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Linear Algebra - Solution Sets of Linear Systems

  1. Dec 30, 2011 #1
    See Attachment 2 for question or read below
    Describe and compare the solution sets of X_1 + 5 X_2 - 3 X_3 = 0 and X_1 + 5 X_2 - 3 X_3 = -2.

    See Attachment 1 for answer from back of book

    I do not understand how the answer in the back of the book answers the question or were to even begin to get that answer. I do not see how I am suppose to come up with the solution to an equation with three variables and one equation. Am I suppose to assume that the X_1 X_2 X_3 are the same in both equations?
    [1 5 -3 0
    1 5 -3 -2]
    ~
    [1 5 -3 0
    0 0 0 -2]
    system is inconsistent

    Not exactly sure what I'm suppose to do to answer this question or what exactly I'm being asked. Thanks for any help anyone can provide.
     

    Attached Files:

  2. jcsd
  3. Dec 30, 2011 #2

    Dick

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    Yes, you have to assume X1, X2 and X3 represent the same variables in the two equations. It might have been better if they had labeled them (x,y,z) instead of (X1,X2,X3). x+5y-3z=0 is a plane through the origin in (x,y,z) coordinates. x+5y-3z=(-2) is a plane that doesn't pass through the origin but is parallel to the first plane. It's displaced by a constant vector from the first plane, so they don't intersect, as you found. Take another look at the answer now and see if it makes more sense.
     
  4. Dec 30, 2011 #3
    Ok well the concept is making perfect sense to me know but I'm unsure on how to get the vectors u,v,p as they did. The vector p is <-2,0,0>, is it just by coincidence that the vector that the plates are displaced by is the last entry in the second row of as I could reduce the original matrix?

    I'm understand the concept now a bit. The planes are defined by the base vectors u and v. I have to find u and v and the vector connects the two separate parallel planes which are defined by the equations. I'm still unsure how to proceed
     
  5. Dec 30, 2011 #4
    Don't I need more equations to define this form

    Ax = 0
    A = [1 5 -3]
    x = [x y z]

    Ax = b
    A(p + tv) = b
    A = [1 5 -3]
    b = -2

    hmm idk
     
  6. Dec 30, 2011 #5
    oh i got that was a mean problem lol
     
  7. Dec 31, 2011 #6

    Dick

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    The vectors u, v and p that they give are not unique. u and v could be ANY basis for the plane that goes through the origin. And p could be ANY vector that is a difference between points in the two planes. That's maybe what's confusing about the problem.
     
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