Infinitely Many Solutions: Proving or Disproving a Linear System

Click For Summary
SUMMARY

A linear system with more unknowns than equations does not necessarily have infinitely many solutions. The discussion highlights that when two equations contradict each other, such as x+y+z=1 and x+y+z=2, the system has no solutions. The participants demonstrate that subtracting one equation from another can reveal inconsistencies, leading to false statements like 0=1. Therefore, the assertion that more unknowns than equations guarantees infinite solutions is disproven in specific cases.

PREREQUISITES
  • Understanding of linear equations and systems
  • Familiarity with algebraic manipulation techniques
  • Knowledge of the concepts of solutions in linear algebra
  • Ability to identify contradictions in mathematical statements
NEXT STEPS
  • Study the implications of linear independence in systems of equations
  • Learn about the rank of a matrix and its relation to solutions
  • Explore the concept of homogeneous vs. non-homogeneous systems
  • Investigate the conditions for unique solutions in linear systems
USEFUL FOR

Students of mathematics, educators teaching linear algebra, and anyone interested in solving systems of equations and understanding their properties.

Mic :)
Messages
48
Reaction score
0
Member notified on the mandatory use of the homework template.
If a linear system has more unknowns than equations, then it must have infinitely many solutions.

Prove or disprove.

I'm not at all sure what to do with this one.

Thank you very much for any help. :)
 
Physics news on Phys.org
How many are the solutions of the following system of equations? Two equations with three unknowns...

x+y+z = 1
x+y+z = 2
 
ehild said:
How many are the solutions of the following system of equations? Two equations with three unknowns...

x+y+z = 1
x+y+z = 2

It is not specified.
I'm guessing any combination, unless there is one that disproves it.
I'm under the impression that the statement is true.
 
ehild said:
How many are the solutions of the following system of equations? Two equations with three unknowns...

x+y+z = 1
x+y+z = 2

If your'e able to show me how to do it with 2 equations and 3 unknowns, please go ahead so I can get and idea as to what's going on :)
 
Mic :) said:
It is not specified.
I'm guessing any combination, unless there is one that disproves it.
I'm under the impression that the statement is true.
Can you say a single solution?
Subtract one of the equations from the other one. What do you get? can it be true?
 
ehild said:
Can you say a single solution?
Subtract one of the equations from the other one. What do you get? can it be true?

Is there a way to determine the number of solutions by subtracting one from the other?
 
Subtracting two equations is legal. It serves to cancel some of the variables out.

How would you solve the following equations: x+y=5, x-y=1?
What do you get if you add or subtract them?
 
ehild said:
Subtracting two equations is legal. It serves to cancel some of the variables out.

How would you solve the following equations: x+y=5, x-y=1?
What do you get if you add or subtract them?

Add -----> 2x=6 , x=3

Sub. -----> 2y=4, y =2
 
And x=3, y=2 are solutions, as x+y=3+2=5 and x-y=3-2=1.

If you have two equations,
x+y+z=2 and
x+y+z=1,
these are two equations with three unknowns. You state that there must be infinitely many solutions.
Subtracting the second equation from the first (it is allowed) you get the equation x+y+z - (x+y+z) = 1,
The right-hand side is zero. So you get 0=1 which is a false statement. x+y+z can not be 1 and 2 at the same time. So there are no solutions.

Is it true then that all linear system has infinitely many solutions if the number of unknowns is more than the number of equations?
 
  • #10
Mic :) said:
It is not specified.
I'm guessing any combination, unless there is one that disproves it.
I'm under the impression that the statement is true.

Look at the system
X = 1
X = 2
Does this system have (a) no; (b) one; or (c) infinitely many solutions? Now what happens if you set x+y+z=X?
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
1K
Replies
12
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K