Let A and B are square matrices of same order which   commute  whith each other such  that B is orthogonal and $A^{-1}=A, B^{-1}=B.$ Suppose $X=adj{\left(A{B}^T\right)}^n$ then which of the following is false.(Where det (AB)>0)

$B{B}^T=I,B^{-1}=B\Rightarrow B^T=B^{-1}=B$ and $B^2=I,A^2=I$  
$X=adj{\left(A{B}^T\right)}^n=adj{\left(AB\right)}^n=adj\left(A^n{B}^n\right) (\because AandBarecommite)=\left(adj{B}^n\right)\left(adj{A}^n\right)$   
If n is even $\Rightarrow A^n=B^n=I\Rightarrow X=I,$ If n is odd $\Rightarrow A^n=A,B^n=B$  
$X=adjB.adjA=adj\left(AB\right)={\left(AB\right)}^{-1}\left|AB\right|=B^{-1}{A}^{-1}=BA(\left|AB\right|=1>0)$