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Math Help - Linear algebra dependence & matrices

  1. #1
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    Linear algebra dependence & matrices

    Find the value(s) of h for which the vectors below are linearly dependent.

     \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix},  \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}

    Thanks!
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  2. #2
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    Quote Originally Posted by Linnus View Post
    Find the value(s) of h for which the vectors below are linearly dependent.

     \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix},  \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}

    Thanks!
    \det \left[ \begin{array}{ccc}1&3&-1\\-1&-5&5\\4&7&h \end{array}\right] = 0

    Does that make sense?
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  3. #3
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    I'm confused on what exactly is linear independence. I googled it and this but all the explanations were confusing. Thanks!
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  4. #4
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     \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix},  \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}, \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}
    These three vectors are linearly independent if we write:
    a \begin{bmatrix} 1 \\ -1 \\ 4 \end{bmatrix}+b  \begin{bmatrix} 3 \\ -5 \\ 7 \end{bmatrix}+c \begin{bmatrix} -1 \\ 5 \\ h \end{bmatrix}  = \bold{0} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}
    Then a=b=c=0.

    This leads to the system of equations,
    a+3b-c = 0
    -a-5b+5c=0
    4a+7b+hc=0

    You want that this has a trivial solution i.e. a=b=c=0.

    To do that you need to compute the determinant of that system and make it equal to zero. And that will tell you all the values of h which give non-trivial solutions. Once you know that you can find what would give trivial solutions.
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