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Matrice problem

  1. Jan 28, 2009 #1


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    1. The problem statement, all variables and given/known data
    A= [1 1 -1], B=[0 1 2], C=[3 0 1]

    Show that r=s=t=0

    2. Relevant equations

    3. The attempt at a solution
    I said r=s=t=0

    0[ 1 1 -1] + 0[0 1 2] + 0[3 0 1] so
    0 + 0 + 0= 0
    so rA + sB + tC = 0 and r=s=t=0

    Is that right way to do it?
  2. jcsd
  3. Jan 28, 2009 #2


    Staff: Mentor

    Is that the right way to do what? You title is misleading; it is apparently not a matrix (there is no such word as "matrice" in English) problem, but a problem about vectors.

    Are you trying to show that the vectors are linearly independent? If so, you have to show that the equation rA + sB + tC = 0 has only one solution for the constants r, s, and t. The equation rA + sB + tC = 0 always has what is sometimes called the trivial solution (namely r = s = t = 0). Your job is to show that there are no other solutions for these constants.
  4. Jan 28, 2009 #3


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    Could you please help me on how to start the problem because I don't know how to prove it only has one solution etc..
  5. Jan 28, 2009 #4


    Staff: Mentor

    Write the vectors as columns in a matrix, and then row-reduce the matrix. If you end up with three nonzero rows (three rows, each with a nonzero leading entry), then that means that the only solution is r = 0, s = 0, t = 0, and there are no other solutions.

    If you end up with one or more rows that have all zeros, then there are multiple solutions.

    To help you understand what is going on here, think back to we're trying to do, namely find all solutions of the equation rA + sB + tC = 0. Try to picture this equation with the vectors A, B, and C written in vertical form.

    The preceding equation can be written as matrix equation that looks like this:
    [A B C][r s t]^T = [0] (Note: [r s t]^T is a column vector)

    This matrix equation can be written as an augmented matrix like so:
    [1 0 3 | 0]
    [1 1 0 | 0]
    [-1 2 1 | 0]

    Row reduce this matrix as described at the beginning of this post.

    Is that enough for you to start in on?
  6. Jan 28, 2009 #5
    Err. . . vectors are an 1 x n (row) or n x 1 (column) matrix. . .
  7. Jan 28, 2009 #6


    Staff: Mentor

    True enough.
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