If A = (1)/(5 ! 6 ! 7 !)[ cases 5 ! & 6 ! & 7 ! 6 ! & 7 ! & 8 ! 7 !...

Given,

A = \dfrac{1}{5 ! 6 ! 7 !}\left[\begin{cases}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{cases}\right]

⇒ \left|A\right| = \dfrac{1}{5 ! 6 ! 7 !}\left|\begin{cases}5 ! & 6 ! & 7 ! \\ 6 ! & 7 ! & 8 ! \\ 7 ! & 8 ! & 9 !\end{cases}\right|

⇒ \left|A\right| = \dfrac{1}{5 ! 6 ! 7 !}5 ! \times 6 ! \times 7 ! \left|\begin{cases}1 & 6 & 42 \\ 1 & 7 & 56 \\ 1 & 8 & 72\end{cases}\right|

$⇒ | A | = 2$

Now using the property of adjoint of matrix we get,

\left|adj \left(adj \left(A\right)\right)\right| = {\left|A\right|}^{( n - 1{)}^{2}} \text{ and } \left|kA\right| = {k}^{n}\left|A\right|$, where$n$is order of square matrix,</p><p>Now using the above two formula we get,</p><p>\left|adj \left(adj \left(2A\right)\right)\right| = {\left|2A\right|}^{4} = {2}^{12}{\left|A\right|}^{4}$

$⇒ \left|adj \left(adj \left(2A\right)\right)\right| = {2}^{12} \cdot {2}^{4}$

$⇒ \left|adj \left(adj \left(2A\right)\right)\right| = {2}^{16}$