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Math Help - [SOLVED] Vectors - Linear independence

  1. #1
    Member mybrohshi5's Avatar
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    [SOLVED] Vectors - Linear independence

    The vectors
    v =  \begin{bmatrix}0\\3\\-6\\ \end{bmatrix}  u = \begin{bmatrix}-1\\3\\-22+k\\ \end{bmatrix}  w = \begin{bmatrix}-2\\8\\-22\\ \end{bmatrix}


    are linearly independent if and only if k does not equal ______?

    Ok so i set up the matrix

    \begin{bmatrix}0&-1&2\\3&3&8\\-6&-22+k&-22\\ \end{bmatrix}

    then i reduced it down to this

    \begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&-16+k&-6\\ \end{bmatrix}

    Not sure what to do now or if i am even headed in the right direction

    Thanks for any suggestions.
    Last edited by mybrohshi5; March 28th 2010 at 07:27 PM.
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  2. #2
    Newbie vincent's Avatar
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    you're almost there! u, v and w are linearly independant when the equation c_1u+c_2v+c_3v=0 only admits the trivial solution which is when the scalars c_1= c_2= c_3=0. you need to find a k such that your system of equations satisfies this condition. this is the same as saying that your matrix can be reduced to a diagonal matrix!
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by vincent View Post
    you're almost there! u, v and w are linearly independant when the equation c_1u+c_2v+c_3v=0 only admits the trivial solution which is when the scalars c_1= c_2= c_3=0. you need to find a k such that your system of equations satisfies this condition. this is the same as saying that your matrix can be reduced to a diagonal matrix with nonzero diagonal entries!
    A perhaps shorter way is to evaluate the determinant of the matrix whose columns (or rows) are your three vectors, and find the values of k for which this determinant does not vanish. Make sure you understand why this works, however - otherwise it's cheating!
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  4. #4
    Member mybrohshi5's Avatar
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    Quote Originally Posted by vincent View Post
    you're almost there! u, v and w are linearly independant when the equation c_1u+c_2v+c_3v=0 only admits the trivial solution which is when the scalars c_1= c_2= c_3=0. you need to find a k such that your system of equations satisfies this condition. this is the same as saying that your matrix can be reduced to a diagonal matrix!
    ok so i reduced my matrix to a diagonal with the diagonals having nonzero entries

     \begin{bmatrix} 1&0&\frac{2}{3} \\ 0&1&2 \\ 0&0&2-\frac{6}{16+k} \end{bmatrix}

    I am not sure what to do from here though to solve for k?
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    No! First off, you should avoid dividing by an expression containing k because the expression might be zero. In any case, I believe you also made a calculation mistake (the numerator of the fraction should be 16-k and not 16+k).

    When you reach

    <br />
\begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&-16+k&-6\\ \end{bmatrix}<br />

    you can subtract -16+k times the second row from the third, to get

    <br />
\begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&0&-6-2(-16+k)\\ \end{bmatrix} = \begin{bmatrix}1&0&\frac{2}{3}\\0&1&2\\0&0&26-2k\\ \end{bmatrix}.<br />

    No divisions involved!

    Now you want the bottom right corner to be nonzero, i.e. you want k\neq 13.

    Check to see if you get the same answer with the determinant method. I didn't check all your calculations so I can't guarantee you that your answer is correct!
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  6. #6
    Member mybrohshi5's Avatar
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    Thank you. 13 was correct.

    I will do it using determinants and see if i get the same answer

    Thanks again for all the help
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