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Math Help - Linear Algebra: Sub-Space!

  1. #1
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    Linear Algebra: Sub-Space!

    For the vectors in R^3, determine whether the first vector can be expressed as a linear combination of the other two.

    a) (-2,0,3), (1,3,0), (2,4,-1)
    b) (1,2,-3), (-3,2,1), (2,-1,-1)

    If you could please help me solving one of these, I would really appreciate!

    Thanks a lot!
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  2. #2
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    Vectors

    Hello Vedicmaths
    Quote Originally Posted by Vedicmaths View Post
    For the vectors in R^3, determine whether the first vector can be expressed as a linear combination of the other two.

    a) (-2,0,3), (1,3,0), (2,4,-1)
    b) (1,2,-3), (-3,2,1), (2,-1,-1)

    If you could please help me solving one of these, I would really appreciate!

    Thanks a lot!
    (a) If the first vector can be expressed as a linear combination of the other two, then you'll be able to find values of a and b for which

    -2\vec{i} + 0 \vec{j} + 3\vec{k} = a(1\vec{i} + 3\vec{j} + 0\vec{k}) + b(2\vec{i} + 4 \vec{j}  -1\vec{k})

    Comparing coefficients of \vec{i}, \vec{j} and \vec{k}:

    -2 = a + 2b (1)

    0 = 3a + 4b (2)

    3 = -b (3)

    So you've now got three equations in 2 unknowns. Use 2 of them to find values of a and b. Then plug these values into the third. If it works then, yes, you can do it; if it doesn't, you can't.

    So in this case:

    From (3): b = -3

    Subst into (2): 0 = 3a -12

    \Rightarrow a = 4

    And test in (1): a + 2b = 4 -6 = -2, which is correct.

    So, in this case, yes the first vector can be expressed as a linear combination of the other two. It is:

    -2\vec{i} + 0 \vec{j} + 3\vec{k} = 4(1\vec{i} + 3\vec{j} + 0\vec{k}) -3(2\vec{i} + 4 \vec{j}  -1\vec{k})

    Use the same method in part (b).

    Grandad
    Last edited by Grandad; February 6th 2009 at 10:07 PM. Reason: Improve format
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    wow...this is great!
    thanks a lot...great explanation!
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