Mathematics - Matrices Question with Solution | TestHub

$|A|=\left|\begin{array}{ccc}2& 2+p& 2+p+q\\ 4& 6+2p& 8+3p+2q\\ 6& 12+3p& 20+6p+3q\end{array}\right|$
${\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_2-{\mathrm{C}}_1\times \frac{\mathrm{q}}{2}$
Then ${\mathrm{C}}_3\to {\mathrm{C}}_2-{\mathrm{C}}_1\mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)$
$\begin{array}{rl}& \Rightarrow |\mathrm{A}|=\left|\begin{array}{ccc}2& 0& 0\\ 4& 2& 2+\mathrm{p}\\ 6& 6& 8+3\mathrm{p}\end{array}\right|\\ & \Rightarrow |\mathrm{A}|=2(16+6\mathrm{p}-12-6\mathrm{p})=8=2^3\\ & |\mathrm{adj}(\mathrm{adj}(3\mathrm{A}))|=|3\mathrm{A}{|}^{(3-1{)}^2}=|3\mathrm{A}{|}^4\\ & ={\left(3^3|\mathrm{A}|\right)}^4={\left(3^3\times 2^3\right)}^4=2^{12}\times 3^{12}\\ & \Rightarrow \mathrm{m}+\mathrm{n}=24\end{array}$