JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
Let $A$ be a matrix of order $3\times3$ and $|A|=5$. If $|2\operatorname{adj}(3A\operatorname{adj}(2A))|=2^{\alpha}\cdot3^{\beta}\cdot5^{\gamma}$, $\alpha,\beta,\gamma\in\mathbb{N}$, then $\alpha+\beta+\gamma$ is equal to
Determinant
Solution
$\operatorname{adj}(2A)=4\operatorname{adj}A$, so $3A\operatorname{adj}(2A)=12A\operatorname{adj}A=12|A|I=60I$. $\operatorname{adj}(60I)=60^2I=3600I$, so $2\operatorname{adj}(\cdot)=7200I$. $|7200I|=7200^3=(2^5\cdot3^2\cdot5^2)^3=2^{15}3^65^6$. $\alpha+\beta+\gamma=15+6+6=27$. Option (C).