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Solving a linear system of equations

  1. Sep 11, 2007 #1
    Span of vectors in R3

    1. The problem statement, all variables and given/known data
    u1 = (1,0,-1)
    u2 = (1,1,1)
    u3 = (3,1,-1)
    Determine whether the vectors span R^3.


    2. Relevant equations



    3. The attempt at a solution
    I know how to determine if the vectors span R^3(or maybe i dont). In this case checking if I can find a linear combo for (1,0,0)

    Step 1:
    1 = a + b + 3c
    0 = 0 + b + c
    0 = -a + b - c

    Step 2:
    b = -c

    Step 3:
    1 = a - c + 3c
    1 = a + 2c

    Step 4:
    a = b -c
    a = -2c

    Step 5:
    1 = (-2c) + 2c
    1 = 0

    Now with 1 = 0, I would think that the linear system of equations cannot be solved.

    **Solving this problem with my graphing calculator I get the same answer, however my book states that these three vectors do indeed span R^3. Why is that?
     
    Last edited: Sep 11, 2007
  2. jcsd
  3. Sep 11, 2007 #2
    You are looking for whether or not linear combinations of the vectors equal each other. So

    c_1 (1,0,-1)^T + c_2 (1,1,1) + c_3 (3,1,-1) = 0

    Can you find any solution to that other than c_1=c_2=c_3=0? If not the vectors are linearly independent, and you have one condition of span down. What is the other?
     
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