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!

Linear Algebra Solving a System of LE

  1. Jan 24, 2013 #1
    Okay here's my system of equations:

    x − 3y − 2z = 0
    −x + 2y + z = 0
    2x + 4y + 6z = 0

    Solve the following systems using Gaussian elimination

    I put it in a matrix and did Gaussian elimination.

    But I can't find a unique solution and it doesn't end up working out. Is this true? Or am I just making a mistake?
     
  2. jcsd
  3. Jan 24, 2013 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    How can we tell without seeing what you did? Why would you expect a unique solution? Show us your work.
     
  4. Jan 24, 2013 #3

    Mark44

    Staff: Mentor

    As LCKurtz said, show us what you did.

    Also, don't blow away the homework template - it's there for a reason.
     
  5. Jan 24, 2013 #4
    2R2 + R2 = | 1 -3 -2 0 |
    -1 2 1 0
    0 8 8 0

    R1 + R2 = | 0 -3 -2 0 |
    0 -1 -1 0
    0 8 8 0

    8R + R3 = | 1 -3 -2 0 |
    0 -1 -1 0
    0 0 0 0

    So I was left with 2 equations.

    x -3y -2z = 0
    -y -z = 0

    And I'm just stuck.
     
  6. Jan 24, 2013 #5
    Basically when I solve it I get z = 0
    then y = -z
    and x = -z

    There's no unique solution I guess. I was specifically looking for a unique solution. I think that was my problem.
     
  7. Jan 24, 2013 #6

    Mark44

    Staff: Mentor

    So far, so good, but you can do just a bit more to completely reduce your matrix. You should end up here:

    $$ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}$$

    This is shorthand for the system
    x + z = 0
    y + z = 0

    Geometrically, what you have is two planes in space that intersect in a line.
     
  8. Jan 25, 2013 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It really makes no sense to "get z= 0" and then have a solution that involves z!

    It should have been obvious from the start that, since we have "= 0" on the right side of each equation (this was a "homogeneous" system), x= y= z= 0 is a solution. So if there had been a unique solution, it would have been trivial.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook