbb{R}}^2$. However, $\{{\mathbf{u}}_1,{\mathbf{u}}_2\}$ is ql-independent where ${\mathbf{u}}_1=\left(\left[-2,-1\right],\left[0,0\right]\right)$ and ${\mathbf{u}}_2=\left(\left[0,0\right],\left[2,3\right]\right).$ On the other hand, let $\mathbf{u}=\left(\left[-2,2\right],\left[-3,3\right]\right)$; then, the singleton $\left\{\mathbf{u}\right\}$ is ql-dependent in ${\mathbb{I}}_{\mathbb{R}}^2$ since

$$(\left[0,0\right],\left[0,0\right])\subseteq \mathbf{u}.$$

We now introduce the concept of dimensionality in QLS. Our analysis indicates that it should be divided into two distinct notions, namely, the stone dimension and the foam dimension. Before doing so, we first present a variation of a classical definition.

Definition 2.

Let S be a ql-independent subset of the QLS X. S is called maximal ql-independent subset of X whenever S is ql-independent, but any set including S is ql-dependent.

Definition 3

([4]). The stone (foam) dimension of any QLS X is the cardinality of any maximal ql-independent subset of $X_s\left(X_f\right)$. If this number is finite, then X is said to be finite stone (foam)-dimensional; otherwise, it is said to be infinite stone (foam)-dimensional. The stone dimension is denoted by s-$dimX$ and the foam dimension is denoted by f-$dimX$. If s-$dimX=m$ and f-$dimX=n$, then we say that X is an $\left(m_s,n_f\right)$-dimensional QLS where m and n are natural numbers or ∞.

The above definition means that s-$dimX$ is the classical definition of a dimension in the linear space $X_s.$ So, s-$dimX=dim{X}_s.$ Notice that a non-trivial foam subspace of a QLS cannot be a linear space. Further, we can easily see that any QLS is $\left(n_s,0_f\right)$-dimensional if and only if it is n-dimensional linear space. In this respect, the trivial linear space $\left\{0\right\}$ is a $\left(0_s,0_f\right)$-dimensional QLS. We see in the following example that there are some $\left(0_s,0_f\right)$-dimensional QLSs other than the trivial space $\left\{0\right\}.$

Example 4.

Let us consider the quasilinear space $X={\left({\mathbb{I}}_{\mathbb{R}}^2\right)}_f\cup \{(\left[x,x\right],\left[0,0\right]):x\in \mathbb{R}\}.$ Then, X is a subspace of ${\mathbb{I}}_{\mathbb{R}}^2$, and $X_s=\{(\left[x,x\right],\left[0,0\right]):x\in \mathbb{R}\}$, and $X_f={\left({\mathbb{I}}_{\mathbb{R}}^2\right)}_f$. Furthermore, $X_s$ is just the real axis of the plane or linear space ${\mathbb{R}}^2.$ Hence, $s-dimX=1$. Now let us determine $f-dimX.$ Let us consider elements

$$v_1=\left(\left[0,0\right],\left[1,2\right]\right)$$

and

$$v_2=\left(\left[1,3\right],\left[0,0\right]\right)$$

in $X_f.$ Then, $\{v_1,v_2\}$ is ql-independent. Furthermore, in the space $X_f$, it is not difficult to see that a set with three or more elements will be ql-dependent. This proves that $f-dimX=2$. Hence, X is a $\left(1_s:2_f\right)$-dimensional QLS.

On the other hand, ${\left({\mathbb{I}}_{\mathbb{R}}^2\right)}_s$ is a $\left(2_s:0_f\right)$-dimensional QLS while ${\left({\mathbb{I}}_{\mathbb{R}}^2\right)}_f$ is $\left(0_s:2_f\right)$-dimensional. But, the balanced subspace

$${\left({\mathbb{I}}_{\mathbb{R}}^2\right)}_b=\{(\left[-t,t\right],\left[-s,s\right]):t,s\in \mathbb{R}\}$$

is a $\left(0_s:0_f\right)-$ dimensional QLS.

Remark 1.

In general, we can easily see that any set including a balanced element must be ql-dependent in any QLS. Hence, the balanced subspace of any QLS is $\left(0_s:0_f\right)-$ dimensional. Further stone subspace of ${\mathbb{I}}_{\mathbb{R}}^n$ is $\left(n_s,0_f\right)-$ dimensional while its foam subspace is $\left(0_s,n_f\right)-$ dimensional.

Example 5.

Consider the QLS $X={\Omega }_C\left(c_0\right)$: the family of all closed, bounded and convex subsets of $c_0,$ the space of all real sequences converging to zero. $X_s\cong c_0,$ and so $s-dimX=\infty$. Let us define the set

$$\Pi =\left\{\right\{(t,0,0,...):1\le t\le 4\},\{(0,t,0,...):1\le t\le 4\},...\}.$$

using another and more flexible notation $\Pi =\{[1,4]\odot e_1,[1,4]\odot e_2,...\},$ where $e_k$’s are unit coordinate vectors of $c_0.$Π is ql-independent in $X_f$. Therefore, $f-dimX=\infty$ and so $X={\Omega }_C\left(c_0\right)$ is an $\left({\infty }_s:{\infty }_f\right)-$ dimensional QLS. In general, an infinite-dimensional linear space E is an $\left({\infty }_s:0_f\right)-$ dimensional QLS while ${\Omega }_C\left(E\right)$ is an $\left({\infty }_s:{\infty }_f\right)-$ dimensional QLSs.

4. Rank and Determinant of an Interval Matrix

This section includes some new definition and results on interval matrices and on the solution of some linear interval equations. We have been frequently benefited from the source [33] for classical linear algebra facts. First of all, let us fix some notation for an interval matrix $\mathcal{A}$ with columns and rows. When we consider an interval matrix $\mathcal{A}=\left({\mathcal{A}}_{ij}\right)$ where ${\mathcal{A}}_{ij}=\left[\underline{a_{ij}},\overline{a_{ij}},\right]$ then we can write $\mathcal{A}=(\left({\mathcal{A}}_{1k}\right),\left({\mathcal{A}}_{2k}\right),...,\left({\mathcal{A}}_{mk}\right))$ for $k=1,2,...,n.$ To get a solution of a system of linear interval equations if it exists, we think that we should first define linear algebra-like tools such as rank and inverse of an interval matrix.

First of all, let us give some concepts and results on quasilinear operators given by Aseev.

Definition 4.

([3]). Let X and Y be quasilinear spaces. A mapping $T:X\to Y$ is called a quasilinear operator if it satisfies the following conditions:

$$T(x_1+x_2)⪯T\left(x_1\right)+T\left(x_2\right),$$

$$T\left(\alpha x\right)=\alpha T\left(x\right)\mathit{for}\mathit{any}\alpha \in \mathbb{R},$$

$$\mathit{if}{x}_1⪯x_2,\mathit{then}T\left(x_1\right)⪯T\left(x_2\right).$$

In this definition, the last two conditions remain the same, and if we tighten the first condition a little more so that $T(x_1+x_2)=T\left(x_1\right)+T\left(x_2\right)$, we get the definition of a linear operator between quasilinear spaces.

Theorem 1.

An $m\times n$ interval matrix $\mathcal{A}$ defines a quasilinear operator from ${\mathbb{I}}_{\mathbb{R}}^n$ into ${\mathbb{I}}_{\mathbb{R}}^m$ from the interval matrix-product $\mathcal{A}\mathbf{x}=\mathbf{b},$ explicitly:

$$\underset{\mathcal{A}}{\left(\begin{array}{ccc}\left[\underline{a_{11}},\overline{a_{11}}\right]& \dots & \left[\underline{a_{1n}},\overline{a_{1n}}\right]\\ \left[\underline{a_{21}},\overline{a_{21}}\right]& \dots & \left[\underline{a_{2n}},\overline{a_{2n}}\right]\\ ⋮& \ddots & ⋮\\ \left[\underline{a_{m1}},\overline{a_{m1}}\right]& \dots & \left[\underline{a_{mn}},\overline{a_{mn}}\right]\end{array}\right)}\underset{\mathbf{x}}{\left(\begin{array}{c}\left[\underline{x_1},\overline{x_1},\right]\\ \left[\underline{x_2},\overline{x_2}\right]\\ ⋮\\ \left[\underline{x_n},\overline{x_n}\right]\end{array}\right)}=\underset{\mathbf{b}}{\left(\begin{array}{c}\sum _{j=1}^n\left[\underline{x_j},\overline{x_j},\right]\left[\underline{a_{1j}},\overline{a_{1j}}\right]\\ \sum _{j=1}^n\left[\underline{x_j},\overline{x_j},\right]\left[\underline{a_{2j}},\overline{a_{2j}}\right]\\ ⋮\\ \sum _{j=1}^n\left[\underline{x_j},\overline{x_j},\right]\left[\underline{a_{mj}},\overline{a_{mj}}\right]\end{array}\right)}$$

(15)

where $\mathbf{x}=\left(\left[\underline{x_1},\overline{x_1},\right],\dots ,\left[\underline{x_n},\overline{x_n}\right]\right)$ and $\mathbf{b}=\left(\left[\underline{b_1},\overline{b_1}\right],\dots ,\left[\underline{b_m},\overline{b_m}\right]\right)$ such that

$$\left[\underline{b_i},\overline{b_i}\right]=\sum _{j=1}^n\left[\underline{x_j},\overline{x_j},\right]\left[\underline{a_{ij}},\overline{a_{ij}}\right],\mathrm{for}i=1,2,...,m,$$

and the product in the summation is the multiplication between intervals.

Proof. 

