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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Matrices

Let A be a 3 × 3 matrix of non-negative real elements such that A[...

Let AA be a 3×33 \times 3 matrix of non-negative real elements such that A\left[111\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=3\left[111\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]. Then the maximum value of det⁡(A)\operatorname{det}(\mathrm{A}) is ______

Correct answer

0

Step-by-step solution

 Let A=[a1a2a3b1b2b3c1c2c3]A[111]=3[111]\begin{aligned} \text{ Let } A & =\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right] \\ A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] & =3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\end{aligned}\Rightarrow a_1+a_2+a_3=3...(1)...(1)\Rightarrow b_1+b_2+b_3=3...(2)...(2)\Rightarrow c_1+\mathrm{ca}_2+\mathrm{c}_3=3...(3)Now,...(3) Now,∣A∣=(a1b2c3+a2b3c1+a3b1c2)−(a3b2c1+a2b1c3+a1b3c2)\begin{aligned} & |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) \end{aligned}∴\therefore From above in formation, clearly ∣A∣max⁡=27|\mathrm{A}|_{\max }=27, when a1=3,b2=3,c3=3a_1=3, b_2=3, c_3=3
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