Answer:
Step-by-step explanation:
We know that for two similar matrices [tex]A[/tex] and [tex]B[/tex] exists an invertible matrix [tex]P[/tex] for which
[tex][tex]B = P^{-1} AP[/tex][/tex]
∴ [tex]B^{T} = (P^{-1})^{T} A^{T} P^{T} \\[/tex]
Also [tex]P^{-1}P = I\\[/tex]
and [tex](P^{-1})^{T} = (P^{T})^{-1}[/tex]
∴[tex](P^{-1})^{T}P^{T} = I[/tex]
so, [tex]B^{T} = (P^{-1})^{T} A^{T} P^{T} = (P^{T})^{-1}A^{T} P^{T}\\B^{T} = A^{T} I\\B^{T} = A^{T}[/tex]
