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System of equations: finding a plane

  1. Jul 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the following system of equations:

    2x1 - 2x2 -3x3 = -2
    3x1 -3x2 -2x3 + 5x4 = 7
    x1 - x2 -2x3 -x4 = -3


    2. Relevant equations



    3. The attempt at a solution
    Ok so I solved the system and got:

    x1 -x2 = 5
    x3 = 4
    x4 = 0

    so I've got a point (5,0,4,0) but the answer is r(1,1,0,0) + s(-3,0,-2,1) + (5,0,4,0)

    How do I get the other vectors?
     
  2. jcsd
  3. Jul 21, 2012 #2

    tiny-tim

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    how? :confused:
     
  4. Jul 21, 2012 #3

    HallsofIvy

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    x1= 5, x2= 0 is one pair of numbers that satisfy x1- x2= 5.

    However, your "solution" is wrong. For example, x1= 2, x2= 0, x3= 2, x4= 1 satisfy the initial equation but none of x1- x2= 5, x3= 4, or x4= 0 are satisfied.
     
  5. Jul 21, 2012 #4
    2x1 - 2x2 -3x3 = -2
    3x1 -3x2 -2x3 + 5x4 = 7
    x1 - x2 -2x3 -x4 = -3

    x1 - x2 -2x3 -x4 = -3
    4x3 + 8x4 = 16
    x3 + 2x4= 4

    x1 - x2 -2x3 -x4 = -3
    x3 + 2x4 = 4

    x1 - x2 +3x4 = 5
    x3 + 2x4 = 4

    x1 = x2 - 3x4 + 5
    x3 = -2x4 + 4

    Looks like I missed adding the bolded term. This is correct now I hope.
     
  6. Jul 23, 2012 #5
    How do I find r(1,1,0,0) and s(-3,0,-2,1)?
     
  7. Jul 23, 2012 #6

    tiny-tim

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    not following that :confused:
     
  8. Jul 23, 2012 #7

    HallsofIvy

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    Yes, that is correct. And now you are saying that any (x1, x2, x3, x4) satisfying those four equations can be written as (x1, x2, x3, x4)= (x2- 3x4+ 5, x2, -2x4+ 4, x4)= (x2, x2, 0, 0)+ (-3x4, 0, -2x4, x4)+ (5, 0, 4, 0)= x2(1, 1, 0, 0)+ x4(-3, 0, -2, 1)+ (5, 0, 4, 0).
     
  9. Jul 23, 2012 #8
    AHHH thank you so much, I was pulling my hair out over here. I knew it had to be something pretty straight forward.
     
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