We are going to only prove that $\mathcal{A}(\mathbf{x}+\mathbf{z})\subseteq \mathcal{A}\left(\mathbf{x}\right)+\mathcal{A}\left(\mathbf{z}\right)$ since verifying the other conditions is routine. We can easily write using interval arithmetic (see [2], p. 99) that

$$\begin{array}{ccc}\hfill \mathcal{A}(\mathbf{x}+\mathbf{z})& =& {\left(\sum _{j=1}^n\left(\left[\underline{x_j},\overline{x_j},\right]+\left[\underline{z_j},\overline{z_j},\right]\right)\left[\overline{a_{ij}},\underline{a_{ij}}\right]\right)}_{i=1}^m\hfill \\ & \subseteq & {\left(\sum _{j=1}^n\left[\underline{x_j},\overline{x_j},\right]\left[\overline{a_{ij}},\underline{a_{ij}}\right]\right)}_{i=1}^m+{\left(\sum _{j=1}^n\left[\underline{z_j},\overline{z_j},\right]\left[\overline{a_{ij}},\underline{a_{ij}}\right]\right)}_{i=1}^m\hfill \\ & =& \mathcal{A}\left(\mathbf{x}\right)+\mathcal{A}\left(\mathbf{z}\right).\hfill \end{array}$$

□

Now, for such an interval matrix $\mathcal{A}:{\mathbb{I}}_{\mathbb{R}}^n\to {\mathbb{I}}_{\mathbb{R}}^m$ and for any interval vectors $\mathbf{b},$ from a system of linear interval equations

$$\mathcal{A}\mathbf{x}=\mathbf{b}$$

we mean a family of all linear equation systems $Ax=b$ such that $A\in \mathcal{A}$, $x\in \mathbf{x}$ and $b\in \mathbf{b}.$ If (15) has a solution, then the solution set is written as

$$\mathit{Y}=\left\{\mathbf{x}\in {\mathbb{I}}_{\mathbb{R}}^n:Ax=b\mathrm{for}\mathrm{some}A\in \mathcal{A},x\in \mathbf{x}\mathrm{and}b\in \mathbf{b}\right\}.$$

However, determining the solution sets of such equations is an extremely difficult problem. In fact, a much simpler form of such systems of equations arises when ${\mathbb{I}}_{\mathbb{R}}^n$ is replaced by its linear subspace ${\mathbb{R}}^n$. In this case, the interval matrix $\mathcal{A}$ again defines a quasilinear operator from ${\mathbb{R}}^n$ into ${\mathbb{I}}_{\mathbb{R}}^m$ and the interval vector $\mathbf{x}$ becomes a classical real n-tuple x. Moreover, the solution set of the simpler case of equation (15) is then expressed as

$$\mathit{Y}=\left\{x\in {\mathbb{R}}^n:Ax=b\mathrm{for}\mathrm{some}A\in \mathcal{A}\mathrm{and}b\in \mathbf{b}\right\}.$$

Even in this simple case, the solution set is very difficult; in fact, it is an NP-hard problem. In the literature, this simpler case is known as the system of linear interval equation. An earlier and fundamental result on the description of the solution set of simpler case is given in [21]. Some further investigations in this manner are presented in [20,22,23]. In fact, we aim to develop solution techniques similar to the classical case for the simpler case of (15).

From this point forward, we consider the simpler case of the equation $\mathcal{A}\mathbf{x}=\mathbf{b}.$ That is, from this point on, we will deal with the slightly simpler linear interval equation

$$\mathcal{A}x=\mathbf{b}$$

where $\mathcal{A}$ is an interval matrix, $x=\left(x_1,x_2,\dots ,x_n\right)\in {\mathbb{R}}^n$ and

$$\mathbf{b}=\left(\left[\underline{b_1},\overline{b_1}\right],\dots ,\left[\underline{b_m},\overline{b_m}\right]\right)\in {\mathbb{I}}_{\mathbb{R}}^m.$$

Remark 2.

In general, the solution set $\mathbf{Y}$ of $\mathcal{A}x=\mathbf{b}$ does not appear as an interval vector. A simple example of the shape of such a set $\mathbf{Y}$ can be seen in ([2], p. 99). Determining the exact solution set of this type of problem is known as an NP-hard problem. Instead, in many cases, it is sufficient to determine a sufficiently narrow envelope $\mathbf{X}$ containing the solution set $\mathbf{Y}.$ Further, a solution of $\mathcal{A}x=\mathbf{b}$ is not an element x satisfying the equality $\mathcal{A}x=\mathbf{b}$, but an element x satisfying the (classical) linear equation $Ax=b$ for any $A\in \mathcal{A}$ and for any $b\in \mathbf{b}$. Let us illustrate this with a simple example. Consider the system of linear interval equations $[1,2]x=4$. Here, $\mathcal{A}=\left(\right[1,2\left]\right),$$x\in \mathbb{R}$ and $\mathbf{b}=[4,4]$. We know from interval arithmetic that there is no real number x satisfying this equality. If the solution set were defined in this way, we would say that this equation has no solution. However, this is not the case. According to the definition above, for $A=(3/2)\in \mathcal{A}$, the system $Ax=4$ has a solution and $x=8/3$ is the solution. Furthermore, if we had chosen another classical matrix $A=\left(2\right)$ from the interval matrix $\mathcal{A}$, then the solution to the system $Ax=4$ would be $x=2$. Similarly, for every $A\in \mathcal{A}$, there exists a solution of the system $Ax=4$ and the solution set $\mathbf{Y}$ of $\mathcal{A}x=\mathbf{b}$ is just $[2,4].$ In this simple example, the solution set $\mathbf{Y}=[2,4]$ is a one-dimensional interval vector. However, when the system of equations moves to a higher dimension, the solution set is generally not an interval vector, unlike in this simple case.

Definition 5.

Let $\mathcal{A}$ be an $m\times n$ interval matrix. Then, $\mathcal{A}x=\mathbf{b}$ is called quasi-homogeneous whenever $0\subseteq \mathbf{b}.$ The dimension of the solution space of such a quasi-homogeneous system is called the quasi-nullity of $\mathcal{A}$.

Remark 3.

Now, since an interval matrix defines a quasilinear operator between quasilinear spaces, we will first define the rank of a quasilinear operator. First of all, it should be noted that a quasilinear operator may not be represented as an interval matrix even if its domain and range are finite $\left(n_s:n_f\right)$-dimensional for any natural number n. Furthermore, although the domain and range of a linear operator are linear spaces, the range of a quasilinear operator may not be a quasilinear spaces. For example, $T:\mathbb{R}\to {\mathbb{I}}_{\mathbb{R}}$, $Tx=x[0,2]$ for $x\in \mathbb{R},$ is a quasilinear operator but the range $\mathcal{R}\left(T\right)=\left\{x[0,2]:x\in \mathbb{R}\right\}$ is not a subspace of ${\mathbb{I}}_{\mathbb{R}}$ since $[0,2]-[0,2]=[-2,2]\notin \mathcal{R}\left(T\right)$. Therefore, we will use the quasispan of $\mathcal{R}\left(T\right)$, that is, $Qsp\left(\mathcal{R}\left(T\right)\right)$ for the rank definition. If a quasilinear operator T were defined between linear spaces—in which case it would be a linear operator—then $\mathcal{R}\left(T\right)$ would be a linear space and $Qsp\left(\mathcal{R}\left(T\right)\right)=Sp\left(\mathcal{R}\left(T\right)\right)=\mathcal{R}\left(T\right)$.

Since the notion of dimension is defined above as a pair of natural numbers, the notion of rank will also appear as a pair of natural numbers.

Definition 6.

Let X and Y be quasilinear spaces. Rank of a quasilinear operator $T:X\to Y$ is defined as the dimension of the quasispan of the range of T in $Y,$ that is, $RankT=dimQsp\left(\mathcal{R}\left(T\right)\right)$.

Definition 7.

Let $\mathcal{A}$ be an $m\times n$ interval matrix. A quasilinear space which is quasispanned by row (column) vectors of $\mathcal{A}$ is called row (column) space of $\mathcal{A}$. The dimension of the row (column) space of $\mathcal{A}$ is called the row (column) rank of $\mathcal{A}$. We denote row and column ranks of $\mathcal{A}$ by $R-rank\mathcal{A}$ and $C-rank\mathcal{A}$, respectively. We will use the symbol $Rank\mathcal{A}=\left(m_s,n_f\right)$ if $R-rank\mathcal{A}=C-rank\mathcal{A}=\left(m_s,n_f\right)$ where m and n are natural numbers.

We will see in the sequel, unlike classical matrices, that the row and column ranks may not be equal in some interval matrices.

Let us give a first example from $1\times 1$ interval matrices.

Example 6.

Consider interval matrices $\mathcal{A}=\left(\left[-1,2\right]\right),$$\mathcal{B}=\left(\left[1,2\right]\right)$ and $\mathcal{C}=\left(\left[2,2\right]\right)$. Their row and column vectors are the same. The row (column) vector of $\mathcal{A}$ is $\left[-1,2\right]$. First, we must find

$$Qsp\left\{\left[-1,2\right]\right\}=\{\left[\overline{a},\underline{a}\right]\in {\mathbb{I}}_{\mathbb{R}}:\lambda \left[-1,2\right]\subseteq \left[\overline{a},\underline{a}\right],\lambda \in \mathbb{R}\}.$$

Obviously, $Qsp\left\{\left[-1,2\right]\right\}$ never contains a degenerate interval (stone) other than zero. Hence, the stone subspace of $Qsp\left\{\left[-1,2\right]\right\}$ is $\left\{\left[0,0\right]\right\},$ the trivial subspace. So, the stone dimension of $Qsp\left\{\left[-1,2\right]\right\}$ is just zero. Moreover, the foam subspace of $Qsp\left\{\left[-1,2\right]\right\}$ is itself, and every subset of $Qsp\left\{\left[-1,2\right]\right\}$ is ql-dependent. This assertion is clear from the definition of $Qsp\left\{\left[-1,2\right]\right\}$ since $\left[0,0\right]\subseteq \left[-1,2\right]$ and so $\left[0,0\right]\subseteq \lambda \left[-1,2\right]\subseteq \left[\overline{a},\underline{a}\right]$ for some $\lambda \ne 0.$ This means that the foam dimension of $Qsp\left\{\left[-1,2\right]\right\}$ is also zero. Eventually, we conclude that $R-rank\mathcal{A}=C-rank\mathcal{A}=\left(0_s,0_f\right)$ so that $Rank\mathcal{A}=\left(0_s,0_f\right)$.

Now, the row and column spaces of $\mathcal{B}$ are the same and

$$Qsp\left\{\left[1,2\right]\right\}=\{\left[\overline{a},\underline{a}\right]\in {\mathbb{I}}_{\mathbb{R}}:\lambda \left[1,2\right]\subseteq \left[\overline{a},\underline{a}\right],\lambda \in \mathbb{R}\}.$$

The stone subspace of $Qsp\left\{\left[1,2\right]\right\}$ is again the trivial subspace. Therefore, its stone dimension is zero. On the other hand, the foam subspace of $Qsp\left\{\left[1,2\right]\right\}$ is again equal to itself. $\left\{\left[1,2\right]\right\}$ is ql-independent in this space, which tells us that the foam dimension of $Qsp\left\{\left[1,2\right]\right\}$ is 1 or a greater integer. Further, two elements in $Qsp\left\{\left[1,2\right]\right\}$ must be ql-dependent by the definition. So, the row and column rank of $\mathcal{B}$ is $\left(0_s,1_f\right)$ and hence $Rank\mathcal{B}=\left(0_s,1_f\right)$.

