MathematicsMedium244×since 2002Q4428Let α∈(0,∞)\alpha \in(0, \infty)α∈(0,∞) and A=[12α101012]A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]A=110201α12. If det(adj(2A−AT)⋅adj(A−2AT))=28\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8det(adj(2A−AT)⋅adj(A−2AT))=28, then (det(A))2(\operatorname{det}(A))^2(det(A))2 is equal to:A16B36C49D1Check answerSkip