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Determinants and Adjoints

  1. Nov 20, 2005 #1
    Just wondering, how would you solve a problem such as this one:

    Suppose A is an 5 x 5 matrix, with det(A) = 2 find the following:

    det(A^-1 + adj(A))

    Thanks in advance.
     
  2. jcsd
  3. Nov 20, 2005 #2

    mathwonk

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    my first instinct is to multiply by A to see what happens, but I don't know the definition of adj(A). so I don't know what adj(A).A is.


    then use the fact that det respects products.
     
  4. Nov 20, 2005 #3
    isn't adj(A) the adjoint matrix?
     
  5. Nov 20, 2005 #4
    if adj means the adjugate, that is the transpose of the matrix of minors (which I guess some people also call the adjoint, but I save that for the Hermitian adjoint), then use the fact that A^-1=adj(A)/det(A)
     
  6. Nov 20, 2005 #5

    cronxeh

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    A*A^-1 = I

    det(A*A^-1)=det(I)=5
    det(A)*det(A^-1)=5
    2*det(A^-1)=5
    det(A^-1)=5/2
    A*Adj(A) = det(A)*I
    det(Adj(A)) = det(A)^(n-1)
    det(adj(A)) = 2^4 = 16
     
  7. Nov 21, 2005 #6
    I'm pretty sure that det(1)=1
     
  8. Nov 21, 2005 #7

    cronxeh

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    :redface:

    of course!
     
  9. Nov 21, 2005 #8
    A^-1 = adj(A)/det(A)
    => adj(A)= det(A)*A^-1
    det(A)=2 given
    adj(A) = 2A^-1
    A^-1 + adj(A) = A^-1 + 2 A^-1
    =3 A^-1
    det (3A^-1) = 3 det(A^-1) = 3 (det(A))^-1
    =3*2^-1
    =3/2

    How's This
     
  10. Nov 21, 2005 #9
    Tough question!

    Actually guys, thanks for considering my question, but Im afraid all of your answers are different from what the actual answer is.

    It says the answer is (3^5)/2

    Any usefull remarks.
     
  11. Nov 21, 2005 #10
    Sorry in my earlier solution there was a problem
    This is the correct solution

    A^-1 = adj(A)/det(A)
    => adj(A)= det(A)*A^-1
    det(A)=2 given
    adj(A) = 2A^-1
    A^-1 + adj(A) = A^-1 + 2 A^-1
    =3 A^-1
    =>det (3A^-1)
    since A is 5*5 matrix
    det(3A^-1)=3^5 det(A^-1)
    =3^5 (det A)^-1
    =3^5 (2)^-1
    =3^5/2

    enjoy
     
  12. Nov 21, 2005 #11
    neo143 says that det (3A^-1) = 3 det(A^-1), which is incorrect. You should be able to fix this easily yourself, though.
     
  13. Nov 21, 2005 #12

    cronxeh

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    A^-1 = (1/2)*adj(A)
    A^-1 + adj(A) = (1/2)*adj(A) + adj(A) = (3/2)*adj(A)

    det(a*A) = (a^n)*det(A)

    so, det(3/2*adj(A)) = (3/2)^5 * det(adj(A))

    A^-1 = adj(A)/det(A) , adj(A)=(1/2)*(A^-1) , A*A^-1 = I , det(A*A^-1)=1=det(A)*det(A^-1) , det(A^-1) = 1/2

    A^-1 = (1/2)*adj(A)
    adj(A) = 2*(A^-1)
    det(adj(A)) = (2^5)*(1/2)

    det(3/2*adj(A)) = ( (3/2)^5 ) * (( 2^5))/2 = 121.5

    Edit: you already got it
     
  14. Nov 21, 2005 #13
    Thanks, I already knew how to do it after some serious thinking about neo143's first post. Thanks again.
     
  15. Nov 22, 2005 #14

    mathwonk

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    i think i completely solved it for you, modulo the definition of adj.

    i.e. since adj(A).det(A) = A^(-1),

    my advice gives det(A)det(A^(-1)+adj(A)) = det(I + 2I) = det(3I) = 3^5.

    hence det(A^(-1)+adj(A)) = 3^5/2.
     
  16. Nov 24, 2005 #15
    how many definitions does adjoint take?

    1) there is the classical adjoint (its exact definition too messy to write) which has the useful relation A^(-1)=Adj(A)/det(A).

    2) then there is the definition of adjoint as the transpose and conjugate of a matrix.

    These two adjoint operation are different. May i know what notation is usually used? Adj?
     
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