If [A=15 !

Question

If $$\mathrm{A}=\frac{1}{5 ! 6 ! 7 !}\left[\[\begin{array}{ccc} 5 ! & 6 ! & 7 ! \ 6 ! & 7 ! & 8 ! \ 7 ! & 8 ! & 9 ! \end{array}$$\right]\], then $$|\operatorname{adj}(\operatorname{adj}(2 \mathrm{~A}))|$$ is equal to :

Accepted Answer

Given that

$$
\begin{aligned}
& A=\frac{1}{5 ! 6 ! 7 !}\left[\[\begin{array}{lll} 5 ! & 6 ! & 7 ! \ 6 ! & 7 ! & 8 ! \ 7 ! & 8 ! & 9 ! \end{array}\end{aligned}$$\right] \\
\& \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !}\left|$$\begin{array}{lll} 5 ! & 6 ! & 7 ! \ 6 ! & 7 ! & 8 ! \ 7 ! & 8 ! & 9 ! \end{array}$$\right| \\
\& \Rightarrow|A|=\frac{1}{5 ! 6 ! 7 !} \times 5 ! 6 ! 7 !\left|$$\begin{array}{lll} 1 & 6 & 7 \times 6 \ 1 & 7 & 8 \times 7 \ 1 & 8 & 9 \times 8 \end{array}$$\right|

\]

$$
\begin{aligned}
& R_3 \rightarrow R_3-R_2 \text { and } R_2 \rightarrow R_2-R_1 \\
& |A|=\left|\[\begin{array}{lll} 1 & 6 & 42 \ 0 & 1 & 14 \ 0 & 1 & 16 \end{array}\end{aligned}$$\right|=2 \\
\& |\operatorname{adj}(\operatorname{adj}(2 A))|=|2 A|\^{(n-1)\^2} \\
\& =|2 A|\^4=\left(2\^3|A|\right)\^4=2\^{12}|A|\^4=2\^{16}

\]

Solutions · 1