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Math Help - linear systems help

  1. #1
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    linear systems help

    Hello,

    Can anyone help me with this problem.

    Find the λ for which the system has a unique solution
    x + y + λz =1
    x +λy + z = 1
    λx +y + z = 1

    Thanks
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  2. #2
    MHF Contributor Calculus26's Avatar
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    The system will have a unique solution for any value of lambda for
    which the 3x3 determinant of the coefficients of the left hand side
    is not 0.

    In your case all lambda but 1 or -2
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  3. #3
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    Quote Originally Posted by Papi3 View Post
    Hello,

    Can anyone help me with this problem.

    Find the λ for which the system has a unique solution
    x + y + λz =1
    x +λy + z = 1
    λx +y + z = 1

    Thanks
    There is no single solution. Any value of \lambda such that the determinant \left|\begin{array}{ccc}1 & 1 & \lambda \\ 1 & \lambda & 1 \\ \lambda & 1 & 1 \end{array}\right| is NOT 0 gives a unique solution.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    in previous post I meant to say the determinant of the left hand side
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  5. #5
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    1 1 lambda
    1 lambda 1
    lambda 1 1

    This matrix has a determinant:

    lambda + lambda + lambda - 1 -1 - lambda3

    This has to be different from 0, to make the matrix non-singular, and thus guarantee that it has only one solution.

    This third-order polynomial has the following solutions:

    -2 and 1

    Any other lambda will give you a system that has a unique solution.
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  6. #6
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    Thanks for your help.

    The determinant gives 3λ-λ^3-2
    How can we find the solutions for λ?

    Thanks again
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  7. #7
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    You can find one solution and then divide the polynomial.

    For example, it is easy to see that 1 is a solution.

    3l - l^3 - 2 = 0

    This means that 3l - l^3 - 2 = p(l ) * (l - 1)

    where p(l ) is a second order polynomial. You can find p(l ) by dividing:

    (3l - l^3 - 2) / (l -1) = -l^2 - l + 2

    and using the solving formula for 2nd order polynomials

    p(l ) = 0 <=> l = -2 V l = 1
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