JEE Main 2024 — 4 April, Shift 1 — Question 27

A=\left[ \begin{array}{*{35}{l}}{{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\{{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\{{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\end{array} \right] \text{A}\left[ \begin{array}{*{35}{l}}1 \\1 \\1 \\\end{array} \right]=3\left[ \begin{array}{*{35}{l}}1 \\1 \\1 \\\end{array} \right]

$\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}=3$

$\Rightarrow \mathrm{b}_{1}+\mathrm{b}_{2}+\mathrm{b}_{3}=3$

$\Rightarrow \mathrm{c}_{1}+\mathrm{ca}_{2}+\mathrm{c}_{3}=3$

Now,

$|A|=\left(a_{1} b_{2} c_{3}+a_{2} b_{3} c_{1}+a_{3} b_{1} c_{2}\right)$ $-\left(a_{3} b_{2} c_{1}+a_{2} b_{1} c_{3}+a_{1} b_{3} c_{2}\right)$

$\therefore$ From above in formation, clearly $|\mathrm{A}|_{\max }=27$,

when $\mathrm{a}_{1}=3, \mathrm{~b}_{2}=3, \mathrm{c}_{3}=3$