What Is the Dimension of a Constrained Set in R3?

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SUMMARY

The dimension of the constrained set S in R3, defined by the inequalities -2x+y+z ≤ 4, x-2y+z ≤ 1, 2x+2y-z ≤ 5, x ≥ 1, y ≥ 2, and z ≥ 3, can be determined through geometric interpretation and algebraic manipulation. The approach involves transforming the inequalities into a system of equations by introducing slack variables, resulting in a system with three equations and six variables. This setup allows for the application of matrix methods to analyze the dimensionality of the set S, which can be 0, 1, 2, or 3.

PREREQUISITES
  • Understanding of linear inequalities and their geometric interpretations.
  • Familiarity with slack variables in linear programming.
  • Knowledge of matrix representation of linear systems.
  • Basic concepts of dimensionality in Euclidean spaces.
NEXT STEPS
  • Learn about slack variables and their role in linear programming.
  • Study matrix methods for solving systems of equations.
  • Explore geometric interpretations of linear inequalities in R3.
  • Research dimensionality concepts in topology and geometry.
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Students studying linear algebra, mathematicians interested in geometric interpretations of inequalities, and anyone working with constrained optimization problems in R3.

FightingWizard
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Homework Statement



Let S denote (x,y,z) in R3 which satisfies the following inequalities:
-2x+y+z <= 4
x-2y+z <= 1
2x+2y-z <= 5
x >=1
y >=2
z >= 3

Homework Equations


How to find the dimension of the set S ?

The Attempt at a Solution


I have tried to transform the inequalities into matrix form but I'm not quite sure that this is even the right way.

Please help
 
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FightingWizard said:

Homework Statement



Let S denote (x,y,z) in R3 which satisfies the following inequalities:
-2x+y+z <= 4
x-2y+z <= 1
2x+2y-z <= 5
x >=1
y >=2
z >= 3

Homework Equations


How to find the dimension of the set S ?

The Attempt at a Solution


I have tried to transform the inequalities into matrix form but I'm not quite sure that this is even the right way.

Please help

Try to think geometrically instead of algebraically. The dimension will either not exist at all, or else will be one of the integers 0, 1, 2 or 3. Is S a 0-dimensional set? (You can use Google to look up what that means.) Is S 1-dimemsional? (What does a bounded 1-dimensional set look like?) Is S 2-dimensional? What does a bounded 2-dimensionl set look like? Does your S look like that?

When you want to approach such problems algebraically, it is more involved and complex than you might think at first. The first step is to eliminate all "algebraic" inequalities, leaving only simple bounds like x >= 1, etc. We do this by introducing so-called slack or surplus variables, one for each inequality. Thus. we re-write ##-2x+y+z \leq 4## as ##-2x+y+z+s_1 = 4##, where ##s_1 \geq 0## is a slack variable. Similarly, ##x-2y+z \leq 1## becomes ##x-2y+z+s_2 = 1##, where ##s_2 \geq 0## is another slack variable. Finally, we re-write ##2x+2y-z \leq 5## as ##2x+2y-z+s_3 = 5##, where ##s_3 \geq 0## is still another slack variable. So, altogether your system becomes
\begin{array}[rcccc]<br /> -2x+y+z&amp;+s_1&amp; &amp; &amp; =4 \\<br /> x-2y+x &amp; &amp;+s_2&amp; &amp;=1 \\<br /> 2x+2y-z &amp; &amp; &amp;+s_3&amp;=5\\<br /> \end{array}\\<br /> x \geq 1, y \geq 2, z \geq 3, s_1, s_2, s_3 \geq 0<br />
Your system of equations has 3 equations and 6 variables, and that is the system on which you would start to use matrix methods.
 

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