Finally, let us consider the interval matrix $\mathcal{C}$. We know that $Qsp\left\{\left[2,2\right]\right\}={\mathbb{I}}_{\mathbb{R}}$, and we know that ${\mathbb{I}}_{\mathbb{R}}$ is $\left(1_s,1_f\right)$-dimensional. So, $Rank\mathcal{C}=\left(1_s,1_f\right).$ On the other hand, the matrix $\mathcal{C}$ is also a classical matrix $\left(2\right)$ and it is a transformation of the linear spaces $\mathbb{R}$. From this point of view, its rank is 1. Every linear space is a quasilinear space and when we consider $\mathcal{C}$ as a quasilinear operator on the quasilinear space $\mathbb{R}$, then $Rank\mathcal{C}=\left(1_s,0_f\right)$. But, $\mathcal{C}$ also defines a quasilinear operator on ${\mathbb{I}}_{\mathbb{R}}$ as before $Rank\mathcal{C}=\left(1_s,1_f\right)$.

Example 7.

Now, let us give examples from $2\times 2$ interval matrices. Consider $\mathcal{A}=\left(\begin{array}{cc}\left[0,1\right]& \left[-1,2\right]\\ \left[1,3\right]& \left[1,2\right]\end{array}\right),$ $\mathcal{B}=\left(\begin{array}{cc}\left[1,2\right]& \left[1,3\right]\\ \left[-4,-2\right]& \left[1,1\right]\end{array}\right)$ and $\mathcal{C}=\left(\begin{array}{cc}\left[1,1\right]& \left[3,3\right]\\ \left[-2,-2\right]& \left[2,2\right]\end{array}\right)\equiv \left(\begin{array}{cc}1& 3\\ -2& 2\end{array}\right)$.

Rows of $\mathcal{A}$ are ${\mathbf{v}}_1=(\left[0,1\right],\left[-1,2\right])$ and ${\mathbf{v}}_2=(\left[1,3\right],\left[1,2\right]).$ Now

$$Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}=\{\mathbf{u}:{\lambda }_1{\mathbf{v}}_1+{\lambda }_2{\mathbf{v}}_2\subseteq \mathbf{u},{\lambda }_1,{\lambda }_2\in \mathbb{R}\}.$$

where $\mathbf{u}=(\left[\overline{a_1},\underline{a_1}\right],\left[\overline{a_2},\underline{a_2}\right])\in {\mathbb{I}}_{\mathbb{R}}^2.$ Again, $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ never contains any degenerate interval pairs other than zero $(\left[0,0\right],\left[0,0\right])$. Hence, the stone subspace of $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ is the trivial subspace $\left\{(\left[0,0\right],\left[0,0\right])\right\}$ of ${\mathbb{I}}_{\mathbb{R}}^2$. So, its stone dimension is zero. Moreover, the foam subspace of $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ is again $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$, and every subset of $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ is ql-dependent. This is clear since $(\left[0,0\right],\left[0,0\right])\subseteq 3{\mathbf{v}}_1-2{\mathbf{v}}_2,$ for example. On the other hand, $\left\{{\mathbf{v}}_2\right\}$ is a ql-independent set in ${\mathbb{I}}_{\mathbb{R}}^2$ and in $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$. This means that the foam dimension of $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ is $1$. Eventually, we conclude that row rank of $\mathcal{A}$ is $\left(0_s,1_f\right),$ that is, $R-rank\mathcal{A}=\left(0_s,1_f\right).$ Let us now determine the column rank of $\mathcal{A}$. Consider the column vectors ${\mathbf{u}}_1=(\left[0,1\right],\left[1,3\right])$ and ${\mathbf{u}}_2=(\left[-1,2\right],\left[1,2\right])$ in ${\mathbb{I}}_{\mathbb{R}}^2$.

$$Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}=\{\mathbf{w}:{\lambda }_1{\mathbf{u}}_1+{\lambda }_2{\mathbf{u}}_2\subseteq \mathbf{w},{\lambda }_1,{\lambda }_2\in \mathbb{R}\}.$$

Again, the stone subspace of $Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}$ is the trivial subspace and so its stone dimension is the zero. Let us now determine foam dimension of $Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}$. Observe that $\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}$ is ql-dependent since $(\left[0,0\right],\left[0,0\right])\subseteq {\mathbf{u}}_1-{\mathbf{u}}_2.$ This means that the foam dimension of $Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}$ cannot be $2.$ Further, we cannot find any non-zero λ such that $(\left[0,0\right],\left[0,0\right])\subseteq \lambda {\mathbf{u}}_1.$ So $\left\{{\mathbf{u}}_2\right\}$ is ql-independent and this implies foam dimension of $Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}$ is $1.$ As a result, we conclude that $C-rank\mathcal{A}=\left(0_s,1_f\right)$. Then, we can write $Rank\mathcal{A}=\left(0_s,1_f\right)$.

Similarly, we can show that $Rank\mathcal{B}=\left(0_s,2_f\right)$.

$\mathcal{C}$ is in fact a classical matrix, and we know that its rank is $2$ as a mapping on (quasi) linear space ${\mathbb{R}}^2.$ So, if we consider C as a quasilinear operator on ${\mathbb{R}}^2,$$Rank\mathcal{C}=\left(2_s,0_f\right)$. Now let us see that its rank is $\left(2_s,2_f\right)$ as a quasilinear operator on ${\mathbb{I}}_{\mathbb{R}}^2$. Consider row vectors ${\mathbf{u}}_1=(\left[1,1\right],\left[3,3\right])\equiv \left(1,3\right)$ and ${\mathbf{u}}_2=\left(\left[-2,-2\right],\left[2,2\right]\right)\equiv \left(-2,2\right)$ in ${\mathbb{I}}_{\mathbb{R}}^2$.

$$\begin{array}{cc} & Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}\hfill \\ =& \{\mathbf{y}:{\lambda }_1{\mathbf{u}}_1+{\lambda }_2{\mathbf{u}}_2\subseteq \mathbf{y},\mathit{for}\mathit{some}{\lambda }_1,{\lambda }_2\in \mathbb{R}\}\hfill \\ =& \{\mathbf{y}:{\lambda }_1-2{\lambda }_2\in {\mathbf{y}}_1,3{\lambda }_1+2{\lambda }_2\in {\mathbf{y}}_2,{\lambda }_1,{\lambda }_2\in \mathbb{R}\}\hfill \end{array}$$

where $\mathbf{y}=\left({\mathbf{y}}_1,{\mathbf{y}}_2\right)\in {\mathbb{I}}_{\mathbb{R}}^2$. Hence, for any $\left(y_1,y_2\right)\in \mathbf{y},$ there exists ${\lambda }_1,{\lambda }_2\in \mathbb{R}$ such that $y_1={\lambda }_1-2{\lambda }_2$ and $y_2=3{\lambda }_1+2{\lambda }_2.$ As the real numbers ${\lambda }_1$ and ${\lambda }_2$ change, interval pairs $\left({\mathbf{y}}_1,{\mathbf{y}}_2\right)$ form ${\mathbb{I}}_{\mathbb{R}}^2$. Now, let us see this. Take an arbitrary $\left({\mathbf{z}}_1,{\mathbf{z}}_2\right)\in {\mathbb{I}}_{\mathbb{R}}^2$. If $\left(z_1,z_2\right)\in \left({\mathbf{z}}_1,{\mathbf{z}}_2\right)$ then $\left(z_1,z_2\right)\in {\mathbb{R}}^2$ and so we can write

$$\left(z_1,z_2\right)={\lambda }_1\left(1,3\right)+{\lambda }_2\left(-2,2\right)$$

for some ${\lambda }_1,{\lambda }_2\in \mathbb{R}$ since $\left(1,3\right)$ and $\left(-2,2\right)$ linearly independent in ${\mathbb{R}}^2.$ This proves the assertion. Hence,

$$Qsp\left\{{\mathbf{u}}_1,{\mathbf{u}}_2\right\}={\mathbb{I}}_{\mathbb{R}}^2.$$

An analogous conclusion can be derived from column vectors of $\mathcal{C}$. As a result, we conclude that the row and column rank of $\mathcal{C}$ are $\left(2_s,2_f\right)$, that is, $Rank\mathcal{C}=\left(2_s,2_f\right)$.

Definition 8.

An interval vector in which each term consists of degenerate intervals is called a degenerate interval vector.

Thus, an interval vector in which at least one term is not a degenerate interval is called a non-degenerate interval vector. It can be easily shown that summation of a degenerate interval vector and a non-degenerate interval vector is a non-degenerate interval vector.

Example 8.

Consider $\mathcal{A}=\left(\begin{array}{ccc}\left[1,1\right]& \left[2,2\right]& \left[1,3\right]\\ \left[-1,-1\right]& \left[4,4\right]& \left[3,3\right]\end{array}\right)\equiv \left(\begin{array}{ccc}1& 2& \left[1,3\right]\\ -1& 4& 3\end{array}\right)$. Rows of $\mathcal{A}$ are ${\mathbf{v}}_1=(1,2,\left[1,3\right])$ and ${\mathbf{v}}_2=(-1,4,3).$ Now

$$\begin{array}{ccc}\hfill Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}& =& \{\mathbf{u}:{\lambda }_1{\mathbf{v}}_1+{\lambda }_2{\mathbf{v}}_2\subseteq \mathbf{u},{\lambda }_1,{\lambda }_2\in \mathbb{R}\}\hfill \\ & =& \{\mathbf{u}:{\lambda }_1-{\lambda }_2\in \left[\underline{a},\overline{a}\right],2{\lambda }_1+4{\lambda }_2\in \left[\underline{b},\overline{b}\right],{\lambda }_1\left[1,3\right]+3{\lambda }_2\subseteq \left[\underline{c},\overline{c}\right]\}.\hfill \end{array}$$

where $\mathbf{u}=(\left[\underline{a},\overline{a}\right],\left[\underline{b},\overline{b}\right],\left[\underline{c},\overline{c}\right])\in {\mathbb{I}}_{\mathbb{R}}^3.$ For ${\lambda }_1=0,$$\left(-{\lambda }_2,4{\lambda }_2,3{\lambda }_2\right)$ constitutes the stone subspace of $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$, and it is the span of ${\mathbf{v}}_2$. For ${\lambda }_1\ne 0,$ $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ never contains stones, that is, the stone subspace of $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ is $span\left\{{\mathbf{v}}_2\right\}$. Now, let us look at the foam part of the row space of $\mathcal{A}$. The foam part is just $Qsp\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$ as well, and thus $R-rank\mathcal{A}=\left(1_s,2_f\right)$ since it contains maximum two ql-independent vectors, namely, $\left\{{\mathbf{v}}_1,{\mathbf{v}}_2\right\}$. Let us now investigate columns ${\mathbf{w}}_1=(\left[1,1\right],\left[-1,-1\right])$, ${\mathbf{w}}_2=(\left[2,2\right],\left[4,4\right])$ and ${\mathbf{w}}_3=(\left[1,3\right],\left[3,3\right])$ of $\mathcal{A}$. Assume

$$(\left[0,0\right],\left[0,0\right])\subseteq {\lambda }_1{\mathbf{w}}_1+{\lambda }_2{\mathbf{w}}_2+{\lambda }_3{\mathbf{w}}_3.$$

