JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Determinants

For an $n\times n$ matrix, $\det (\mathrm{adj}A)=(\det A{)}^{n-1}$; for $n=3$, $\det (\mathrm{adj}A)=(\det A{)}^2$.

Applying $\mathrm{adj}$ repeatedly, each application squares the determinant's exponent: after $k$ applications, $\det \left({\mathrm{adj}}^{(k)}A\right)=(\det A{)}^{2^k}$.

With $\det A=2$ and $k=2024$:

$n=2^{2^{2024}}$.

To find $n\mathrm{mod}9$, first find the order of $2$ modulo $9$: $2^1,2^2,\dots ,2^6\equiv 2,4,8,7,5,1(\mathrm{mod}9)$, so the order is $6$.

So I need $2^{2024}\mathrm{mod}6$. Since $2^k\mathrm{mod}6$ cycles as $2,4,2,4,\dots$ for $k=1,2,3,4,\dots$ (odd $k$ gives 2, even $k$ gives 4), and $2024$ is even,

$2^{2024}\equiv 4(\mathrm{mod}6)$, so $2^{2024}=6m+4$ for some integer $m$.

$n\mathrm{mod}9=2^{6m+4}\mathrm{mod}9=(2^6{)}^m\cdot 2^4\mathrm{mod}9=1^m\cdot 16\mathrm{mod}9=16\mathrm{mod}9=7$.