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Math Help - Linearly dependent vectors and spans

  1. #1
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    Linearly dependent vectors and spans

    Show that the vectors (1,2,-1),(3,1,1),(1,1,0)(1,-3,3) are linearly dependent by writing the zero vector as a nontrivial combination of them.


    My try:

    c1(1,2,-1)+c2(3,1,1)+c3(1,1,0)+c4(1,-3,3)=0

    [1 3 1 1]
    [2 1 1 -3]
    [-1 1 0 3]

    Would solving this give
    [1 0 1 -2]
    [0 1 0 1]
    [0 0 1 0]

    If this is right, what do I do from here?

    When is the vector (x,y,z) int he span of {(1,2,-1),(3,1,1),(1,-3,3)}?
    I'm suppose to get a equation of a plane here.


    (x,y,z)=k1(1,2,-1)+k2(3,1,1),+k3(1,-3,3)

    The augmented matrix of this is

    [1 3 1]
    [2 1 -3]
    [-1 1 3]

    Pretty confused on where to go from here to get the equation of the plane...

    Thanks
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  2. #2
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    Just pick one of the vectors (say, the last one) and set the corresponding coefficient equal to 1. Now solve for c1, c2, c3. This is easy to do since you have three linear equations in three variables.

    This approach will work unless the first three vectors also happen to be linearly dependent.
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  3. #3
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    [1 3 1 ]
    [2 1 1 ]
    [-1 1 0 ]
    =
    [1 0 1]
    [0 1 0]
    [0 0 1]

    Is that right? Since they all have a pivot, it's dependent?

    What about the second question about the span, and getting the equation of the plane there.
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  4. #4
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    No... how did you get that equation? Where are the unknowns?

    Carefully write down the three equations, then try to turn them into a matrix equation of the form Ax = b. If you show your steps I'll try to help in more detail.

    The second question is very vague... are you looking for a geometric interpretation? An algebraic one?
    The geometric picture will depend on the rank of the matrix formed by the three given vectors. You can determine the rank by trying to put it into row-echelon form. The span of the vectors will then form a linear subspace whose dimension is equal to the rank of the matrix: rank 1 gives you a line, rank 2 a plane, and full rank the entire space R^3. (x,y,z) is in the span if it lies in that subspace.
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  5. #5
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    Quote Originally Posted by evouga View Post
    No... how did you get that equation? Where are the unknowns?

    Carefully write down the three equations, then try to turn them into a matrix equation of the form Ax = b. If you show your steps I'll try to help in more detail.

    The second question is very vague... are you looking for a geometric interpretation? An algebraic one?
    The geometric picture will depend on the rank of the matrix formed by the three given vectors. You can determine the rank by trying to put it into row-echelon form. The span of the vectors will then form a linear subspace whose dimension is equal to the rank of the matrix: rank 1 gives you a line, rank 2 a plane, and full rank the entire space R^3. (x,y,z) is in the span if it lies in that subspace.
    c1(1,2,-1)+c2(3,1,1)+c3(1,1,0)=0

    c1+3c2+c3=0
    2c1+c2+c3=0
    -c1+c2 =0

    [1 3 1][c1]=[a]
    [2 1 1][c2]=[b]
    [-1 1 0][c3]=[c]

    1 3 1
    0 -5 -1
    0 4 1

    1 3 1
    0 1 1/5
    0 0 1/5

    Is that somewhat closer now?

    For question 2:

    [1 3 1]
    [2 1 -3]
    [-1 1 3]

    In row-echelon form would be

    1 3 1
    0 -5 -5
    0 4 4

    1 3 1
    0 1 1
    0 0 0

    Correct? If so, what can I get out of this? Like it says, I need a equation of a plane here
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