1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

If a linear system has more unknowns than equations, then it must have inifnitely many solutions

  1. Nov 17, 2014 #1
    • 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. :)
     
  2. jcsd
  3. Nov 17, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    How many are the solutions of the following system of equations? Two equations with three unknowns....

    x+y+z = 1
    x+y+z = 2
     
  4. Nov 17, 2014 #3
    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.
     
  5. Nov 17, 2014 #4
    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 :)
     
  6. Nov 17, 2014 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Can you say a single solution?
    Subtract one of the equations from the other one. What do you get? can it be true?
     
  7. Nov 17, 2014 #6
    Is there a way to determine the number of solutions by subtracting one from the other?
     
  8. Nov 17, 2014 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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?
     
  9. Nov 17, 2014 #8
    Add -----> 2x=6 , x=3

    Sub. -----> 2y=4, y =2
     
  10. Nov 17, 2014 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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?
     
  11. Nov 17, 2014 #10

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: If a linear system has more unknowns than equations, then it must have inifnitely many solutions
Loading...