If A=(1)/(5!6!7!)[[|gathered 5!

Question

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

Accepted Answer

Given that

$\begin{gathered} A=\frac{1}{5!6!7!}[[|\begin{array}{ccc}5! \quad  6! \quad  7! \ 6! \quad  7! \quad  8! \ 7! \quad  8! \quad  9!\end{array}]]| \ \Rightarrow |A|=\frac{1}{5!6!7!}|\begin{array}{ccc}5! \quad  6! \quad  7! \ 6! \quad  7! \quad  8! \ 7! \quad  8! \quad  9!\end{array}| \ \Rightarrow |A|=\frac{1}{5!6!7!}	imes 5!6!7!|\begin{array}{ccc}1 \quad  6 \quad  7	imes 6 \ 1 \quad  7 \quad  8	imes 7 \ 1 \quad  8 \quad  9	imes 8\end{array}| \end{gathered}$

$\begin{gathered} {R}_{3}	o {R}_{3}-{R}_{2}{R}_{2}	o {R}_{2}-{R}_{1} \ |A|=|\begin{array}{ccc}1 \quad  6 \quad  42 \ 0 \quad  1 \quad  14 \ 0 \quad  1 \quad  16\end{array}|=2 \ |adj(adj(2A))|=|2A{|}^{(n-1{)}^{2}} \ =|2A{|}^{4}={({2}^{3}|A|)}^{4}={2}^{12}|A{|}^{4}={2}^{16} \end{gathered}$

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