Least value of product xyz for which the determinant $left| {begin{array}{*{20}{c}} x&am

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3 and 4 .Determinants and Matrices

hard

The least value of the product $xyz$ for which the determinant $\left| {\begin{array}{*{20}{c}}
x&1&1 \\
1&y&1 \\
1&1&z
\end{array}} \right|$ is non-negative, is

A

$-2\sqrt 2$

B

$-1$

C

$-16\sqrt 2$

D

$-8$

(JEE MAIN-2015)

Solution

$\begin{array}{*{20}{c}}
x&1&1\\
1&y&1\\
1&1&z
\end{array} \ge 0$

$xyz – x – y – z + 2 \ge 0$

$xyz + 2 \ge x + y + z \ge 3{\left( {xyz} \right)^{1/3}}$

$xyz + 2 – 3{\left( {xyz} \right)^{1/3}} \ge 0$

at $\left( {xyz} \right) = {t^3}$

${t^3} – 3t + 2 \ge 0$

$\left( {t + 2} \right){\left( {t – 1} \right)^2} \ge 0$

$\left[ {t = – 2} \right]{t^3} = – 8$

Standard 12

Mathematics

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