If A = frac{1}{5! 6! 7!} begin{bmatrix} 5! & | vedclass

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(D) Given $A = \frac{1}{5! 6! 7!} \begin{bmatrix} 5! & 6! & 7! \ 6! & 7! & 8! \ 7! & 8! & 9! \end{bmatrix}$.Factoring out $5!, 6!, 7!$ from rows $1,

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$2^{16}$

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(D) Given $A = \frac{1}{5! 6! 7!} \begin{bmatrix} 5! & 6! & 7! \ 6! & 7! & 8! \ 7! & 8! & 9! \end{bmatrix}$.Factoring out $5!, 6!, 7!$ from rows $1, 2, 3$ respectively:$A = \frac{5! 6! 7!}{5! 6! 7!} \begin{bmatrix} 1 & 6 & 6 	imes 7 \ 1 & 7 & 7 	imes 8 \ 1 & 8 & 8 	imes 9 \end{bmatrix} = \begin{bmatrix} 1 & 6 & 42 \ 1 & 7 & 56 \ 1 & 8 & 72 \end{bmatrix}$.Calculating $|A|$:$|A| = 1(7 	imes 72 - 8 	imes 56) - 6(1 	imes 72 - 1 	imes 56) + 42(1 	imes 8 - 1 	imes 7)$$|A| = 1(504 - 448) - 6(16) + 42(1) = 56 - 96 + 42 = 2$.We need to find $|\operatorname{adj}(\operatorname{adj}(2A))|$.Using the property $|\operatorname{adj}(\operatorname{adj}(M))| = |M|^{(n-1)^2}$,where $n=3$ is the order of matrix $A$:$|\operatorname{adj}(\operatorname{adj}(2A))| = |2A|^{(3-1)^2} = |2A|^4$.Since $|2A| = 2^3 |A| = 8 	imes 2 = 16 = 2^4$:$|\operatorname{adj}(\operatorname{adj}(2A))| = (2^4)^4 = 2^{16}$.

If $A = \frac{1}{5! 6! 7!} \begin{bmatrix} 5! & 6! & 7! \\ 6! & 7! & 8! \\ 7! & 8! & 9! \end{bmatrix}$,then $|\operatorname{adj}(\operatorname{adj}(2A))|$ is equal to:

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