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Which makes A to be nonsingular

  1. Oct 2, 2004 #1
    Ques: Show that if A is singular then adj A is also singular. Given that I = [1/ det(A)]*A*adj (A)
    I tried to prove it by assuming adj(A) is nonsingular. so, [1/det(A)]*adj(A) is nonsingular, as well as I. Which makes A to be nonsingular. I feel like I am forcing out this proof. So, do you guys have any good suggestion to this?
     
  2. jcsd
  3. Oct 3, 2004 #2

    matt grime

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    If A is invertible, then there is a B such that AB=1, that alos implies that B is invertible....
     
  4. Oct 3, 2004 #3

    shmoe

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    When you write [1/det(A)]*adj(A) you are assuming that det(A) is not zero and A is non-singular.

    In all cases A*adj(A)=det(A)*I. If A is singular, then det(A)=0. In other words, A*adj(A)=0. What does this say about adj(A)?
     
  5. Oct 3, 2004 #4
    thanks guys, but i still have a ques.. (sure this is easy for you guys) so you are saying if A is singular and A*adj(A)=0 then it implies that adj(A) is also singular? how can that be?
     
  6. Oct 3, 2004 #5

    shmoe

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    Suppose BC=0, where B and C are both square and non-zero.

    If C is non-singular, then multiply by C^(-1) to get:

    B*C*C^(-1)=0*C^(-1)

    or

    B=0, contradicting assumption of B non-zero

    If B is nonsingular multiply by B^(-1) on the left and get C=0.

    Therefore both B and C must be singular.
     
  7. May 10, 2005 #6
    Hi,

    Maybe a late reply, but since I was also looking for a proof to this (got stuck at the BC = 0 thing), just wanted to say thanks! :rofl:

    Btw, for those interested: the opposite is also true (Adj A is singular => A is singular) and much easier to prove obviously.
     
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