# Finding the dimension of S

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1. Jun 29, 2016

### FightingWizard

1. The problem statement, all variables and given/known data

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
2. Relevant equations
How to find the dimension of the set S ?

3. 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.

2. Jun 29, 2016

### Ray Vickson

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
$$\begin{array}[rcccc] -2x+y+z&+s_1& & & =4 \\ x-2y+x & &+s_2& &=1 \\ 2x+2y-z & & &+s_3&=5\\ \end{array}\\ x \geq 1, y \geq 2, z \geq 3, s_1, s_2, s_3 \geq 0$$
Your system of equations has 3 equations and 6 variables, and that is the system on which you would start to use matrix methods.