Algebraic Structure to solve a linear system of 2 variables?

In summary, the conversation discussed the methods of "substitution" and "elimination" for solving systems of equations and the importance of choosing the correct algebraic structure, such as a group, ring, field, or integral domain. The example of solving a system with two equations was given and the steps for each method were shown using properties and theorems from the chosen structure. The conversation also touched on the use of vector spaces over fields in solving systems of equations and how some systems can be solved using only the ring structure of Z. Finally, the conversation mentioned the concept of right-angled Artin groups and their connection to solving equations.
  • #1
ksmith630
13
0
Hi folks, I'm hoping to get your opinions about something.

If i were given two linear equations and needed to solve for x and y using the methods of "substitution" and "elimination", what algebraic structure should i use? I can solve the systems just fine, but I'm trying to "explain" each step using properties of groups, rings, fields etc... and it's difficult without deciding on the structure.

1. I must pick from (group, ring, field, integral domain) to solve a system.
2. Once i have chosen the algebraic structure, i'll cite theorems/properties for EACH step of the solutions.


Take this system, which yields one solution (3, 2).

x + y = 5
x - y = 1

Here's the work anyone would do to solve the system in the two methods:

***************************************************************
For the "elimination" method:

x+y=5
x-y=1

1(1x+1y)=(5)1 Multiply the top equation (both sides) by 1
-1(1x11y)=(1)(-1) Multiply the bottom equation (both sides) by -1

note 1+(-1)=0

add eqns together

(1x-1x)+(1y+1y)=5-1
(101)x+(1+1)y=5-1
1+(-1)x+(1+1)y=5-1
0+(1+1)y=5-1
2y=4

divide both sides by 2...
2y/2=4/2
y=2

plug into eqn x+y=5
1x+1(2)=5
1x=5-2 subtract 2 from both sides
1x=3
(1/1)(1)x=(3)(1/1) mult both sides by (1/1)
x=3

So solution is (3,2)

********************************************

By the "substitution" method:

x+y=5
x-y=1

solve for y in eq

1y=5-1x Subtract 1x from both sides
y=(5-1x) Divide both sides by 1
y=5-1x

substitute

1x+(-1)(5-1x)=1
1x-1(5)-1(-1)x=1 Distribute -1 to 5-1x
1x-5+1x=1

1x-5+1x+5=1+5 Add 5 to both sides
1x+1x=6
2x=6

(1/2)(2/1)x=(6/1)(1/2) Mult both sides by (1/2)
x=3

Substitute into eq

1(3)-1y=1
3-1y=1
-1y=1-3 Subtract both sides by 3
-1y=-2
(1/(-1))(-1)y=(-2/1)(1/(-1)) Mult both sides by 1/(-1)
y=-2/(-1)
y=2

So solution is yet again (3,2)

**********************************************

If i had to solve an equation with 1 variable and 1 solution x, like x + 2 = 5 i would solve it using properties in a "group" <Z,+> where + is the normal binary op of addition in Z.

But since (as my work shows above for each method of solving a system with 2 eqns) i now have two binary operations (and rational numbers) so i cannot solve the system in a "group" of integers.
So when solving a system like x + y = 5 and x - y = 1 for x and y, would i then be in a field (of rationals) since i have two binary operations and fractions?

Once i know what structure I'm working in, i can then cite specific properties and theorems from that structure at each step of my work. Note i need to do this twice; once for the "elimination" method and once for the "substitution" method.

The start of my solution will start with a line that looks like:

To solve this system of linear equations, i will be in the (group, ring, field, integral domain) represented by <Z,R,Q,+,x,?,?> where + and x are the ordinary binary ops of the (integers, real numbers, rational numbers).

Can anyone help me fill in the blanks so it flows better? I'm new to this material so I'm not fluent just yet...


Then once the structure is decided, i will cite theorem from that structure that look like:

For any a and b in a (group, ring, field, integral domain) <Z,R,Q,+,x,?,?> if we know a=b, then for any c we know that a*c=b*c...

Can anyone tell me if I'm on the right track? I'm thinking field of rational numbers, but I'm not sure if that's the easiest... Thanks in advance!
 
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  • #2
This may not be official enough, but with systems of equations, you're typically talking about inverting operators on vector spaces over fields. So I would pick the field as the fundamental algebraic structure.
 
  • #3
Ahh perfect. I'll choose the field then- but how do i write the notation? I know if i was working in a group i'd write:

"To solve, i will be in the group <Z,+> where + is the ordinary binary operation in Z."

So for this system in the field with rationals as well as integers, would i write:

"To solve, i will be in the field <Q,+,x> where + and x are the ordinary binary operations in Q." ?
 
  • #4
it turns out that in this case, only the ring structure of Z is needed. in fact, since Z admits a rather obvious description as a Z-module, in truth, only the abelian group structure of Z is needed.

in general, however, most systems of Z-linear equations require the field Q to solve (this will be the case if the solutions are not in Z itself). the reason we can get away with solving these in Z, is that Z is an integral domain, so we can apply the cancellation laws in lieu of dividing.

one caveat: it is not explicitly stated what system 5 and 1 reside in. the integers are probably what was intended.
 
  • #5
A statement like $2x+2y+2z=0$ in $\langle\mathbb{Z}, +\rangle$ is equivalent to $x^2y^2z^2=1$ when we write the group multiplicatively. Equations like this have been much studied, and quite frankly the solutions to these problems, in the 'easiest' case (of free groups), go way over my head! (The solutions use, I believe, Sela's $\mathbb{R}$-trees, whatever they are!)

However, it has been shown recently that if your group is something called a right-angled Artin group (which includes both free groups and free abelian groups) then if $x^py^2q^r=1$ then there exists some $g\in G$ such that $x, y, z\in\langle{g}\rangle$. Which is pretty neat! (The case of free groups and $p=q=r=2$ has been known since 1959.)
 
  • #6
So I'm only working in the ring <Z,x,+> of integers?
 

FAQ: Algebraic Structure to solve a linear system of 2 variables?

What is an algebraic structure?

An algebraic structure is a set of elements with operations defined on them, such as addition, subtraction, multiplication, and division. These operations follow specific rules, known as axioms, and allow for the manipulation and solving of mathematical equations.

How does algebraic structure help solve a linear system of 2 variables?

Algebraic structure provides a framework for the systematic manipulation of equations in a linear system of 2 variables. By applying the rules of algebraic operations, we can isolate and solve for the unknown variables in the system.

What are the basic steps for solving a linear system of 2 variables using algebraic structure?

The basic steps for solving a linear system of 2 variables using algebraic structure are:
1. Identify the variables and write out the equations for the system.
2. Use algebraic operations to manipulate the equations and eliminate one variable.
3. Solve for the remaining variable.
4. Substitute the value of the solved variable into one of the original equations to solve for the other variable.
5. Verify the solution by plugging in the values for the variables into both equations.

Can algebraic structure be used to solve systems with more than 2 variables?

Yes, algebraic structure can be used to solve systems with any number of variables. The same basic steps can be applied, but the complexity of the equations and operations may increase with more variables.

Are there any limitations to using algebraic structure to solve linear systems?

Algebraic structure is a powerful tool for solving linear systems, but it may not always provide a solution. In some cases, the equations may not have a unique solution or may have no solution at all. Additionally, the process of solving a linear system using algebraic structure can become more complex and time-consuming as the number of variables increases.

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