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Math Help - Checking for value that makes vectors linearly dependent

  1. #1
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    Checking for value that makes vectors linearly dependent

    Problem on the past qualifier asks to find the value of h for which the vectors are linearly dependent:

    v_1=\begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix},v_2=\begin{pmatrix} -5 \\ 7 \\ 8 \end{pmatrix},v_3=\begin{pmatrix} 1 \\ 1 \\ h \end{pmatrix}

    For vectors to be linearly dependent they need to be products of each other multiplied by some coefficient. By mental math I found that with  6*v_1 + v_2 = v_3 vectors are linearly dependent if h = -10.

    Would someone please comment on the correctness and point me to the formulaic way to find such solutions?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Checking for value that makes vectors linearly dependent

    I would use:

    A\begin{pmatrix} 1 \\ -1 \\ -3 \end{pmatrix}+B\begin{pmatrix} -5 \\ 7 \\ 8 \end{pmatrix}+C\begin{pmatrix} 1 \\ 1 \\ h \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

    Where A,B,C\ne0

    Giving the system:

    (1) A-5B+C=0

    (2) -A+7B+C=0

    (3) -3A+8B+hC=0

    Adding (1) and (2) we find:

    B=-C

    Substituting for B into (1) we find:

    A=-6C

    Substituting for A and B into (3) we find:

    C(h+10)=0\:\therefore\:h=-10
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Checking for value that makes vectors linearly dependent

    Another way: v_1,v_2, v_3\;\textrm{l.d.}\Leftrightarrow \textrm{rank}\;[v_1,v_2,v_3]<3\Leftrightarrow \det\;[v_1,v_2,v_3]=0\Leftrightarrow h=-10
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