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Math Help - Solution to matrix with row of zeros!

  1. #1
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    Solution to matrix with row of zeros!

    I have reduced a matrix to the following:

     \left[ \begin{array}{rrr|r} -1&1&0&0 \\ 1&-1&0&0 \\ 0&0&0&0 \end{array} \right]

    Obviously x_1 = x_2.
    How do I figure out what x_3 is here?
    I did this a few years ago but am lost right now... I have the answer for this problem but I don't understand how to reach it myself.

    Cheers to anybody interested!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Solution to matrix with row of zeros!

    Quote Originally Posted by howdigethere View Post
    Obviously x_1 = x_2 How do I figure out what x_3 is here?
    Right, x_1=x_2 and there are no more conditions so, x_3 can take any real value. The solutions of the system are x_1=\lambda,\;x_2=\lambda,\;x_3=\mu with \lambda,\mu \in \mathbb{R} .
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  3. #3
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    Re: Solution to matrix with row of zeros!

    Thanks!

    I am actually finding eigenvectors here and I guess my question should have been: Why would I state the eigenvectors as:
    \lambda _1 = \lambda _2 = \begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}
    as opposed to any other vectors?

    I clearly don't understand eigenvectors...
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  4. #4
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    Re: Solution to matrix with row of zeros!

    What FernandoRevilla told you was that the general solution to your original problem is of the form \left<\lambda, \lambda, \mu\left>. That can be written as \lambda\left<1, 1, 0\right>+ \mu\left<0, 0, 1\right>. The two eigenvectors are \left<1, 1, 0\right> and \left<0, 0, 1\right>. A three by three matrix may have up to three independent eigenvectors- whether that is true in this case depends upon the original matrix. What did you get for the eigenvalues?

    What are you given as the definition of "eigenvector"?
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    Re: Solution to matrix with row of zeros!

    Sorry for late response - and thanks for adding a reply.

    The question I was trying to get my head around was this:

    Find eigenvalues and eigenvectors of \begin{bmatrix} 2&1&0\\ 1&2&0 \\ 0&0&3 \end{bmatrix}

    To which I found \lambda _1 = \lambda _2 = 3, \lambda _3 = 1 .

    Then, for (\lambda _1 = \lambda _2 = 3),  \begin{bmatrix} 2&1&0\\ 1&2&0 \\ 0&0&3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \rightarrow \left[ \begin{array}{rrr|r} -1&1&0&0 \\ 1&-1&0&0 \\ 0&0&0&0 \end{array}\right] \rightarrow x_1=x_2, \ x_3 = \ ?

    So, and I am trying to remember here (!), we describe the vector x in terms of parameters. Which I should have done as
     x_1=x_2=s, \ x_3 = t \rightarrow \langle s, s, t \rangle \rightarrow s \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} (is this right ?!)
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  6. #6
    Ant
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    Re: Solution to matrix with row of zeros!

    That's right.

    Note that the vector x represents an eigenvector (for eigenvalue 3).

    So any eigenvector (with eigenvalue 3) can be expressed as a linear combination of (1,1,0) and (0,0,1) and in particular two linearly independent eigenvectors will be (1,1,0) and (0,0,1) (and any scalar multiplies of these two vectors!)
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