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(Please help ) If A² = I, prove det A = ±1

  1. May 23, 2008 #1
    (Please help!!) If A² = I, prove det A = ±1

    If A² = I, show that det A = ±1

    This book is very unclear, but I am assuming by "I" they mean the identity matrix with a size of 2x2. I have tried putting in A for row 1 column 1 - B for 1,2 - C for 2,1 and D for 2,2 multiplying and setting the results equal to the values of the identity matrix. I thought I was close, but now am doubting that I am going about this the right way. Any help is MUCH appreciated! Thank you!
     
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  3. May 23, 2008 #2

    Hurkyl

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    Well, in this case, doing matrix arithmetic is going to be a lot easier than doing scalar arithmetic on the entries of the matrix.
     
  4. May 23, 2008 #3

    tiny-tim

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    Welcome to PF!

    Hi chapone! Welcome to PF! :smile:

    What makes you think they mean 2x2?

    This theorem is true for any n x n matrix. :smile:

    Do you know any formulas for determinants (for example, for det (AB))?
     
  5. May 26, 2008 #4
    Take the determinant of both sides of A[tex]^{2}[/tex] = I.

    (Any don't make any assumptions beyond what the book gives you.)
     
  6. Oct 14, 2010 #5
    Re: (Please help!!) If A² = I, prove det A = ±1

    a² = 1
    a² - 1 = 0
    (a - 1) (a + 1) = 0
    a=1 and a = -1
    then a = ±1
     
  7. Oct 14, 2010 #6

    Mark44

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    Re: (Please help!!) If A² = I, prove det A = ±1

    This is all well and good for a real number a, but the OP is working with a matrix A, not a scalar. As such, A [itex]\neq[/itex] 1.
     
  8. Oct 14, 2010 #7

    Fredrik

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    Re: (Please help!!) If A² = I, prove det A = ±1

    chapone got 3 good answers already, so I'll just add to what Mark44 said (also a good post, but not an answer for chapone) by providing a counterexample that shows that John's argument gets the wrong result for matrices:

    [tex]\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}[/tex]
     
  9. Oct 23, 2010 #8
    Re: (Please help!!) If A² = I, prove det A = ±1

    If A^2=1, then A is a matrix of order 2, which means that A is invertible.
     
    Last edited by a moderator: Oct 23, 2010
  10. Oct 23, 2010 #9

    D H

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    Re: (Please help!!) If A² = I, prove det A = ±1

    There are multiple meanings of the term "order", and your meaning here is not the meaning usually meant for matrices.
     
    Last edited: Oct 23, 2010
  11. Oct 24, 2010 #10
    Re: (Please help!!) If A² = I, prove det A = ±1

    What do you mean? Order in the sense of "group order", meaning A generates a group of order 2. Implying A must be invertible
     
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