This means

$$\begin{array}{ccc}\hfill 0& \in & {\lambda }_1+2{\lambda }_2+{\lambda }_3\left[1,3\right],\hfill \\ \hfill 0& =& -{\lambda }_1+4{\lambda }_2+3{\lambda }_3.\hfill \end{array}$$

Then, for ${\lambda }_1=1,$${\lambda }_2=1$ and ${\lambda }_3=-1,$ the above inclusion system is satisfied. This shows $\left\{{\mathbf{w}}_1,{\mathbf{w}}_2,{\mathbf{w}}_3\right\}$ is ql-dependent in $Qsp\left\{{\mathbf{w}}_1,{\mathbf{w}}_2,{\mathbf{w}}_3\right\}$. On the other hand, $\left\{{\mathbf{w}}_1,{\mathbf{w}}_2\right\}$ is ql-independent in $Qsp\left\{{\mathbf{w}}_1,{\mathbf{w}}_2,{\mathbf{w}}_3\right\}$ because $\left\{{\mathbf{w}}_1,{\mathbf{w}}_2\right\}$ is already linearly independent. The stone subspace of $Qsp\left\{{\mathbf{w}}_1,{\mathbf{w}}_2,{\mathbf{w}}_3\right\}$ is just $span\left\{{\mathbf{w}}_1,{\mathbf{w}}_2\right\}$. Hence, the column rank of $\mathcal{A}$ is $\left(2_s,2_f\right)$, that is, $C-rank\mathcal{A}=\left(2_s,2_f\right).$ So, we conclude by this example that: unlike classical matrices, the row and column ranks in interval matrices may not be equal.

If we examine the rank of the (interval) matrix $\mathcal{A}=\left(\begin{array}{ccc}1& 2& 1\\ -1& 4& 3\end{array}\right)$ as a quasilinear operator from ${\mathbb{I}}_{\mathbb{R}}^3$ into ${\mathbb{I}}_{\mathbb{R}}^2$. Then, $Rank\mathcal{A}=\left(2_s,2_f\right)$.

Conclusion 1.

Unlike classical matrices, the row and column ranks of some interval matrices containing a non-degenerate interval may not be the same.

Proposition 2.

Any classical real $m\times n$ matrix A with rank $r,$ which is an operator from ${\mathbb{R}}^n$ to ${\mathbb{R}}^m$, is also an interval matrix from ${\mathbb{I}}_{\mathbb{R}}^n$ to ${\mathbb{I}}_{\mathbb{R}}^m$ for which the (row and column) rank is $\left(r_s,r_f\right)$.

The partial order in the $n\times n$ square interval matrix space ${\mathcal{IM}}^n$ is just defined as

$$\mathcal{A}\subseteq \mathcal{B}iff{\mathcal{A}}_{ij}\subseteq {\mathcal{B}}_{ij},foreachi,j.$$

Further, let us say that the (interval) matrix

$$I_n=\left(\begin{array}{cccc}\left[1,1\right]& \left[0,0\right]& \dots & \left[0,0\right]\\ \left[0,0\right]& \left[1,1\right]& \dots & \left[0,0\right]\\ ⋮& ⋮& \ddots & ⋮\\ \left[0,0\right]& \left[0,0\right]& \dots & \left[1,1\right]\end{array}\right)\equiv \left(\begin{array}{cccc}1& 0& \dots & 0\\ 0& 1& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 1\end{array}\right)$$

is the multiplicative unit in square interval matrix space ${\mathcal{IM}}^n$. It is not difficult to define the multiplication operation between two interval matrices using the multiplication between two intervals, because the multiplication rule here is the same as in classical matrices.

Definition 9.

For any $\mathcal{A}=\left({\mathcal{A}}_{ij}\right)\in {\mathcal{IM}}^n$, determinant of $\mathcal{A}$ is an interval-valued function such that

$$det\mathcal{A}=\sum \left(\pm \right){\mathcal{A}}_{1j_1}{\mathcal{A}}_{2j_2}...{\mathcal{A}}_{n{j}_n}$$

where the sum is taken on all $j_1{j}_2...j_n$ permutations of the set $\left\{1,2,...,n\right\}$. If the $j_1{j}_2...j_n$ permutation is even, then $\left(\pm \right)=+$; if it is odd, then $\left(\pm \right)=-.$

Example 9.

For $\mathcal{A}\in {\mathcal{IM}}^2$,

$$\begin{array}{ccc}\hfill det\mathcal{A}& =& det\left(\begin{array}{cc}\left[\underline{a_{11}},\overline{a_{11}}\right]& \left[\underline{a_{12}},\overline{a_{12}}\right]\\ \left[\underline{a_{21}},\overline{a_{21}}\right]& \left[\underline{a_{22}},\overline{a_{22}}\right]\end{array}\right)\hfill \\ & =& \left[\underline{a_{11}},\overline{a_{11}}\right]\left[\underline{a_{22}},\overline{a_{22}}\right]-\left[\underline{a_{12}},\overline{a_{12}}\right]\left[\underline{a_{21}},\overline{a_{21}}\right]\hfill \\ & =& \left[min{S}_1,max{S}_1\right]-\left[min{S}_2,max{S}_2\right]\hfill \\ & =& \left[min{S}_1-max{S}_2,max{S}_1-min{S}_2\right].\hfill \end{array}$$

where $S_1=\left\{\overline{a_{11}}\overline{a_{22}},\underline{a_{11}}\underline{a_{22}},\overline{a_{11}}\underline{a_{22}},\underline{a_{11}}\overline{a_{22}}\right\}$ and $S_2=\left\{\overline{a_{21}}\overline{a_{12}},\underline{a_{21}}\underline{a_{12}},\underline{a_{12}}\overline{a_{21}},\overline{a_{12}}\underline{a_{21}}\right\}$.

Example 10.

Let $\mathcal{A}=\left(\begin{array}{cc}\left[1,3\right]& \left[-1,2\right]\\ \left[0,2\right]& \left[-2,2\right]\end{array}\right)\in {\mathcal{IM}}^2,$

$$\begin{array}{ccc}\hfill det\mathcal{A}& =& \left[1,3\right]\left[-2,2\right]-\left[0,2\right]\left[-1,2\right]\hfill \\ & =& \left[min\left\{1\left(-2\right),3\left(-2\right),2,3\left(2\right)\right\},max\left\{1\left(-2\right),3\left(-2\right),2,3\left(2\right)\right\}\right]\hfill \\ & & -\left[min\left\{0\left(-1\right),0\left(2\right),2\left(-1\right),2\left(2\right)\right\},max\left\{0\left(-1\right),0\left(2\right),2\left(-1\right),2\left(2\right)\right\}\right]\\ & =& \left[-6,6\right]-\left[-2,4\right]=\left[-6,6\right]+\left[-4,2\right]=\left[-10,8\right].\hfill \end{array}$$

Theorem 2.

For any $\mathcal{A}=\left(\begin{array}{ccc}\left[\overline{a_{11}},\underline{a_{11}}\right]& \left[\overline{a_{12}},\underline{a_{12}}\right]& \left[\overline{a_{13}},\underline{a_{13}}\right]\\ \left[\overline{a_{21}},\underline{a_{21}}\right]& \left[\overline{a_{22}},\underline{a_{22}}\right]& \left[\overline{a_{23}},\underline{a_{23}}\right]\\ \left[\overline{a_{31}},\underline{a_{31}}\right]& \left[\overline{a_{32}},\underline{a_{32}}\right]& \left[\overline{a_{33}},\underline{a_{33}}\right]\end{array}\right)\in {\mathcal{IM}}^3,$

$$\begin{array}{cc} & det\mathcal{A}\hfill \\ =& \left[\overline{a_{11}},\underline{a_{11}}\right]\left[\overline{a_{22}},\underline{a_{22}}\right]\left[\overline{a_{33}},\underline{a_{33}}\right]+\left[\overline{a_{12}},\underline{a_{12}}\right]\left[\overline{a_{23}},\underline{a_{23}}\right]\left[\overline{a_{31}},\underline{a_{31}}\right]\hfill \\ & +\left[\overline{a_{13}},\underline{a_{13}}\right]\left[\overline{a_{21}},\underline{a_{21}}\right]\left[\overline{a_{32}},\underline{a_{32}}\right]-\left[\overline{a_{13}},\underline{a_{13}}\right]\left[\overline{a_{22}},\underline{a_{22}}\right]\left[\overline{a_{31}},\underline{a_{31}}\right]\hfill \\ & -\left[\overline{a_{11}},\underline{a_{11}}\right]\left[\overline{a_{23}},\underline{a_{23}}\right]\left[\overline{a_{32}},\underline{a_{32}}\right]-\left[\overline{a_{12}},\underline{a_{12}}\right]\left[\overline{a_{21}},\underline{a_{21}}\right]\left[\overline{a_{33}},\underline{a_{33}}\right].\hfill \end{array}$$

The rule given in this theorem is called the interval Sarrus rule.

Example 11.

$$\begin{array}{cc} & det\left(\begin{array}{ccc}\left[1,1\right]& \left[2,3\right]& \left[1,3\right]\\ \left[-1,1\right]& \left[0,2\right]& \left[-1,3\right]\\ \left[0,0\right]& \left[-2,2\right]& \left[-3,-1\right]\end{array}\right)\hfill \\ =& \left[-6,0\right]+\left[-6,6\right]-\left[-6,6\right]-\left[-9,9\right]\hfill \\ =& \left[-6,0\right]+\left[-6,6\right]+\left[-6,6\right]+\left[-9,9\right]=\left[-27,21\right].\hfill \end{array}$$

Remark 4.

Since $det\mathcal{A}$ is an interval, we can write it with lower and upper bounds as $det\mathcal{A}=\left[\underline{det\mathcal{A}},\overline{det\mathcal{A}}\right]$. If $0\notin det\mathcal{A}$, then ${\displaystyle \frac{1}{det\mathcal{A}}}$ can be calculated as ${\displaystyle \frac{1}{det\mathcal{A}}}=\left[{\displaystyle \frac{1}{\overline{det\mathcal{A}}}},{\displaystyle \frac{1}{\underline{det\mathcal{A}}}}\right]$ from the interval calculus (see [2]). We can easily see that if $0\notin det\mathcal{A}$ then $detA\ne 0$ for each $A\in \mathcal{A}$.

