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Invertible Matrices

  1. Oct 26, 2006 #1
    If [tex] A = [a_{ij}]^{n\times n} [/tex] is invertible, show that [tex] (A^{2})^{-1} = (A^{-1})^{2} [/tex] and [tex] (A^{3})^{-1} = (A^{-1})^{3} [/tex]

    So basicaly we have a square matrix with elements [tex] a_{ij} [/tex]. This looks slightly familar to [tex] (A^{T})^{-1} = (A^{-1})^{T} [/tex]. Are [tex] A^{2} [/tex] and [tex] A^{3} [/tex] meant to be the elements of the matrix raised to those respective powers? Or does it mean that the matrix is [tex] 2\times 2 [/tex] or [tex] 3\times 3 [/tex]?
     
    Last edited: Oct 26, 2006
  2. jcsd
  3. Oct 27, 2006 #2

    quasar987

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    The matrix is nxn. A^2 means AA.

    As 90% of linear algebra proofs, these problems are solvable in 2-3 lines. If you really don't find it, I can start you and you will find it imidiately. Laying the first equation is the hardest.
     
  4. Oct 27, 2006 #3
    thanks. I got it, just wasn't clear about the notation.
     
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