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Math Help - row echelong reduction qusetion-for examination

  1. #1
    n22
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    row echelong reduction qusetion-for examination

    Hi
    After after doing row reduction on matrix[1 3; 2 2]
    what would you get?
    Im checking if method is completely misguided.
    Thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: row echelong reduction qusetion-for examination

    Quote Originally Posted by n22 View Post
    Hi
    After after doing row reduction on matrix[1 3; 2 2]
    what would you get?
    Im checking if method is completely misguided.
    Thanks.
    Let's try this. Show us what and how you got it and we'll help from there.

    -Dan
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  3. #3
    n22
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    Re: row echelong reduction qusetion-for examination

    Quote Originally Posted by topsquark View Post
    Let's try this. Show us what and how you got it and we'll help from there.

    -Dan
    this matrix was formed in order to find an eigenvector or basis using a given eigenvalue .
    got[1 3;0 -4] after some reductions ...which involved ensuring theres a zero beneath the first leading entry..
    row2/-4 then gives [ 1 3 ;0 1]
    now assigning parameters y=t
    x=-3t;y=t
    then [x ;y]=t[-3;1]
    eigenvector then in [-3;1]
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  4. #4
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    Re: row echelong reduction qusetion-for examination

    Yes, that is correct for "row-reduction" but I don't know what you mean about "finding eigenvectors". You cannot find eigenvectors (or eigenvalues) by row reduction!

    In this particular problem, it is easy to see that \begin{bmatrix}1 & 3\\ 0 & 1\end{bmatrix}\begin{bmatrix}-3 \\ 1\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix} which is NOT a multiple of ([3, 1]. [-3, 1] is NOT an eigenvector for this matrix!

    (Here, the eigenvalues are 1+\sqrt{3} and 1- \sqrt{3}. Eigenvectors corresponding to eigenvalue 1+ \sqrt{3} are multiples of [3, \sqrt{3}] and eigenvectors corresponding to eigenvalue 1- \sqrt{3} are multiples of [3, -\sqrt{3}]).
    Last edited by HallsofIvy; June 13th 2013 at 05:52 AM.
    Thanks from topsquark and n22
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