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Linear dependence question

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/P9udvTs

    2. Relevant equations



    3. The attempt at a solution

    So I set the scalar multiples of a, b, and c as x,y,z

    so i had 3 equations

    4x-4y+4z=0
    -4x+4y-4z=0
    -2x-4y-5z=0

    i tried solving it numerous times each time trying to use a different x value, but i cannot get it. Did I even do it correctly up to the point where i get my 3 equations?

    Thanks for any help
     
  2. jcsd
  3. Jan 27, 2013 #2
    The equations are correct. Notice the first and second equations are equivalent so you'll only really use the second and third equation and will get an undetermined solution.
    Write x and y as functions of z for, example. You will hopefully arrive at a solution where (3/2)*a + (1/2)*b - 1*c = 0.
     
  4. Jan 29, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The second equation is just -1 times the first so you really have just two equations.

    "get it"? What are you trying to get? What do you mean by "use a different x value"?

    As said, you really just have two equations: 4x- 4y+ 4z= 0 and -2x- 4y- 5z= 0.
    If you subtract the second equation from the first, you eliminate y leaving 6x+ 9z= 0 so that 9z= -6z and then z= -(2/3)x. Putting that back into the first equation 4x- 4y+ 4(-(2/3)x)= 4x- 4y- (8/3)x= (4/3)x- 4y= 0 so that 4y= (4/3)x and then y= (1/3)x.

    That is, for any value of x, y= (1/3)x and z= -(2/3)x is a solution. x= 3, y= 1, z= -2 is one solution, x= 6, y= 2, z= -4 is another. The set of all solutions to this system of equations forms a one-dimensional vector space.
     
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