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

Jamison Estrada

Jamison Estrada

Answered question

2022-06-03

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?

Answer & Explanation

kerutak0emro

kerutak0emro

Beginner2022-06-04Added 3 answers

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.

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