Let $A$ be a $3 imes 3$ matrix of non-negative real elements such that $A\left[\begin{array}{l}1 \ 1 \ 1\end{array} ight]=3\left[\begin{array}{l}1 \ 1 \ 1\end{array} ight]$. Then the maximum value of $\operatorname{det}(\mathrm{A})$ is ______

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Let $A$ be a $3 	imes 3$ matrix of non-negative real elements such that $A\left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]=3\left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]$. Then the maximum value of $\operatorname{det}(\mathrm{A})$ is ______

Aicharya · Accepted Answer

$\begin{aligned} 	ext { Let } A & =\left[\begin{array}{lll}a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3\end{array}ight] \ A\left[\begin{array}{l}1 \ 1 \ 1\end{array}ight] & =3\left[\begin{array}{l}1 \ 1 \ 1\end{array}ight]\end{aligned}$ $\Rightarrow a_1+a_2+a_3=3$ ...(1) $\Rightarrow b_1+b_2+b_3=3$ ...(2) $\Rightarrow c_1+\mathrm{ca}_2+\mathrm{c}_3=3$ ...(3) Now, $\begin{aligned} & |A|=\left(a_1 b_2 c_3+a_2 b_3 c_1+a_3 b_1 c_2ight) \ & -\left(a_3 b_2 c_1+a_2 b_1 c_3+a_1 b_3 c_2ight) \end{aligned}$ $	herefore$ From above in formation, clearly $|\mathrm{A}|_{\max }=27$, when $a_1=3, b_2=3, c_3=3$

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\left[\begin{array}{l}1 \\ 1 \\…

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