JEE 2025 — Mathematics PYQ

|A| = –2 Order of matrix A = 3

$$\because adj(KA)=k^{n-1}adjA$$

$$\therefore |3adj(-6adj3A)|$$

$$|3adj(-6\times 3^{3-1}adjA)|$$

$$=|3(-6\times 3^2{)}^{3-1}adjadjA|$$

$$=|3\times 6^2\times 3^{4}adjadjA|$$

$$=3^3\times (6^2{)}^3\times (3^4{)}^3|(adjadjA)|\because |kA|=k^n|A|$$

$$=3^3\times 2^6\times 3^6\times 3^{12}\cdot |A{|}^{(n-1{)}^2}$$

$$=2^6\times 3^{21}\times (-2{)}^4$$

$$=2^{10}\times 3^{21}$$

$$\therefore m+n=10$$

mn = 21 \text{ and } m > n

$$\Rightarrow m=7\text{ and }n=3$$

$$4m+2n=28+6=34$$