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KatClaire Group Title

Prove that if A is a square matrix then (A^T)^-1=(A^-1)^T

  • one year ago
  • one year ago

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  1. KatClaire Group Title
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    \[(A^{T})^{-1}=(A^{-1})^{T}\]

    • one year ago
  2. TuringTest Group Title
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    since A is invertible, it can be broken up into the product of two matrices \(A=XY\) combine the properties \((AB)^T=B^TA^T\) and \((AB)^{-1}=B^{-1}A^{-1}\)

    • one year ago
  3. KatClaire Group Title
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    This is what I wrote down, does this make sense haha \[CA^{T}=A^{T}C=I\] \[(A^{-1})^{T}A^{T}=(A(A^{-1}))^{T}=(AA^{-1})^{T}=I^{T} = I\] and \[A^{T}(A^{-1})^{T}=((A^{-1})A)^{T}=(A^{-1}A)^{T}= I^{T} = I\]

    • one year ago
  4. KatClaire Group Title
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    so they both equal I

    • one year ago
  5. TuringTest Group Title
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    I guess that last part is tautological :/

    • one year ago
  6. TuringTest Group Title
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    hey, I like your way better!

    • one year ago
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