JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Matrices

For a $3\times 3$ matrix: $\det (kM)=k^3\det M$, $\det (\mathrm{adj}M)=(\det M{)}^2$, $\mathrm{adj}(kM)=k^2\mathrm{adj}M$, and $\mathrm{adj}(M^{-1})=M/\det M$. With $\det A=3$: $(2A{)}^{-1}=\frac{1}{2}{A}^{-1}$, so $\mathrm{adj}((2A{)}^{-1})=\frac{1}{4}\mathrm{adj}(A^{-1})=\frac{1}{4}\cdot \frac{A}{3}=\frac{A}{12}$. Multiplying by 3: $\frac{A}{4}$. Taking adj again (using $\mathrm{adj}(cA)=3c^2{A}^{-1}$): $\mathrm{adj}(A/4)=\frac{3}{16}{A}^{-1}$; multiplying by $-3$: $-\frac{9}{16}{A}^{-1}$. Taking adj again (using $\mathrm{adj}(c{A}^{-1})=\frac{c^2}{3}A$): $\frac{27}{256}A$; multiplying by $-4$: $-\frac{27}{64}A$. Finally $\mathrm{adj}\left(-\frac{27}{64}A\right)=3{\left(\frac{27}{64}\right)}^2{A}^{-1}=\frac{2187}{4096}{A}^{-1}$. Then so $m=-36,n=20$, and $m+2n=-36+40=4$.