Remark 5.

Another important work on the determinant of square interval matrices is given in [34], where the determinant of an interval matrix is also defined as an interval. In that work, the important result characterizing the determinant calculus is presented as Proposition 3.1. With our definition, the determinant of an interval matrix includes the determinant given by the other definition, but it is not the same.

We found that the interval-valued determinant has similar properties to the classical determinant of a matrix.

Theorem 3.

For a square interval matrix $\mathcal{A},$

1.

$det\mathcal{A}=det\left({\mathcal{A}}^T\right)$ where ${\mathcal{A}}^T$ denotes the transpose of the interval matrix $\mathcal{A}$, and the transpose is defined as in classical matrices.

2.

If a square interval matrix $\mathcal{B}$ is obtained from $\mathcal{A}$ by interchanging two rows (columns) of $\mathcal{A}$, then $det\mathcal{B}=-det\mathcal{A}.$

3.

If two rows (columns) of $\mathcal{A}$ are equal, then $det\mathcal{A}$ must be a symmetric interval.

4.

If all the elements in a row (column) of $\mathcal{A}$ are zero, that is, the interval $\left[0,0\right]$, then $det\mathcal{A}=\left[0,0\right]$

Proof. 

Only claim 3 seems different from similar results in classical matrices. Here we will only prove claim 3. Other proofs are easily done similarly to those in classical matrices. But, it is not very difficult to prove this because we can easily reach the result $det\mathcal{A}=-det\mathcal{A}$ from the second claim. This means that $det\mathcal{A}$ is a symmetric interval. □

Remark 6.

In classical matrices, $det\mathcal{A}=-det\mathcal{A}$ implies $det\mathcal{A}=0.$ But, for interval matrices the assumption can only say $det\mathcal{A}$ is a symmetric interval. Any symmetric interval is a balanced element in ${\mathbb{I}}_{\mathbb{R}}$ and also can be seen as a balanced $1\times 1$ interval matrix.

Definition 10.

Let $\mathcal{A}=\left({\mathcal{A}}_{ij}\right)$ be an $n\times n$ interval matrix. Let ${\mathcal{B}}_{ij}$ be a sub-interval matrix of type $\left(n-1\right)\times \left(n-1\right)$ obtained by deleting the elements in the $j^{th}$ column and $i^{th}$ row of $\mathcal{A}$. Then, $det{\mathcal{B}}_{ij}$ is called the minor of ${\mathcal{A}}_{ij}.$ Further, the cofactor of ${\mathcal{A}}_{ij}$ is again an interval $C_{ij}$ such that $C_{ij}={\left(-1\right)}^{i+j}det{\mathcal{B}}_{ij}.$

We found that the interval-valued determinant function has similar properties to the classical determinant function.

Theorem 4.

Let $\mathcal{A}=\left(A_{ij}\right)$ be an $n\times n$ square interval matrix. Then,

$$det\mathcal{A}={\mathcal{A}}_{i1}{C}_{i1}+{\mathcal{A}}_{i2}{C}_{i2}+...+{\mathcal{A}}_{in}{C}_{in}$$

where each ${\mathcal{A}}_{ik}{C}_{ik}$ is just interval multiplication.

This theorem is an interval generalization of the classical case and the proof can be derived from the former theorem and from the proof of its classical counterpart (see [33]).

Example 12.

From this theorem, for $\mathcal{A}=\left(\begin{array}{ccc}\left[1,3\right]& \left[-1,2\right]& \left[0,2\right]\\ \left[-2,2\right]& \left[-1,2\right]& 2\\ -1& \left[-1,2\right]& \left[-1,3\right]\end{array}\right),$

$$\begin{array}{ccc}\hfill det\mathcal{A}& =& \left[1,3\right]det\left(\begin{array}{cc}\left[-1,2\right]& 2\\ \left[-1,2\right]& \left[-1,3\right]\end{array}\right)+\left[-1,2\right]\left(-1\right)det\left(\begin{array}{cc}\left[-2,2\right]& 2\\ -1& \left[-1,3\right]\end{array}\right)\hfill \\ & & +\left[0,2\right]det\left(\begin{array}{cc}\left[-2,2\right]& \left[-1,2\right]\\ -1& \left[-1,2\right]\end{array}\right)\hfill \\ & =& \left[1,3\right]\left[-7,8\right]+\left[-2,1\right]\left[-4,8\right]+\left[0,2\right]\left[-5,6\right]\hfill \\ & =& \left[-21,24\right]+\left[-16,8\right]+\left[-10,12\right]\hfill \\ & =& \left[-47,44\right].\hfill \end{array}$$

Remark 7.

We know from interval calculus that if $0\notin \left[\underline{a},\overline{a}\right]$, then ${\displaystyle \frac{\left[\underline{a},\overline{a}\right]}{\left[\underline{a},\overline{a}\right]}}$ is an interval and always includes 1. Furthermore, $\left[\underline{a},\overline{a}\right]-\left[\underline{a},\overline{a}\right]$ is always a balanced element that is a symmetric interval and so always $0\in \left[\underline{a},\overline{a}\right]-\left[\underline{a},\overline{a}\right]$.

Now, let us provide a definition that will play an important role in our work.

Definition 11.

Let $\mathcal{A}$ be an interval matrix in ${\mathcal{IM}}^n.$ Any element $\mathcal{B}$ of ${\mathcal{IM}}^n$ is called a right quasi-inverse of $\mathcal{A}$ if it satisfies the condition $I_n\subseteq \mathcal{AB}.$ Similarly, Any element $\mathcal{B}$ of ${\mathcal{IM}}^n$ is called a left quasi-inverse of $\mathcal{A}$ if it satisfies the condition $I_n\subseteq \mathcal{BA}.$ An interval matrix $\mathcal{B}$ satisfying the condition $I_n\subseteq \mathcal{AB}=\mathcal{BA}$ is called a quasi-inverse of $\mathcal{A}$. Any $\mathcal{B}$ satisfying the condition

$$\mathcal{AB}=\mathcal{BA}\subseteq I_n\subseteq \mathcal{AB}=\mathcal{BA},\mathrm{or}\mathrm{equivalently},I_n=\mathcal{AB}=\mathcal{BA},$$

is called an inverse of $\mathcal{A}$ and then $\mathcal{B}$ is denoted by ${\mathcal{A}}^{-1}.$

Of course, any right (left) inverse of $\mathcal{A}$ is a right (left) quasi-inverse, but not conversely.

Remark 8.

Here, it is possible to give the definition of a right quasi-inverseas “…$\mathcal{AB}\subseteq I_n$…”. But, in this case $\mathcal{AB}$ has to be a classical real-term matrix (stone) because $I_n$ is a minimal element in the partially ordered set $\left({\mathcal{IM}}^n,\subseteq \right)$. Thus, as soon as we write $\mathcal{AB}\subseteq I_n$,we get $\mathcal{AB}=I_n$ immediately. In such a case, we arrive at the definition of the concept of the right inverse of the interval matrix $\mathcal{A}$. An interval matrix may have many (right or left) quasi-inverses. If a quasi-inverse of an interval matrix is an inverse, then it must be a stone. Hence, an inverse of an (interval) matrix must be unique in this case. A foam cannot have an inverse element; it can only have some quasi-inverses. Only stones may have inverses. If we want to introduce an inverse concept for all interval matrices, we have to work with the quasi-inverse concept.

Example 13.

For $\mathcal{A}=\left[1,3\right],$ the interval matrices ${\displaystyle \frac{1}{3}}$ and 1 are both left and right quasi-inverses of $\mathcal{A}.$ Further, $\left[{\displaystyle \frac{1}{3}},1\right]$ is another right (left) quasi-inverse of $\mathcal{A}.$ Any closed interval (matrix) $\mathcal{B}$ for which $\mathcal{B}\subseteq \left[{\displaystyle \frac{1}{3}},1\right]$ is a quasi-inverse of $\mathcal{A}$. If $\mathcal{D}=\left[3,3\right]\equiv 3$, then ${\mathcal{D}}^{-1}$ exits and ${\mathcal{D}}^{-1}=\left[1/3,1/3\right]\equiv {\displaystyle \frac{1}{3}}$. However, the foam $\left[{\displaystyle \frac{1}{3}},1\right]$ is only a quasi-inverse of $\mathcal{D}.$

Definition 12.

(Adjoint) Let $\mathcal{A}=\left({\mathcal{A}}_{ij}\right)$ be an $n\times n$ square interval matrix. Then, the adjoint of $\mathcal{A}$ is written as $adj\mathcal{A}$ and it is defined by

$$adj\mathcal{A}=\left(\begin{array}{cccc}{C}_{11}& C_{21}& \dots & C_{n1}\\ C_{12}& C_{22}& \dots & C_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ C_{1n}& C_{2n}& \dots & C_{nn}\end{array}\right)$$

where $C_{ij}={\left(-1\right)}^{i+j}det{\mathcal{B}}_{ij},$ and ${\mathcal{B}}_{ij}$ is a sub-interval matrix of $\mathcal{A}$ of type $\left(n-1\right)\times \left(n-1\right)$ obtained by deleting the elements in the $j^{th}$ column and $i^{th}$ row of $\mathcal{A}$.

Just as we can multiply a real number by a matrix, we can similarly multiply an interval by an interval matrix. Of course, this multiplication is performed by multiplying an interval by each term of the interval matrix, i.e., $\left[\underline{a},\overline{a}\right]{\left(\left[\underline{A_{ji}},\overline{A_{ji}}\right]\right)}_{i,j}={\left(\left[\underline{a},\overline{a}\right]\left[\underline{A_{ji}},\overline{A_{ji}}\right]\right)}_{i,j}$. From this multiplication, let us now give a main result.

Theorem 5.

Let $\mathcal{A}=\left(A_{ij}\right)$ be an $n\times n$ square interval matrix and let us assume that $\left[0,0\right]⊈det\mathcal{A}$, that is, $0\notin det\mathcal{A}$. Then ${\displaystyle \frac{1}{det\mathcal{A}}}.adj\mathcal{A}$ is a quasi-inverse of $\mathcal{A}$.

Proof. 

