Q46 / 75 · JEE Main 2025 (22 Jan Shift 1) · Mathematics
Let $A$ be a square matrix of order 3 such that $\operatorname{det}(A)=-2$ and $\operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=2^{\mathrm{m}+\mathrm{n}} \cdot 3^{\mathrm{mn}}, \mathrm{m}\gt\mathrm{n}$. Then $4 \mathrm{~m}+2 \mathrm{n}$ is equal to _______
Correct answer: 34
Solution
$\begin{aligned}
& \text { As } A \operatorname{adj} A=|A| l, \operatorname{det}(\lambda A)=\lambda^n \operatorname{det} A \\
& \operatorname{det}(3 \operatorname{adj}(-6 \operatorname{adj}(3 A)))=3^3 \operatorname{det}(\operatorname{adj}(-6 \operatorname{adj}(3 A))) \\
& =3^3(-6 \operatorname{adj}(3 A))^2 \\
& =3^3(-6)^6|3 A|^4 \\
& =3^9 2^6 \cdot 3^{12} \cdot(-2)^4 \\
& =3^{21} \cdot 2^{10}
\end{aligned}$
Now comparing with given condition
$\begin{aligned}
& 2^{m+n} 3^{m n}=2^{10} \cdot 3^{21} \\
& m+n=10, m n=21 \\
& \Rightarrow \quad m=7, n=3(m>n) \\
& \therefore \quad 4 m+2 n=28+6=34
\end{aligned}$
Hence, the correct answer is 34.