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Math Help - Basis of these vectors

  1. #1
    Member zzizi's Avatar
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    Basis of these vectors

    Im trying to figure out the following: can someone tell me the process and I will try and work it on my own. I am just not sure how to work it out.


    Thanks

    Let S be the subspace of R^4 with the vectors as basis:

    (1,1,0,-1) (1,3,0,1) (-1,0,1,8)

    a) Show that the (6,9,-9,-9)^T is in S
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  2. #2
    MHF Contributor

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    Re: Basis of these vectors

    solve the equation a(1,1,0,-1) + b(1,3,0,1) + c(-1,0,1,8) = (6,9,-9,-9) for a,b, and c.

    this leads to: (a+b-c,a+3b,c,-a+b+8c) = (6,9,-9,-9) or the 4 linear equations:

    a+b-c = 6
    a+3b = 9
    c = -9
    -a+b+8c = -9

    or, if you prefer, the matrix equation:

    \begin{bmatrix}1&1&-1\\1&3&0\\0&0&1\\-1&1&8 \end{bmatrix} \begin{bmatrix}a\\b\\c \end{bmatrix} = \begin{bmatrix}6\\9\\-9\\-9 \end{bmatrix}

    i can tell you right now, however, that a better question to ask yourself is NOT:

    "how do i prove (6,9,-9,-9) is in S?"

    but rather:

    "how do i tell IF (6,9,-9,-9) is in S?"

    S is SUBSPACE of R4, of dimension 3 < 4, so there are going to be LOTS of vectors that AREN'T in S.
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