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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
Let $A$ be a $3 \times 3$ matrix of non$-$negative real elements such that $A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$. Then the maximum value of $\operatorname{det}(\mathrm{A})$ is $.............$
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Answer

$\begin{aligned} \text { Let } A & =\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right] \\ A\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] & =3\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\end{aligned}$
$ \Rightarrow a_1+a_2+a_3=3 $
$ \Rightarrow b_1+b_2+b_3=3 $
$ \Rightarrow c_1+c_2+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 $a_1=3, b_2=3, c_3=3$

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