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