1.If A=(5!6!7!5!⋅6!⋅7!6!⋅7!⋅8!7!⋅8!⋅9!7!⋅8!⋅9!8!⋅9!⋅10!9!⋅10!⋅11!)A = \begin{pmatrix} 5! & 6! & 7! \\ 5!\cdot 6!\cdot 7! & 6!\cdot 7!\cdot 8! & 7!\cdot 8!\cdot 9! \\ 7!\cdot 8!\cdot 9! & 8!\cdot 9!\cdot 10! & 9!\cdot 10!\cdot 11! \end{pmatrix}A=5!5!⋅6!⋅7!7!⋅8!⋅9!6!6!⋅7!⋅8!8!⋅9!⋅10!7!7!⋅8!⋅9!9!⋅10!⋅11!, then ∣adj(adj(2A))∣\left|\text{adj}\bigl(\text{adj}(2A)\bigr)\right|adj(adj(2A)) is equal to:a.2202^{20}220b.282^{8}28c.2122^{12}212d.2162^{16}216Login to continueOnly logged in users canattempt or see the solution.
