JEE Main 2026 Jan 23 Shift 1, Mathematics Q21: Algebra Of Determinants

Use the identities for a $3\times 3$ matrix: $|\text{adj}(M)|=|M{|}^2$, $\text{adj}(kM)=k^2\text{adj}(M)$.

Inside out:

$\text{adj}(2A)=4\text{adj}(A)\Rightarrow A^2\cdot \text{adj}(2A)=4A^2\cdot \text{adj}(A)=4A\cdot (A\cdot \text{adj}(A))=4A\cdot |A|I=24A$.

$3\text{adj}(24A)=3\cdot 24^2\text{adj}(A)=3\cdot 576\text{adj}(A)=1728\text{adj}(A)$.

$\text{adj}(1728\text{adj}(A))=(1728{)}^2\text{adj}(\text{adj}(A))$.

For $n=3$: $\text{adj}(\text{adj}(A))=|A{|}^{n-2}A=|A|A=6A$.

$\therefore \text{adj}(1728\text{adj}(A))=1728^2\cdot 6A$. Taking determinant ($3\times 3$):

$|1728^2\cdot 6A|=(1728^2\cdot 6{)}^3\cdot |A|=1728^6\cdot 216\cdot 6$

$1728=2^6\cdot 3^3$, so $1728^6=2^{36}\cdot 3^{18}$. Also $216\cdot 6=1296=6^4=2^4\cdot 3^4$.

$\therefore |\cdots |=2^{40}\cdot 3^{22}$. So $m=40$, $n=22$, and $m+n=62$.