If the value of the determinant a 1 11 b 11 1 c is positive, then

A
$a b c > 1$
B
$a b c > - 8$
C
$a b c < - 8$
D
$a b c > - 2$

Solution:

We have,

$Δ = |\begin{array}{lll} a 1 1 \\ 1 b 1 \\ 1 1 c \end{array}| = a b c - (a + b + c) + 2$

$\begin{array}{ll} ∴ Δ > 0 \\ \Rightarrow a b c + 2 > a + b + c \\ \Rightarrow a b c + 2 > 3 (a b c)^{1 / 3} [∵ A.M. > G.M. \Rightarrow \frac{a + b + c}{3} > (a b c)^{1 / 3}] \end{array}$

$\begin{array}{l} x^{3} + 2 > 3 x, where x = (a b c)^{1 / 3} \\ \Rightarrow x^{3} - 3 x + 2 > 0 \\ \Rightarrow (x - 1)^{2} (x + 2) > 0 \\ \Rightarrow x + 2 > 0 \Rightarrow x > - 2 \Rightarrow (a b c)^{1 / 3} > - 2 \Rightarrow a b c > - 8 \end{array}$

If the value of the determinant $|\begin{array}{lll} a 1 1 \\ 1 b 1 \\ 1 1 c \end{array}|$ is positive, then

Solution:

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