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Linear Algebra: given adj(A) find A

  1. Dec 17, 2007 #1
    [SOLVED] Linear Algebra: given adj(A) find A

    1. The problem statement, all variables and given/known data

    If [tex]adj\mathbb A = \left(\begin{array}{ccc}1&0&1\\1&-1&0\\0&2&1\end{array}\right)[/tex], find A. Briefly justify your algorithm.

    2. Relevant equations

    [tex]adj\mathbb A=(det\mathbb A)(\mathbb A^{-1})[/tex]

    3. The attempt at a solution

    [tex]adj\mathbb A=(det\mathbb A)(\mathbb A^{-1})[/tex]
    invert both sides to get:
    [tex](adj\mathbb A)^{-1} = [(det\mathbb A)(\mathbb A^{-1})]^{-1}[/tex]
    [tex](adj\mathbb A)^{-1} = (\mathbb A^{-1})^{-1}(det\mathbb A)^{-1}[/tex]
    [tex](adj\mathbb A)^{-1} = (\mathbb A)(\frac{1}{det\mathbb A})[/tex]
    [tex]\mathbb A = (det\mathbb A)(adj\mathbb A)^{-1}[/tex]

    My, isn't that nice.

    I computed [tex](adj\mathbb A)^{-1}[/tex], and found it to be
    [tex]\left(\begin{array}{ccc}-1&2&1\\-1&1&1\\2&-2&-1\end{array}\right)[/tex]
    But I have no clue how to find detA, so I'm stuck with one equation:
    [tex]\mathbb A = (det\mathbb A)(adj\mathbb A)^{-1}[/tex]
    and two unknowns:
    [tex]\mathbb A[/tex] and [tex]det\mathbb A[/tex]
     
    Last edited: Dec 17, 2007
  2. jcsd
  3. Dec 17, 2007 #2

    Hurkyl

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    You have an equation involving A, and you need an equation involving det A... that should suggest something...
     
  4. Dec 18, 2007 #3

    HallsofIvy

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    Are you aware that "adjoint" is a "dual" property?
    that is, that the "adjoint of the adjoint of A" is A itself. You are given the adjoint of A and are asked to find A- just find the adjoint of the matrix you are given.
     
  5. Dec 18, 2007 #4
    Hurkyl, did you mean for me to use HOI's formula, or did you have something else in mind? (and no, I'm not trying to mooch the answer :smile:)

    Thanks, HallsofIvy! No, I had no idea that adj(adjA)=A. Did I ever know that...? Nope; can I derive it?
    :much kerfuffle, then gives up and looks through notes:
    Well, fancy that! I have played with adj(adjA) before, and found that it equals [tex](det\mathbb A)^{n-2}\mathbb A[/tex] (if you'd like to see the whole derivation, I'll type it up, but I did it a very long way and then we proved it differently in class, and got the same result)
    Playing with it again (using [tex]adj\mathbb A=(det\mathbb A)(\mathbb A^{-1})[/tex]), I can't get adj(adjA)=A. Can it be derived using only that formula?
    Thanks!
     
  6. Dec 19, 2007 #5

    HallsofIvy

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    What exactly is your definition of adj(A)? (There are several equivalent definitions.) How you would prove that depends strongly on your definition of adjoint.
     
  7. Dec 19, 2007 #6
    We defined adj(A) as the transpose of the cofactor matrix of A.
     
  8. Dec 31, 2007 #7
    OK, I asked my professor about adj(adjA), and he said it only equals A in special cases. The general formula is [tex]adj(adj\mathbb A)=(det\mathbb A)^{n-2}\mathbb A[/tex], so you can be sure that it equals A when A is a 2x2 matrix.

    Right, so basically, what I'm asking is: is there another method of finding A given adjA?

    Thanks!
     
  9. Dec 31, 2007 #8

    Hurkyl

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    There are other times you can solve that equation for A...
     
  10. Jan 3, 2008 #9
    Hurkyl, I so have not been ignoring your help.
    I just spent a couple of days randomly interrupting conversations with, "Hang on! I think I got it!" only to find that I hadn't got it, after all.
    But now... da dum!
    I was playing with my calculator, and I accidentally found the numeric value of det(adjA).
    One thing led to another, and now: BEHOLD!!! (where's that emoticon with a brass brand when you need one?)

    Given that for my matrix, det(adjA)=1.

    I happen to know (well, my notes know it, but I could theoretically re-derive it, right?) that [itex]det(adj\mathbb A)=(det\mathbb A)^{n-1}[/itex].

    So... [itex]1=(det\mathbb A)^{3-1}[/itex], or more simply [itex]det A=\pm 1[/itex]

    At last, I can use that equation that I'm in love with, and say
    [tex]\mathbb A^{-1}=\frac{adj\mathbb A}{det\mathbb A}[/tex]

    [tex]\mathbb A=(adj\mathbb A)^{-1}det\mathbb A[/tex]

    [tex]\mathbb A=(adj\mathbb A)^{-1}(\pm 1)[/tex]

    [tex]\mathbb A=(\pm1)\left(\begin{array}{ccc}-1&2&1\\-1&1&1\\2&-2&-1\end{array}\right)[/tex]

    Which would explain why my professor gave me so much credit on the exam for finding the inverse of adjA and mumbling stuff.

    Yay!!!

    Thanks, people! :smile:
     
    Last edited: Jan 3, 2008
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