From the assumption that ${\displaystyle \frac{1}{det\mathcal{A}}}$ exists and

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{det\mathcal{A}}}.adj\mathcal{A}& =& \left[1/\overline{det\mathcal{A}},1/\underline{det\mathcal{A}}\right]\left(\begin{array}{cccc}{C}_{11}& C_{21}& \dots & C_{n1}\\ C_{12}& C_{22}& \dots & C_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ C_{1n}& C_{2n}& \dots & C_{nn}\end{array}\right)\hfill \\ & =& {\left(\left[1/\overline{det\mathcal{A}},1/\underline{det\mathcal{A}}\right]\left[\underline{C_{ji}},\overline{C_{ji}}\right]\right)}_{i,j}\hfill \\ & =& {\left(\left[minS,maxS\right]\right)}_{i,j}\hfill \end{array}$$

where $S=\left\{{\displaystyle \frac{\underline{C_{ji}}}{\overline{det\mathcal{A}}}},{\displaystyle \frac{\underline{C_{ji}}}{\underline{det\mathcal{A}}}},{\displaystyle \frac{\overline{C_{ji}}}{\overline{det\mathcal{A}}}},{\displaystyle \frac{\overline{C_{ji}}}{\underline{det\mathcal{A}}}}\right\}$, let us prove $1\in \left[minS,maxS\right]\mathcal{A}$ for $i=j$ and $0\in \left[minS,maxS\right]\mathcal{A}$ for $i\ne j.$ It is sufficient to prove the assertion for $n=3$. For $n>3,$ the proof of the assertion is similar, and it can be derived using induction. For

$$\mathcal{A}=\left(\begin{array}{ccc}{\mathcal{A}}_{11}& {\mathcal{A}}_{12}& {\mathcal{A}}_{13}\\ {\mathcal{A}}_{21}& {\mathcal{A}}_{22}& {\mathcal{A}}_{23}\\ {\mathcal{A}}_{31}& {\mathcal{A}}_{32}& {\mathcal{A}}_{33}\end{array}\right),$$

$$adj\mathcal{A}=\left(\begin{array}{ccc}det{\mathcal{B}}_{11}& -det{\mathcal{B}}_{21}& det{\mathcal{B}}_{31}\\ -det{\mathcal{B}}_{12}& det{\mathcal{B}}_{22}& -det{\mathcal{B}}_{32}\\ det{\mathcal{B}}_{13}& -det{\mathcal{B}}_{23}& det{\mathcal{B}}_{33}\end{array}\right).$$

and hence

$$\begin{array}{cc} & \left(adj\mathcal{A}\right)\mathcal{A}\hfill \\ =& \left(\begin{array}{ccc}det{\mathcal{B}}_{11}& -det{\mathcal{B}}_{21}& det{\mathcal{B}}_{31}\\ -det{\mathcal{B}}_{12}& det{\mathcal{B}}_{22}& -det{\mathcal{B}}_{32}\\ det{\mathcal{B}}_{13}& -det{\mathcal{B}}_{23}& det{\mathcal{B}}_{33}\end{array}\right)\left(\begin{array}{ccc}{\mathcal{A}}_{11}& {\mathcal{A}}_{12}& {\mathcal{A}}_{13}\\ {\mathcal{A}}_{21}& {\mathcal{A}}_{22}& {\mathcal{A}}_{23}\\ {\mathcal{A}}_{31}& {\mathcal{A}}_{32}& {\mathcal{A}}_{33}\end{array}\right)\hfill \\ =& \left(\begin{array}{ccc}{\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k1}{\mathcal{A}}_{k1}& {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k1}{\mathcal{A}}_{k2}& {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k1}{\mathcal{A}}_{k3}\\ {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k}det{\mathcal{B}}_{k2}{\mathcal{A}}_{k1}& {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k}det{\mathcal{B}}_{k2}{\mathcal{A}}_{k2}& {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k}det{\mathcal{B}}_{k2}{\mathcal{A}}_{k3}\\ {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k1}& {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k2}& {\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k3}\end{array}\right).\hfill \end{array}$$

Observe that diagonal elements (intervals) in

$${\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathcal{A}$$

include $1,$ and other elements include 0 because each of the diagonal elements in interval matrix $\left(adj\mathcal{A}\right)\mathcal{A}$ is a determinant expansion of $\mathcal{A}$. That is,

$$\begin{array}{ccc}\hfill \sum _{k=1}^3{\left(-1\right)}^{k+1}\left(det{\mathcal{B}}_{k1}\right){\mathcal{A}}_{k1}& =& \sum _{k=1}^3{\left(-1\right)}^k\left(det{\mathcal{B}}_{k2}\right){\mathcal{A}}_{k2}\hfill \\ & =& \sum _{k=1}^3{\left(-1\right)}^{k+1}\left(det{\mathcal{B}}_{k3}\right){\mathcal{A}}_{k3}\hfill \\ & =& det\mathcal{A}.\hfill \end{array}$$

Hence, each diagonal elements in ${\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathcal{A}$ is ${\displaystyle \frac{det\mathcal{A}}{det\mathcal{A}}}$ and since $0\notin det\mathcal{A}$, ${\displaystyle \frac{det\mathcal{A}}{det\mathcal{A}}}$ exists and of course includes 1 from Remark 7. For non-diagonal elements in ${\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathcal{A}$, consider, for example, ${\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k1}$ and observe that

$$\begin{array}{cc} & \sum _{k=1}^3{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k1}\hfill \\ =& \left(det{\mathcal{B}}_{13}\right){\mathcal{A}}_{11}-\left(det{\mathcal{B}}_{23}\right){\mathcal{A}}_{21}+\left(det{\mathcal{B}}_{33}\right){\mathcal{A}}_{31}\hfill \\ =& {\mathcal{A}}_{11}\left({\mathcal{A}}_{21}{\mathcal{A}}_{32}-{\mathcal{A}}_{22}{\mathcal{A}}_{31}\right)-{\mathcal{A}}_{21}\left({\mathcal{A}}_{11}{\mathcal{A}}_{32}-{\mathcal{A}}_{12}{\mathcal{A}}_{31}\right)\hfill \\ & +{\mathcal{A}}_{31}\left({\mathcal{A}}_{11}{\mathcal{A}}_{22}-{\mathcal{A}}_{21}{\mathcal{A}}_{12}\right)\hfill \\ =& {\mathcal{A}}_{11}{\mathcal{A}}_{21}{\mathcal{A}}_{32}-{\mathcal{A}}_{11}{\mathcal{A}}_{22}{\mathcal{A}}_{31}-{\mathcal{A}}_{21}{\mathcal{A}}_{11}{\mathcal{A}}_{32}+{\mathcal{A}}_{21}{\mathcal{A}}_{12}{\mathcal{A}}_{31}\hfill \\ & +{\mathcal{A}}_{31}{\mathcal{A}}_{11}{\mathcal{A}}_{22}-{\mathcal{A}}_{31}{\mathcal{A}}_{21}{\mathcal{A}}_{12}\hfill \\ =& \left({\mathcal{A}}_{11}{\mathcal{A}}_{21}{\mathcal{A}}_{32}+{\mathcal{A}}_{21}{\mathcal{A}}_{12}{\mathcal{A}}_{31}+{\mathcal{A}}_{31}{\mathcal{A}}_{11}{\mathcal{A}}_{22}\right)\hfill \\ & -\left({\mathcal{A}}_{11}{\mathcal{A}}_{21}{\mathcal{A}}_{32}+{\mathcal{A}}_{21}{\mathcal{A}}_{12}{\mathcal{A}}_{31}+{\mathcal{A}}_{31}{\mathcal{A}}_{11}{\mathcal{A}}_{22}\right)\hfill \end{array}$$

We obtain the last equality by changing the order of the multiplication since the interval multiplication is commutative. That is, ${\displaystyle \sum _{k=1}^3}{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k1}$ has the form $\left[\underline{a},\overline{a}\right]-\left[\underline{a},\overline{a}\right]$ and so we can say

$$0\in {\displaystyle \frac{1}{det\mathcal{A}}}\sum _{k=1}^3{\left(-1\right)}^{k+1}det{\mathcal{B}}_{k3}{\mathcal{A}}_{k1}$$

from again Remark 7. Similarly, we can see other non-diagonal terms also includes 0. Hence, we can deduce that

$$I_3\subseteq {\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathcal{A}.$$

Furthermore, it is not hard to see that

$$\left(adj\mathcal{A}\right)\mathcal{A}=\mathcal{A}\left(adj\mathcal{A}\right).$$

This means that ${\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}$ is a quasi-inverse of $\mathcal{A}$. □

With the help of this important result, let us now present another important theorem of our work.

Theorem 6

(Interval Cramer’s rule). Let $\mathcal{A}$ be an $n\times n$ square interval matrix from ${\mathbb{R}}^n$ into ${\mathbb{I}}_{\mathbb{R}}^n$, $\mathbf{b}$ be an n-dimensional interval vector, and let us assume that $0\notin det\mathcal{A}$. Then, the system $\mathcal{A}x=\mathbf{b}$ has a solution set $\mathbf{Y}$ such that

$$\mathbf{X}={\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathbf{b}$$

is an envelope including $\mathbf{Y}$.

Proof. 

Since any system of linear interval equation is a family of systems of linear equations and since $0\notin det\mathcal{A}$ implies $detA\ne 0$ for each $A\in \mathcal{A}$, we can guarantee the existence of a solution set $\mathbf{Y}$ for $\mathcal{A}x=\mathbf{b}$. Further, the assumption $0\notin det\mathcal{A}$ again implies that ${\displaystyle \frac{1}{det\mathcal{A}}}$ exists; so, using $det\mathcal{A}=\left[\underline{det\mathcal{A}},\overline{det\mathcal{A}}\right],$${\displaystyle \frac{1}{det\mathcal{A}}}$ can be determined by

$${\displaystyle \frac{1}{det\mathcal{A}}}=\left[1/\overline{det\mathcal{A}},1/\underline{det\mathcal{A}}\right]$$

to be an interval. Let us consider $adj\mathcal{A}$. Then, ${\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}$ is a quasi-inverse of $\mathcal{A}$ from the Theorem 5. So, we can conclude that

$$I_n\subseteq \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathcal{A}=\mathcal{A}\left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)$$

and so for any $x\in \mathbf{Y}$

$$\begin{array}{ccc}\hfill I_{n}x& \subseteq & \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathcal{A}x\hfill \\ & =& \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathbf{b}.\hfill \end{array}$$

Hence, we have

$$x\subseteq \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathbf{b}.$$

This means explicitly that $x=\left(x_1,x_2,\dots ,x_n\right)\in {\mathbb{R}}^n$ must satisfy the condition

$$x_i\in {\displaystyle \frac{{\mathcal{A}}_{1i}}{det\mathcal{A}}}\left[\overline{b_1},\underline{b_1}\right]+{\displaystyle \frac{{\mathcal{A}}_{2i}}{det\mathcal{A}}}\left[\overline{b_2},\underline{b_2}\right]+\dots +{\displaystyle \frac{{\mathcal{A}}_{ni}}{det\mathcal{A}}}\left[\overline{b_n},\underline{b_n}\right]$$

for each $i=1,2,...,n$ where $\mathbf{b}=\left(\left[\overline{b_1},\underline{b_1}\right],\dots ,\left[\overline{b_n},\underline{b_n}\right]\right)$. Here, writing

$$x\subseteq \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathbf{b}$$

is equivalent to writing

$$x\in \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathbf{b}.$$

because $x\in {\mathbb{R}}^n$ is a degenerate interval vector. Just like in the simple case, writing $\left[2,2\right]\subseteq \left[1,3\right]$ is equivalent to writing $2\in \left[1,3\right]$. As a result

$$\mathbf{X}={\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathbf{b}$$

is the desired set (envelope) containing the solution set $\mathbf{Y}$. □

Remark 9.

