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Math Help - Determining Linear Independence/Dependence using Determinants

  1. #1
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    Determining Linear Independence/Dependence using Determinants

    I'm doing a practice test for my Matrix Algebra exam. One of the questions says to "Use determinants to decide if the set of four vectors shown below is linearly dependent. Explain your reasoning."

    (Sorry, not sure how to type out matrices. These are all 4x1 matrices)

    v1 = [3]
    .......[5]
    .......[2]
    .......[0]

    v2= [0]
    ......[0]
    ......[1]
    ......[1]

    v3 = [0]
    .......[-2]
    .......[-3]
    .......[1]

    v4 = [3]
    .......[3]
    .......[0]
    .......[2]

    I know how to determine linear dependence/independence, but not by using determinants. I can't find any examples in the notes.

    I'm thinking I'd put them all in an augmented matrix and find the determinant of that, but how would I tell if it is linearly dependent?
    Last edited by Lprdgecko; December 14th 2010 at 10:30 PM. Reason: Spelling
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    MHF Contributor FernandoRevilla's Avatar
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    Consider the matrix

    A=(v_1,v_2,v_3,v_4)

    then,
    x_1v_1+x_2v_2+x_3v_3+x_4v_4=0\Leftrightarrow A\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=\b  egin{pmatrix}0\\0\\0\\0\end{pmatrix}

    Now, use Rouche-Fröbenius theorem.

    Fernando Revilla
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  3. #3
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    I'm not sure what the Rouche-Fröbenius theorem is, but, after looking at other sites as well, I took the determinant of the 4x4 matrix [v1 v2 v3 v4] which I got to be -6, which is not equal to 0, so it is linearly dependent? Right?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Lprdgecko View Post
    I'm not sure what the Rouche-Fröbenius theorem is, but, after looking at other sites as well, I took the determinant of the 4x4 matrix [v1 v2 v3 v4] which I got to be -6, which is not equal to 0, so it is linearly dependent? Right?
    No, it is linearly independent. If \det A\neq 0 then, A is an invertible matrix, so:

    \begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}=A^{-1}\begin{pmatrix}0\\0\\0\\0\end{pmatrix}=\begin{pm  atrix}0\\0\\0\\0\end{pmatrix}

    Fernando Revilla
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  5. #5
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    Woopsies, just realized that determinant should be 0. I realized what I did wrong.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Lprdgecko View Post
    Woopsies, just realized that determinant should be 0. I realized what I did wrong.
    All right, if \det A=0 then, the family is linearly dependent.

    Fernando Revilla
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