QJEEAdv18Maeec3: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Maeec3 Solution: Let P be a matrix of order 3 3 such that all the entries in P…

Let $$Det(P)=\left|\begin{array}{ccc}{a}_2& b_2& c_3\\ a_3& b_3& c_3\end{array}\right|$$

$= {a_1}({b_2}{c_3} - {b_3}{c_2}) - {a_2}({b_1}{c_3} - {b_3}{c_1}) + {a_3}({b_1}{c_2} - {b_2}{c_1})$

Now, maximum value of Det (P) = 6

If ${a_1} = 1$ , ${a_2} = - 1$ , ${a_3} = 1$ , ${b_2}{c_3} = {b_1}{c_3} = {b_1}{c_2} = 1$ and ${b_3}{c_2} = {b_3}{c_1} = {b_2}{c_1} = - 1$

But it is not possible as

$({b_2}{c_3})({b_3}{c_1})({b_1}{c_2})$ = $-$ 1

and $({b_1}{c_3})({b_3}{c_2})({b_2}{c_1})$ = 1

i.e., ${b_1}{c_2}{b_3}{c_1}{c_2}{c_3}$ = 1 and $-$ 1

Similar contradiction occurs when

${a_1} = 1$ , ${a_2} = 1$ , ${a_3} = 1$ , ${b_2}{c_1} = {b_3}{c_1} = {b_1}{c_2}$ = 1 and ${b_3}{c_2} = {b_1}{c_3} = {b_1}{c_2} = - 1$

Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5

Now,

$$\left|\begin{array}{ccc}1& 1& 1\\ -1& 1& 1\\ 1& -1& 1\end{array}\right|=4$$

Hence, maximum value is 4