There may be many other interval matrices $\mathcal{C}$ satisfy the condition $\mathbf{Y}\subseteq \mathcal{C}\mathbf{b}$, and it may be a quasi-inverse of $\mathcal{A}$. Already, ${\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)$ is one of the matrices $\mathcal{C}$ that meet this condition, and it is obtained with the help of the interval Cramer’s rule. A narrower $\mathcal{C}$ that satisfies the condition $\mathbf{Y}\subseteq \mathcal{C}\mathbf{b}$ is more valuable, and the solution x obtained from it is a closer and better result. With this method, we do not determine the exact solution set of the interval equation, but we determine an n-dimensional envelope containing the solution set. In many cases, the set

$$\mathbf{X}={\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathbf{b}$$

determined using this method is a sufficiently narrow or acceptable envelope containing the solution. We tested this by solving the same problem as one previously solved using other methods, as shown in the example below. We observed that the result obtained using the method presented here is an acceptably narrow envelope. However, for some problems, the envelope containing the solution may be excessively wide. In such cases, other methods may be preferred. Since our method is derived as a natural extension of classical linear algebra techniques, we believe it is a more systematic approach and one that is conducive to developing similar methods in the future.

Now, let us return the problem given in the introduction, which is given in Example 7.5 in [2]. It is the mesh equation of an electrical circuit system. An envelope for solving this problem is given in [2]. Using the interval Cramer’s method, let us determine an envelope covering the solution and compare the results with those in Example 7.5 [2].

Example 14

([2], Example 7.5). The mesh equations for an electric circuit are expressed as

$$\left(\begin{array}{cc}{R}_1+R_2& -R_2\\ -R_2& R_2+R_3\end{array}\right)\left(\begin{array}{c}{I}_1\\ I_2\end{array}\right)=\left(\begin{array}{c}{V}_1\\ -V_2\end{array}\right)$$

with $V_1=10,$$V_2=5,$ and $R_1=R_2=R_3=1000\pm 10\%.$ Here, $R_i$ denotes resistances, $I_i$ denotes currents and $V_i$ denotes voltages. We find enclosures for $I_1$ and $I_2.$ In [2], Example 7.5., it is expressed that a pair of envelopes of currents which are includes the solution of the mesh equation are given by

$$I_1=\left[0.00433,0.00582\right]\mathrm{and}{I}_2=\left[-0.000419,0.000419\right].$$

Let us now give another envelope for this problem using our interval Cramer’s rule. Let us consider the interval matrix and vectors

$$\begin{array}{ccc}\hfill \mathcal{A}& =& \left(\begin{array}{cc}\left[1800,2200\right]& \left[-1100,-900\right]\\ \left[-1100,-900\right]& \left[1800,2200\right]\end{array}\right),\hfill \\ \hfill x& =& \left(\begin{array}{c}{I}_1\\ I_2\end{array}\right),b=\left(\begin{array}{c}10\\ -5\end{array}\right)=\mathbf{b}\hfill \end{array}$$

respectively. First, we will get an enclosure for the solution set $\mathbf{Y}$ of the linear interval equation

$$\mathcal{A}x=\mathbf{b}$$

where $\mathbf{b}$ is a model (interval) vector. First of all, we must calculate $det\mathcal{A}$. Using the interval calculus we get

$$\begin{array}{ccc}\hfill det\mathcal{A}& =& \left[{\left(1800\right)}^2,{\left(2200\right)}^2\right]-\left[{\left(900\right)}^2,{\left(1100\right)}^2\right]\hfill \\ & =& \left[{\left(1800\right)}^2-{\left(1100\right)}^2,{\left(2200\right)}^2-{\left(900\right)}^2\right]\hfill \\ & =& \left[2030000,403000\right]\hfill \end{array}$$

Since $0\notin det\mathcal{A}$, we can say that $\mathcal{A}x=\mathbf{b}$ has a solution $\mathbf{Y},$ and we can determine an envelope from the interval Cramer’s rule. Again, from this theorem

$$\left(\begin{array}{c}{I}_1\\ I_2\end{array}\right)\subseteq {\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathbf{b}$$

and so first we must calculate $adj\mathcal{A}=\left(\begin{array}{cc}{C}_{11}& C_{21}\\ C_{12}& C_{22}\end{array}\right)$. Using interval calculus and the definition of the adjoint we get

$$adj\mathcal{A}=\left(\begin{array}{cc}\left[1800,2200\right]& \left[900,1100\right]\\ \left[900,1100\right]& \left[1800,2200\right]\end{array}\right).$$

Further

$${\displaystyle \frac{1}{det\mathcal{A}}}=\left[{\displaystyle \frac{1}{4030000}},{\displaystyle \frac{1}{2030000}}\right]$$

and

$$\begin{array}{cc} & {\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\hfill \\ =& \left(\begin{array}{cc}\left[{\displaystyle \frac{1800}{4030000}},{\displaystyle \frac{2200}{2030000}}\right]& \left[{\displaystyle \frac{900}{4030000}},{\displaystyle \frac{1100}{2030000}}\right]\\ \left[{\displaystyle \frac{900}{4030000}},{\displaystyle \frac{1100}{2030000}}\right]& \left[{\displaystyle \frac{1800}{4030000}},{\displaystyle \frac{2200}{2030000}}\right]\end{array}\right)\hfill \\ =& \left(\begin{array}{cc}\left[0.0004466501,0.0010837438\right]& \left[0.0002233251,0.0005418719\right]\\ \left[0.0002233251,0.0005418719\right]& \left[0.0004466501,0.0010837438\right]\end{array}\right).\hfill \end{array}$$

Now

$$\begin{array}{cc} & {\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\mathbf{b}\hfill \\ =& \left(\begin{array}{cc}\left[0.0004466501,0.0010837438\right]& \left[0.0002233251,0.0005418719\right]\\ \left[0.0002233251,0.0005418719\right]& \left[0.0004466501,0.0010837438\right]\end{array}\right)\left(\begin{array}{c}10\\ -5\end{array}\right)\hfill \\ =& (\left[0.004466501,0.010837438\right]+\left[-0.0027093595,-0.00116625\right]\hfill \\ & ,\left[0.002233251,0.005418719\right]+\left[-0.005418719,-0.0022332505\right])\hfill \\ =& \left(\left[0.0017571415,0.009671188\right],\left[-0.003185468,0.0031854685\right]\right)\hfill \end{array}$$

We can conclude again from the theorem (interval Cramer’s rule) that

$$x=\left(\begin{array}{c}{I}_1\\ I_2\end{array}\right)\subseteq \left({\displaystyle \frac{1}{det\mathcal{A}}}adj\mathcal{A}\right)\mathbf{b}$$

for every $x\in \mathbf{Y}.$ So this means

$$\begin{array}{ccc}\hfill I_1& \in & \left[0.0017571415,0.009671188\right]\hfill \\ \hfill \mathrm{and}{I}_2& \in & \left[-0.003185468,0.0031854685\right].\hfill \end{array}$$

Compared to the other result, we can say that interval Cramer’s rule also gives a close and relatively good result. This shows that the method we have presented provides sufficiently good (narrow) and useful intervals for the current values we are looking for.

Now, we will consider a problem mentioned in ([35], Example 3.2). and examined in comparison with some known solution methods. We will then compare the result obtained with our method with the mentioned result. We hope that our solution is consistent with the solution in [35].

Example 15.

Let us consider the interval equation system $\mathcal{A}x=\mathbf{b}$ in ([35], Example 3.2), with

$$\begin{array}{ccc}\hfill \mathcal{A}& =& \left(\begin{array}{ccc}\left[3.7,4.3\right]& \left[-1.5,-0.5\right]& \left[0,0\right]\\ \left[-1.5,-0.5\right]& \left[3.7,4.3\right]& \left[-1.5,-0.5\right]\\ \left[0,0\right]& \left[-1.5,-0.5\right]& \left[3.7,4.3\right]\end{array}\right),\hfill \\ \hfill \mathbf{b}& =& \left(\begin{array}{c}\left[-14,14\right]\\ \left[-9,9\right]\\ \left[-3,3\right]\end{array}\right).\hfill \end{array}$$

The set (envelope or box) containing the solution obtained in this study is just

$${\mathbf{X}}_{Ning}=\left(\begin{array}{c}{\mathbf{x}}_1\\ {\mathbf{x}}_2\\ {\mathbf{x}}_3\end{array}\right)=\left(\begin{array}{c}\left[-6.38,6.38\right]\\ \left[-6.40,6.40\right]\\ \left[-3.40,3.40\right]\end{array}\right).$$

Now, let us determine the envelope $\mathbf{X}$ containing the solution that we will obtain using the interval Cramer’s rule. First, we must verify that $0\notin det\mathcal{A}$, and for this, let us calculate $det\mathcal{A}$. Using interval arithmetic and the determinant calculations mentioned above, we obtain

$$det\mathcal{A}=\left[32.853,77.507\right].$$

Indeed, since $0\notin det\mathcal{A}$, the system has a solution, and using our method, we can determine an ${\mathbf{X}}_{ours}$ interval vector containing the solution. The value of ${\displaystyle \frac{1}{det\mathcal{A}}}$ required for our method is easily calculated as

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{det\mathcal{A}}}& =& \left[0.0129020605,0.0233355891\right]\hfill \\ & \cong & \left[0.0129,0.0233\right]\hfill \end{array}$$

