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Math Help - Another check:

  1. #1
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    Another check:

    Been given the matrix A=[k,k,0;k^2,2,k;0,k,k] ; indicates an new row

    Need to find: a) the determinant of A using cofactor expansion, and
    b) the values of k for which A is invertible...

    a)

    detA=kdet[2,k;k,k]-kdet[k^2,k;0,k]-0det[k^2,2;0,k]
    detA=k(2k-k^2)-k(K^3-0)-0
    detA=-k^4-k^3+2k^2

    b)

    0=-k^4-k^3+2k^2

    k= -2,1,0(repeated)
    Therefore A is invertable where DetA doesn't = zero, which is for values of k>1, 0<k<1, -2<k<0, and k<-2
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  2. #2
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    Hello,

    Quote Originally Posted by deragon999 View Post
    Been given the matrix A=[k,k,0;k^2,2,k;0,k,k] ; indicates an new row

    Need to find: a) the determinant of A using cofactor expansion, and
    b) the values of k for which A is invertible...

    a)

    detA=kdet[2,k;k,k]-kdet[k^2,k;0,k]-0det[k^2,2;0,k]
    detA=k(2k-k^2)-k(K^3-0)-0
    detA=-k^4-k^3+2k^2

    b)

    0=-k^4-k^3+2k^2

    k= -2,1,0(repeated)
    Therefore A is invertable where DetA doesn't = zero, which is for values of k>1, 0<k<1, -2<k<0, and k<-2
    It's OK !
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  3. #3
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    fyi,
    -k4 k + 2k

    = -k( k + k 2)
    = -k(k + 2)(k - 1)

    which is easier to solve in (b) than leaving it as -k4 k + 2k.

    Good luck with the assignment. due tomorrow is it not?
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  4. #4
    Moo
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    Quote Originally Posted by Dr Zoidburg View Post
    fyi,
    -k4 k + 2k

    = -k( k + k 2)
    = -k(k + 2)(k - 1)

    which is easier to solve in (b) than leaving it as -k4 k + 2k.

    Good luck with the assignment. due tomorrow is it not?
    I think he did it, but didn't show it, right ?
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  5. #5
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    most likely. But I figured it looks better assignment-wise to show the factorisation.
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