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Math

posted by on .

If
kx + y + z = 1
x + ky + z = k
x + y + kz = k^2
has unique solution

then k is not equal to 1 and W. What is W?

  • Math - ,

    since the determinant of the system is

    | k 1 1 |
    | 1 k 1 |
    | 1 1 k |
    = k^3 - 3k + 2
    = (k+2)(k-1)^2

    Then k cannot be -1 or 1

  • math - oops - ,

    did I say -1?
    I meant -2

  • Math - ,

    The determinant of the three equation left sides cannot be zero.
    (k^3 -k) +(1-k) +(1-k) NOT=0
    k^3-k NOT= (k-1)
    k(k^2 -1) NOT= k-1
    k(k+1)NOT =1
    k cannot be 0 or -1
    W = -1

  • Math - ,

    Steve is right. I left out a 2

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