Now, let us determine $adj\mathcal{A}$. Remember that

$$adj\mathcal{A}=\left(\begin{array}{ccc}det{\mathcal{B}}_{11}& -det{\mathcal{B}}_{21}& det{\mathcal{B}}_{31}\\ -det{\mathcal{B}}_{12}& det{\mathcal{B}}_{22}& -det{\mathcal{B}}_{32}\\ det{\mathcal{B}}_{13}& -det{\mathcal{B}}_{23}& det{\mathcal{B}}_{33}\end{array}\right).$$

Here

$${\mathcal{B}}_{11}=\left(\begin{array}{cc}\left[3.7,4.3\right]& \left[-1.5,-0.5\right]\\ \left[-1.5,-0.5\right]& \left[3.7,4.3\right]\end{array}\right),{\mathcal{B}}_{21}=\left(\begin{array}{cc}\left[-1.5,-0.5\right]& 0\\ \left[-1.5,-0.5\right]& \left[3.7,4.3\right]\end{array}\right)$$

$$\begin{array}{ccc}\hfill {\mathcal{B}}_{31}& =& \left(\begin{array}{cc}\left[-1.5,-0.5\right]& 0\\ \left[3.7,4.3\right]& \left[-1.5,-0.5\right]\end{array}\right),\hfill \\ \hfill {\mathcal{B}}_{12}& =& \left(\begin{array}{cc}\left[-1.5,-0.5\right]& \left[-1.5,-0.5\right]\\ 0& \left[3.7,4.3\right]\end{array}\right)\hfill \end{array}$$

$${\mathcal{B}}_{22}=\left(\begin{array}{cc}\left[3.7,4.3\right]& 0\\ 0& \left[3.7,4.3\right]\end{array}\right),{\mathcal{B}}_{32}=\left(\begin{array}{cc}\left[3.7,4.3\right]& 0\\ \left[-1.5,-0.5\right]& \left[-1.5,-0.5\right]\end{array}\right)$$

$${\mathcal{B}}_{13}=\left(\begin{array}{cc}\left[-1.5,-0.5\right]& \left[3.7,4.3\right]\\ 0& \left[-1.5,-0.5\right]\end{array}\right),{\mathcal{B}}_{23}=\left(\begin{array}{cc}\left[3.7,4.3\right]& \left[-1.5,-0.5\right]\\ 0& \left[-1.5,-0.5\right]\end{array}\right)$$

$${\mathcal{B}}_{33}=\left(\begin{array}{cc}\left[3.7,4.3\right]& \left[-1.5,-0.5\right]\\ \left[-1.5,-0.5\right]& \left[3.7,4.3\right]\end{array}\right).$$

Now let us calculate these determinants.

$$\begin{array}{ccc}\hfill det{\mathcal{B}}_{11}& =& det{\mathcal{B}}_{33}=\left[11.44,18.24\right]\hfill \\ \hfill det{\mathcal{B}}_{32}& =& det{\mathcal{B}}_{23}=det{\mathcal{B}}_{21}=det{\mathcal{B}}_{12}=\left[-6.45,-1.85\right]\hfill \\ \hfill det{\mathcal{B}}_{13}& =& det{\mathcal{B}}_{31}=\left[0.25,2.25\right]\mathrm{and}det{\mathcal{B}}_{22}=\left[13.69,18.49\right].\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}\hfill adj\mathcal{A}& =& \left(\begin{array}{ccc}\left[11.44,18.24\right]& -\left[-6.45,-1.85\right]& \left[0.25,2.25\right]\\ -\left[-6.45,-1.85\right]& \left[13.69,18.49\right]& -\left[-6.45,-1.85\right]\\ \left[0.25,2.25\right]& -\left[-6.45,-1.85\right]& \left[11.44,18.24\right]\end{array}\right)\hfill \\ & =& \left(\begin{array}{ccc}\left[11.44,18.24\right]& \left[1.85,6.45\right]& \left[0.25,2.25\right]\\ \left[1.85,6.45\right]& \left[13.69,18.49\right]& \left[1.85,6.45\right]\\ \left[0.25,2.25\right]& \left[1.85,6.45\right]& \left[11.44,18.24\right]\end{array}\right).\hfill \end{array}$$

Now, if we calculate the quasi-inverse of $\mathcal{A}$, which is ${\displaystyle \frac{1}{det\mathcal{A}}}.adj\mathcal{A}$, we obtain the interval matrix

$$\left(\begin{array}{ccc}[0.147576,0.424992]& [0.023865,0.150285]& [0.003225,0.052425]\\ [0.023865,0.150285]& [0.176361,0.430357]& [0.023865,0.150285]\\ [0.003225,0.052425]& [0.023865,0.150285]& [0.147576,0.424992]\end{array}\right)$$

with an extremely small error. As the final step, let us determine

$${\mathbf{X}}_{ours}=\left(\begin{array}{c}{\mathbf{x}}_1\\ {\mathbf{x}}_2\\ {\mathbf{x}}_3\end{array}\right)={\displaystyle \frac{1}{det\mathcal{A}}}\left(adj\mathcal{A}\right)\mathbf{b},$$

considering the interval vector

$$\mathbf{b}=\left(\begin{array}{c}\left[-14,14\right]\\ \left[-9,9\right]\\ \left[-3,3\right]\end{array}\right).$$

When we apply the above interval matrix to $\mathbf{b}$, the matrix multiplication results in each component being

$$\begin{array}{ccc}\hfill {\mathbf{x}}_1& \cong & [-5.949888,5.949888]+[-1.352565,1.352565]+[-0.157275,0.157275]\hfill \\ & \cong & [-7.46,7.46]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\mathbf{x}}_2& \cong & [-2.1040,2.1040]+[-3.873213,3.873213]+[-0.450855,0.450855]\hfill \\ & \cong & [-6.42,6.42]\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\mathbf{x}}_3& \cong & [-0.733950,0.733950]+[-1.352565,1.352565]+[-1.274976,1.274976]\hfill \\ & \cong & [-3.36,3.36].\hfill \end{array}$$

Finally, we get

$${\mathbf{X}}_{ours}\cong \left(\begin{array}{c}[-7.46,7.46]\\ [-6.42,6.42]\\ [-3.36,3.36]\end{array}\right).$$

Remember that

$${\mathbf{X}}_{Ning}=\left(\begin{array}{c}\left[-6.38,6.38\right]\\ \left[-6.40,6.40\right]\\ \left[-3.40,3.40\right]\end{array}\right).$$

We can see that there are no significant differences between our results. Sometimes our results are better (a narrower envelope), and sometimes they may be slightly worse (a wider envelope). From this, for this example, we can say that our method is consistent with other methods.

Let us now give another main result.

Theorem 7.

Let $\mathcal{A}$ be an $m\times n$ interval matrix and consider a system of linear interval equation $\mathcal{A}x=\mathbf{b}$.

(1) 

If $\mathcal{A}x=\mathbf{b}$ has a solution, then $C-rank\mathcal{A}=C-rank\left[\mathcal{A}:\mathbf{b}\right]$

(2) 

If there exists an interval vector $\overline{\mathbf{b}}$ such that $\mathbf{b}\subseteq \overline{\mathbf{b}}$ and $C-rank\mathcal{A}=C-rank\left[\mathcal{A}:\overline{\mathbf{b}}\right]$, then the system $\mathcal{A}x=\mathbf{b}$ has at least one (possibly many) solution $x=\left(x_1,x_2,\dots ,x_n\right)\in {\mathbb{R}}^n.$

Proof. 

The proof of (1) is similar to classical case, because if $\mathbf{b}$ is a linear combination of the column vectors of $\mathcal{A}$, it is of course a quasilinear combination. Therefore, let us just prove (2). Assume $C-rank\mathcal{A}=C-rank\left[\mathcal{A}:\overline{\mathbf{b}}\right]$. In this case, $\overline{\mathbf{b}}$ is in the column space of $\mathcal{A}$ and so it is a ql-combination of column vectors of $\mathcal{A}$. This means from ql-combination definition that there exist real numbers $x_1,x_2,\dots x_n$ such that

$$x_1\left(\begin{array}{c}\left[\overline{a_{11}},\underline{a_{11}}\right]\\ \left[\overline{a_{21}},\underline{a_{21}}\right]\\ ⋮\\ \left[\overline{a_{m1}},\underline{a_{m1}}\right]\end{array}\right)+\dots +x_n\left(\begin{array}{c}\left[\overline{a_{1n}},\underline{a_{1n}}\right]\\ \left[\overline{a_{2n}},\underline{a_{2n}}\right]\\ ⋮\\ \left[\overline{a_{mn}},\underline{a_{mn}}\right]\end{array}\right)\subseteq \overline{\mathbf{b}}.$$

By writing

$$\mathbf{b}=x_1\left(\begin{array}{c}\left[\overline{a_{11}},\underline{a_{11}}\right]\\ \left[\overline{a_{21}},\underline{a_{21}}\right]\\ ⋮\\ \left[\overline{a_{m1}},\underline{a_{m1}}\right]\end{array}\right)+\dots +x_n\left(\begin{array}{c}\left[\overline{a_{1n}},\underline{a_{1n}}\right]\\ \left[\overline{a_{2n}},\underline{a_{2n}}\right]\\ ⋮\\ \left[\overline{a_{mn}},\underline{a_{mn}}\right]\end{array}\right)$$

we get $x=\left(x_1,x_2,\dots x_n\right)\in {\mathbb{R}}^n$ is the solution of $\mathcal{A}x=\mathbf{b}.$ □

Remark 10.

According to this theorem, when we find an interval vector $\overline{\mathbf{b}}$ with $\overline{\mathbf{b}}\supseteq \mathbf{b}$ and with the condition $C-rank\mathcal{A}=C-rank\left[\mathcal{A}:\overline{\mathbf{b}}\right]$, we guarantee an envelope containing the solution of the system $\mathcal{A}x=\mathbf{b}$.

5. Conclusions

There are different definitions of the rank of interval matrices than ours [11,36]. In general, these definitions are important and based on the ranks of classical matrices with real terms that are elements of the interval matrix. Our definition of rank comes from the definition of the rank of a quasilinear operator by first considering that an interval matrix is a quasilinear operator. We defined the rank of a quasilinear operator as the dimension of the space quasispanned by the range of a quasilinear operator. Accordingly, we gave the definition of the rank of the interval matrix. The classical definition of the rank of a matrix is also defined depending on linear operators. So, we think that the definition of rank that we give is more suitable for quasilinear algebra. Furthermore, the notion of quasi-inverse is an extension of the notion of inverse and the interval Cramer’s rule is a result obtained with the help of this definition. We believe that the quasilinear algebra developed in this way can provide a systematic linear algebra-like approach to the solution of other problems related to the solution of linear interval equations and further interval matrix problems.