JEE Main 2025 — Mathematics Algebra

$$|A|=\left|\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right|$$∣A∣=​246​2+p6+2p12+3p​2+p+q8+3p+2q20+6p+3q​​

$$\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}$$C3​→C3​−C2​−C1​×2q​

Then $\mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)$C3​→C2​−C1​X(1+2p​)

$\begin{aligned}
& \Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc}
2 & 0 & 0 \
4 & 2 & 2+\mathrm{p} \
6 & 6 & 8+3 \mathrm{p}
\end{array}\right| \
& \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \
& |\operatorname{adj}(\operatorname{adj}(3 \mathrm{A}))|=|3 \mathrm{A}|^{(3-1)^2}=|3 \mathrm{A}|^4 \
& =\left(3^3|\mathrm{A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \
& \Rightarrow \mathrm{~m}+\mathrm{n}=24
\end{aligned}$