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Math Help - How can I solve this system of linear equations?

  1. #1
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    How can I solve this system of linear equations?

    Using either Gaussian or Gauss-Jordan elimination, find the value(s) of k, if any, for which the following system will have
    (i) no solution
    (ii) a unique solution, and
    (iii) infinitely many solutions

    x − 3z = −3
    −2x − λy + z = 2
    x + 2y + λz = 1

    Thanks for any help
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  2. #2
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    Re: How can I solve this system of linear equations?

    There is no "k" in your equations. I assume you're looking for values of lambda, not "k", right?

    Can you set this up as a matrix? Can you row-reduce that matrix?
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  3. #3
    Senior Member jakncoke's Avatar
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    Re: How can I solve this system of linear equations?

    if  \lambda = -5 (no solution) or 2 (infinitely many soln), for every other value, a unique solution exists. . I cheated and used determinant.
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    Re: How can I solve this system of linear equations?

    Sorry, do you mind briefly explaining how you got these answers, just I get what I'm asked to do, just confused by the process..
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  5. #5
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    Re: How can I solve this system of linear equations?

    I took the determinant of
     A = \begin{bmatrix} 1 & 0 & -3 \\ -2 & -\lambda & 0 \\ 1 & 2 & \lambda \end{bmatrix}
    to get Det(A) =  - \lambda^{2} - 3 \lambda + 10 and i used the quadratic formula to get the roots,  \lambda = -5, 2 . A system has zero or infinitely many solution only when Det(A) = 0. so Det(A) = 0, when  \lambda = -5, 2 . I then found that for  \lambda = -5 using row reducing, that the system didint have a solution, and again for  \lambda = 2 , i found out by row reducing that it had infinite soln.
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  6. #6
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    Re: How can I solve this system of linear equations?

    Makes so much sense, thank you !
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  7. #7
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    Re: How can I solve this system of linear equations?

    Sorry Jakncoke,
    I was just confirming, would the values for λ be 5, -2 ?

    because of the negative in front of -λ^2 - 3λ + 10
    -(λ - 5)(λ + 2)
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  8. #8
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    Re: How can I solve this system of linear equations?

    Quote Originally Posted by cellae View Post
    Sorry Jakncoke,
    I was just confirming, would the values for λ be 5, -2 ?

    because of the negative in front of -λ^2 - 3λ + 10
    -(λ - 5)(λ + 2)
     (\lambda - 5)(\lambda + 2) = \lambda^{2} + 2 \lambda - 5 \lambda - 10 = \lambda^{2} + -3 \lambda - 10 negative of that is  -\lambda^{2} + 3 \lambda + 10 which is not the determinant (the 3 has to be -3) so root (  \lambda is as stated -5, and 2.
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  9. #9
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    Re: How can I solve this system of linear equations?

    Thanks again !!
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