Q39 / 75 · JEE Main 2025 (8 Apr Shift 2) · Mathematics
Let $A=\left[\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right]$.
If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A})))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathrm{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to
If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A})))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathrm{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to
- A22
- Correct answerB24
- C26
- D20
Solution
$|A|=\left|\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right|$
$\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}$
Then $\mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)$
$\begin{aligned}<br/>& \Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc}<br/>2 & 0 & 0 \\ 4 & 2 & 2+\mathrm{p} \\ 6 & 6 & 8+3 \mathrm{p}<br/>\end{array}\right| \\ & \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \\ & |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \\ & =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=24<br/>\end{aligned}$
$\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2}$
Then $\mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right)$
$\begin{aligned}<br/>& \Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc}<br/>2 & 0 & 0 \\ 4 & 2 & 2+\mathrm{p} \\ 6 & 6 & 8+3 \mathrm{p}<br/>\end{array}\right| \\ & \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \\ & |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \\ & =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \\ & \Rightarrow \mathrm{~m}+\mathrm{n}=24<br/>\end{aligned}$
Hence, the correct answer is option: 'B'.