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Linear algebra : calculating the determinant

  1. Jan 22, 2009 #1

    fluidistic

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    -1. The problem statement, all variables and given/known data
    Let A and B be 2x2 matrices such that A²B=3I and A^T*B³=-I, where A^T is the transposed matrix of A, and I is the identity matrix.
    Calculate det(A).
    0. The attempt at a solution
    I know I don't know how to approach correctly the problem, but I've tried something.
    I know that the determinant of A squared times the determinant of B is equal to 9 and that minus the determinant of A times the determinant of B cubed is worth 1. From this, I can't reach the answer. And I've no idea about a different approach.
    If you have an idea, I'd be glad to be its tester.
     
    Last edited: Jan 22, 2009
  2. jcsd
  3. Jan 22, 2009 #2

    gabbagabbahey

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    What exactly is the question?:confused:
     
  4. Jan 22, 2009 #3

    fluidistic

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    Oops you're right. The question is to find out det (A).
    I edit the first post to make it clearer.
     
  5. Jan 22, 2009 #4

    gabbagabbahey

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    Okay, in that case you are on the right track ; basically you have 2 equations:

    [tex](\det A)^2\det B=9[/tex] and [tex](\det A)(\det B)^3=1[/tex]

    with two unknowns; [itex]\det A[/itex] and [itex]\det B[/itex]

    How would you normally go about solving a system of two (non-linear) equations and two unknowns? Try solving the first for [itex]\det B[/itex] and substituting it into the second.
     
  6. Jan 22, 2009 #5

    fluidistic

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    That's what I've tried to do, but I gave up because I made an error.
    Well now you gave me confidence and I could reach a result : det (A)=[tex]\sqrt [4] {9^3}[/tex]. Thank you very much!
     
  7. Jan 22, 2009 #6

    gabbagabbahey

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    Unfortunately you still have an error; you should get [tex]\det A =9^{3/5}[/tex]
     
  8. Jan 22, 2009 #7

    fluidistic

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    Oh well, that's maybe a bit too late for me right now. I just saw my error! Unforgivable, I wrote that [tex](det(A)^2)^3=det(A)^5[/tex] instead of [tex]det(A)^6[/tex].
    I got it now. Thank you!
     
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