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KatClaire

  • one year ago

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

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  1. KatClaire
    • one year ago
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    \[(A^{T})^{-1}=(A^{-1})^{T}\]

  2. TuringTest
    • one year ago
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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 \((AB)^{-1}=B^{-1}A^{-1}\)

  3. KatClaire
    • one year ago
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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 \[A^{T}(A^{-1})^{T}=((A^{-1})A)^{T}=(A^{-1}A)^{T}= I^{T} = I\]

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

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

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

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