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KatClaireBest ResponseYou've already chosen the best response.1
\[(A^{T})^{1}=(A^{1})^{T}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

KatClaireBest ResponseYou've already chosen the best response.1
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

KatClaireBest ResponseYou've already chosen the best response.1
so they both equal I
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
I guess that last part is tautological :/
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
hey, I like your way better!
 one year ago
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