Given that , "A^{-1}=B" .
1) We known that,
"\\text{adj}(PQ)=(\\text{adj}Q)(\\text{adj}P) , \\text{for any square matrix } P,Q \\ \\text{of the same order}."
And "\\text{adj}(I)=I" ,where "I" is the identity matrix.
Since "AB=I=BA"
Therefore, "\\text{adj}(AB)=adj(I)=adj(BA)."
"\\implies (adjB)(adjA)=I=(adjA)(adjB)"
"\\implies (adjB)=(adjA)^{-1}" .
2) We known that "(PQ)^T=Q^TP^T" for any square matrix "P,Q" of same order .
Since ,"AB=I=BA"
"\\implies (AB)^T=I^T=(BA)^T"
"\\implies B^TA^T=I=A^TB^T," as "I^T=I"
"\\implies B^T={(A^T)}.^{-1}"
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