If AAA is an n×nn \times nn×n matrix, then det(A)=det(AT)\det(A) = \det(A^T)det(A)=det(AT). Does this imply adj(A)=adj(AT)\text{adj}(A) = \text{adj}(A^T)adj(A)=adj(AT)?
Yes, always
No, adj(A)=adj(AT)T\text{adj}(A) = \text{adj}(A^T)^Tadj(A)=adj(AT)T
Only if AAA is symmetric
Only if AAA is invertible