Here's the question you clicked on:
KatClaire
Prove that if A is a square matrix then (A^T)^-1=(A^-1)^T
\[(A^{T})^{-1}=(A^{-1})^{T}\]
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 \((AB)^{-1}=B^{-1}A^{-1}\)
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 \[A^{T}(A^{-1})^{T}=((A^{-1})A)^{T}=(A^{-1}A)^{T}= I^{T} = I\]
I guess that last part is tautological :/
hey, I like your way better!