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KatClaire
 2 years ago
Best ResponseYou've already chosen the best response.1\[(A^{T})^{1}=(A^{1})^{T}\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1since 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 \((AB)^{1}=B^{1}A^{1}\)

KatClaire
 2 years ago
Best ResponseYou've already chosen the best response.1This 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 \[A^{T}(A^{1})^{T}=((A^{1})A)^{T}=(A^{1}A)^{T}= I^{T} = I\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1I guess that last part is tautological :/

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1hey, I like your way better!
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