The least value of the product xyz for which the determinant x 1 11 y 11 1 z is non-negative is

x 1 11 y 11 1 z=xyz−(x+y+z)+2Since A.M ≥ G.M ⇒x+y+z3≥(xyz)1/3⇒x+y+z≥3(xyz)1/3For least value of xyz, xyz−3 (xyz)1/3+2≥0⇒t3−3t+2≥0 Put t=(xyz)1/3⇒(t+2)t2−2t+1≥0⇒t=−2,1 So, least value of t3=xyz is −8

The least value of the product xyz for which the determinant x 1 11 y 11 1 z is non-negative is

x 1 11 y 11 1 z=xyz−(x+y+z)+2Since A.M ≥ G.M ⇒x+y+z3≥(xyz)1/3⇒x+y+z≥3(xyz)1/3For least value of xyz, xyz−3 (xyz)1/3+2≥0⇒t3−3t+2≥0 Put t=(xyz)1/3⇒(t+2)t2−2t+1≥0⇒t=−2,1 So, least value of t3=xyz is −8

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The least value of the product $x y z$ for which the determinant $|\begin{array}{lll} x 1 1 \\ 1 y 1 \\ 1 1 z \end{array}|$ is non-negative is

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$- 2 \sqrt{2}$

$- 16 \sqrt{2}$

$- 8$

$- 1$

answer is C.

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Detailed Solution

$|\begin{array}{lll} x 1 1 \\ 1 y 1 \\ 1 1 z \end{array}| = x y z - (x + y + z) + 2$

Since A.M $\geq$ G.M

$\begin{aligned} \Rightarrow \frac{x + y + z}{3} \geq (x y z)^{1 / 3} \\ \Rightarrow x + y + z \geq 3 (x y z)^{1 / 3} \end{aligned}$

For least value of $x y z, x y z - 3 (x y z)^{1 / 3} + 2 \geq 0$

$\begin{array}{l} \Rightarrow t^{3} - 3 t + 2 \geq 0 (Put t = (x y z)^{1 / 3}) \\ \Rightarrow (t + 2) (t^{2} - 2 t + 1) \geq 0 \\ \Rightarrow t = - 2, 1 \end{array}$