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Linear algebra proof

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that if the det(A)=1
    then adj(adj(A))=A

    Given Goal
    det(A)=1 adj(adj(A))=A

    Using the following formula A-1=adj(A)/det(A)

    if det(A)=1 then A-1=adj(A)

    likewise A=adj(A-1)/det(A-1)

    if det(A)=1 then det(A-1)=1

    Thus A=adj(A-1)
    A=adj(A-1)=adj(A)adj=adj(adj(A))

    What i'm confused about is

    Not sure if this proof is correct, TA explained it rather quickly so im running on pure memory.

    I'm not sure about a few things

    1.Why can we assume that det(A

    2. why can we use the opposite relation to the equation A=adj(A-1)/det(A-1)

    That's where I get a bit lost on the logic.

    thanks

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 13, 2013 #2
    1) ??

    2) You start off with ##\displaystyle A^{-1} = \frac{adj(A)}{det(A)}##

    Let the matrix ##\displaystyle B = A^{-1}##. Plug B into the above formula and you have

    ##\displaystyle B^{-1} = \frac{adj(B)}{det(B)}##

    ##B## is just ##A^{-1}## so

    ##\displaystyle (A^{-1})^{-1} = \frac{adj(A^{-1})}{det(A^{-1})}##, and ##(A^{-1})^{-1}## is simply ##A##. Therefore, ##\displaystyle A = \frac{adj(A^{-1})}{det(A^{-1})}##
     
  4. Feb 13, 2013 #3
    sorry i did not know that 1 got cut off I meant why can we assume that the det(A)=det(ainverse)

    Also would you agree that my proof is correct?
     
  5. Feb 13, 2013 #4
    In this case, the two determinants are the same. But in general, they are not. There's a relation between the determinant of ##A## and ##A^{-1}##, and I'll let you figure out what that is. You have probably seen it already.

    You'll understand your proof more once you figure out that.
     
  6. Feb 13, 2013 #5
    Only relationship I know that involves A and Ainverse is that (A-1)-1 =A

    is that the path you're leading me to?
     
  7. Feb 13, 2013 #6

    jbunniii

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    det(AB) = det(A)det(B). What does this tell you if you put B = A-1?
     
  8. Feb 13, 2013 #7
    We would get det(A*A^-1) which equals 1
     
  9. Feb 13, 2013 #8
    Yes, it equals 1 but more importantly, you can split the det(A*A^-1) as jubunniii suggested.
     
  10. Feb 13, 2013 #9
    Still pretty confused on how that would show that the adj(adj(A))=A

    Where would that relationship help with the proof?

    Right now I know det(A)=1

    1.A-1=adj(A)/det(A)

    and 2. A=adj(A-1)/det(A-1)

    From 1 and 2 I get A-1=adj(A)

    and A=adj(A-1)

    Here is where I get completely lost.
     
  11. Feb 13, 2013 #10
    Try manipulating the A^-1 and A formulas involving the adjoint in note #2 of the 2nd post in this thread. Then, take the adjoint of both sides of one formula and use the other formula to conclude that adj(adj(A)) = A. Remember that you're given det(A) = 1 which makes both formulas simple.
     
  12. Feb 13, 2013 #11
    I figured it out, it was pretty obvious haha what had to be done. Thanks
     
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