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Math Help - System of linear equations - types of solutions.

  1. #1
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    System of linear equations - types of solutions.



    Will put up what i have done in a couple of mins
    Last edited by mr fantastic; August 28th 2010 at 01:51 PM. Reason: Re-titled.
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  2. #2
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  3. #3
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    Have i done the echelon system reduction correct?
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  4. #4
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    I follow you up until this point (it's your third line):

    \left[\begin{matrix}1 &2\alpha &1\\ 0 &1 &\alpha+1\\ 0 &\alpha-3 &\alpha-5\end{matrix}\left|\begin{matrix}1\\ 2\\ 3\alpha-4\end{matrix}\right].

    What row operation did you do next?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    I follow you up until this point (it's your third line):

    \left[\begin{matrix}1 &2\alpha &1\\ 0 &1 &\alpha+1\\ 0 &\alpha-3 &\alpha-5\end{matrix}\left|\begin{matrix}1\\ 2\\ 3\alpha-4\end{matrix}\right].

    What row operation did you do next?
    3rd - (2nd x ( \alpha -3))
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  6. #6
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    Hmm. Two comments:

    1. Your fourth line doesn't look like that at all.
    2. I don't think that's the right ERO for this stage in the game. I would do 3rd - (2nd x ( \alpha-3)).

    Come to think of it, that's what your fourth line looks like. I got your numeral 1's mixed up with your parentheses (maybe you should make your parentheses more curved, so they don't get confused with 1's). You actually did what I suggested in my second comment.

    [EDIT]: I think you edited your post. Pick up my comments from here on down.

    So I now follow you down to the fifth line. Your row reduction is correct, I believe.

    Your logic is incorrect in your answer to i. In order for that entry in the matrix to be nonzero, you must have both \alpha\not=2 and \alpha\not=1, (not or, as you wrote).

    I would agree that for ii, it's impossible. And I agree with your answer for iii.

    Does that help?
    Last edited by Ackbeet; August 26th 2010 at 09:50 AM. Reason: Edited post.
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Hmm. Two comments:

    1. Your fourth line doesn't look like that at all.
    2. I don't think that's the right ERO for this stage in the game. I would do 3rd - (2nd x ( \alpha-3)).

    Come to think of it, that's what your fourth line looks like. I got your numeral 1's mixed up with your parentheses (maybe you should make your parentheses more curved, so they don't get confused with 1's). You actually did what I suggested in my second comment.

    [EDIT]: I think you edited your post. Pick up my comments from here on down.

    So I now follow you down to the fifth line. Your row reduction is correct, I believe.

    Your logic is incorrect in your answer to i. In order for that entry in the matrix to be nonzero, you must have both \alpha\not=2 and \alpha\not=1, (not or, as you wrote).

    I would agree that for ii, it's impossible. And I agree with your answer for iii.

    Does that help?
    Im just worried because iv asks for a solution for ii, but i have got it to be impossible.
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  8. #8
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    Hmm. Well, that is a bit puzzling. However, I think your row reduction is correct. Perhaps you could say the empty set?
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  9. #9
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    Im still a bit worried that what ive done is wrong as iv asks for ii. But i've got it not to work. Can anyone else check please?
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  10. #10
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    Quote Originally Posted by adam_leeds View Post
    Im still a bit worried that what ive done is wrong as iv asks for ii. But i've got it not to work. Can anyone else check please?
    I think that your work is correct. The problem is not stated correctly.
    If instead of 3\alpha+1 there were just \alpha then for \alpha =1 parts ii) & iv) have answers.
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