I want to determine the number of solutions of a system of linear inequalities, and I was wondering if there was a simple way to to that. I know that linear programming is often used to check whether there are a zero or non-zero number of solutions, i.e. if the system/bounds is/are feasible, but is it possible to distinguish between there being a finite amount of unique solutions or infinitely many solutions? For instance, the system x ≤ y x ≥ y x + y ≤ 1 x + y ≥ 1 has 1 unique solution, namely x = y = 1 2 , while the system x ≥ y x + y ≤ 1 x + y ≥ 1 has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?

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I want to determine the number of solutions of a system of linear inequalities, and I was wondering if there was a simple way to to that. I know that linear programming is often used to check whether there are a zero or non-zero number of solutions, i.e. if the system/bounds is/are feasible, but is it possible to distinguish between there being a finite amount of unique solutions or infinitely many solutions? For instance, the system                    x                    ≤                    y                            x                    ≥                    y                            x        +        y                    ≤                    1                            x        +        y                    ≥                    1            has 1 unique solution, namely   x  =  y  =      1    2  ,  while the system                    x                    ≥                    y                            x        +        y                    ≤                    1                            x        +        y                    ≥                    1            has infinitely many solutions. Is there a away to find out how many solutions a system of linear inequalities has, if any?

kerutak0emro · Accepted Answer

Firstly, let us present the inequality system in the unified form. For example,                    (1)                              {                                                    −                x                +                y                ≥                0                                                                    x                −                y                ≥                0                                                                    −                x                −                y                +                1                ≥                0                                                                    x                +                y                −                1                ≥                0                                                                    x                +                3                y                −                2                ≥                0                ,                                                                        L    k    (  x  ,  y  ,  1  )  ≥  0  ,    k  =  1  ,  2  ,  …  ,  5.Easily to see, that-       L    1    +      L    2    =  0  ,   i.e. the sum of non-negative values equals to zero. Then should       L    1    =      L    2    =  0.  . Therefore, we have the equation instead the pair of the inequalities.- Similarly       L    3    +      L    4    =  0  ,    ⇒        L    3    =      L    4    =  0.-       L    2    +  2      L    3    +      L    5    =  0  ,   i.e. the positive linear combination of non-negative values equals to zero. Then should overdefined       L    2    =      L    3    =      L    5    =  0  , and really we have two independent equalities instead of three inequalities.-       L    5    =      L    1    +  2      L    4    ,   i.e. inequality (1.5) follows from the pair (1.1),(1.4) and can be eliminated.Finally, we have the system         L    1    =      L    3    =  0  ,   with the rank 2 and the single solution.Since the system (1) is presented in the homogenius form, then solutions can exist only if its matrix has rank 2 or less, and any three expressions         L    k     are linearly dependent. The similar situation takes place also in the common case.

I want to determine the number of solutions of a system of linear inequalities, and I was wondering

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