Q47 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
Let $|\mathrm{A}|=6$, where A is a $3 \times 3$ matrix. If $\left|\operatorname{adj}\left(3 \operatorname{adj}\left(\mathrm{~A}^{2} \cdot \operatorname{adj}(2 \mathrm{~A})\right)\right)\right|=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathbf{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to $\_\_\_\_$.
Correct answer: 62
Solution
For $3 \times 3$ matrix: $|\text{adj}(M)| = |M|^2$, $|kM| = k^3|M|$.
$|2A| = 8 \cdot 6 = 48$, $|\text{adj}(2A)| = 48^2 = 2^8 \cdot 3^2$.
$|A^2 \cdot \text{adj}(2A)| = |A|^2 \cdot |\text{adj}(2A)| = 2^2 \cdot 3^2 \cdot 2^8 \cdot 3^2 = 2^{10} \cdot 3^4$.
Let $B = A^2 \cdot \text{adj}(2A)$. $|\text{adj}(B)| = |B|^2 = 2^{20} \cdot 3^8$.
$|3\,\text{adj}(B)| = 3^3 \cdot 2^{20} \cdot 3^8 = 2^{20} \cdot 3^{11}$.
$|\text{adj}(3\,\text{adj}(B))| = (2^{20} \cdot 3^{11})^2 = 2^{40} \cdot 3^{22}$.
$m = 40, n = 22 \Rightarrow m + n = 62$.
$|2A| = 8 \cdot 6 = 48$, $|\text{adj}(2A)| = 48^2 = 2^8 \cdot 3^2$.
$|A^2 \cdot \text{adj}(2A)| = |A|^2 \cdot |\text{adj}(2A)| = 2^2 \cdot 3^2 \cdot 2^8 \cdot 3^2 = 2^{10} \cdot 3^4$.
Let $B = A^2 \cdot \text{adj}(2A)$. $|\text{adj}(B)| = |B|^2 = 2^{20} \cdot 3^8$.
$|3\,\text{adj}(B)| = 3^3 \cdot 2^{20} \cdot 3^8 = 2^{20} \cdot 3^{11}$.
$|\text{adj}(3\,\text{adj}(B))| = (2^{20} \cdot 3^{11})^2 = 2^{40} \cdot 3^{22}$.
$m = 40, n = 22 \Rightarrow m + n = 62$.
Hence, the correct answer is 62.