Question

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

• A. $-16\sqrt{2}$
• B. $-2\sqrt{2}$
• C. -1
• D. -8

Answer 1

Answer: (D)

-8

$\left|\begin{array}{ccc}x& 1& 1\\ 1& y& 1\\ 1& 1& z\end{array}\right|$ is non-negative

$\Rightarrow x(yz-1)-(z-1)+(1-y)\ge 0$

$\Rightarrow xyz-(x+y+z)+2\ge 0$

$\Rightarrow xyz-3(xyz{)}^{1/3}+2\ge 0$ ($\because AM\ge GM$)

let $(xyz{)}^{1/3}=a$

$\Rightarrow a^3-3a+2\ge 0$

$\Rightarrow (a-1{)}^2(a+2)\ge 0$

$\Rightarrow a\ge -2$

$\therefore xyz\ge -8$

i.e Least value of $xyz$ is $-8$

Hence, option D.