If the value of the determinant a111b111c is positive then

a111b111c[C1→C1 −aC3;C2→C2 −C3 ] Δ=|0011−ab−111−ac1−cc|=(1−a)(1−c)−(b−1)(1−ac)=1−a−c+ac−b+abc+1−ac=(2+abc)−(a+b+c)D>0 i.e. abc+2>a+b+c;a+b+c>3(abc)1/3; abc+2>3(abc)1/3⇒x3−3x+2>0⇒(x−1)2(x+2)>0,i.e.x>−2 i.e. x3=abc>−8

If the value of the determinant a111b111c is positive then

a111b111c[C1→C1 −aC3;C2→C2 −C3 ] Δ=|0011−ab−111−ac1−cc|=(1−a)(1−c)−(b−1)(1−ac)=1−a−c+ac−b+abc+1−ac=(2+abc)−(a+b+c)D>0 i.e. abc+2>a+b+c;a+b+c>3(abc)1/3; abc+2>3(abc)1/3⇒x3−3x+2>0⇒(x−1)2(x+2)>0,i.e.x>−2 i.e. x3=abc>−8

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If the value of the determinant $|\begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix}|$ is positive then

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$a b c > 0$

$a b c = 0$

$a b c > - 8$

$a b c < - 2$

answer is C.

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Detailed Solution

$|\begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix}|$

$[C_{1} \to C_{1} - a C_{3}; C_{2} \to C_{2} - C_{3}] Δ = | \begin{matrix} 0 & 0 & 1 \\ 1 - a & b - 1 & 1 \\ 1 - a c & 1 - c & c \end{matrix} |$

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