If $a,b,c$ are unequal what is the condition that the value of the determinant, ${\rm{\Delta }} \equiv \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{{a^3} + 1}\\ b&{{b^2}}&{{b^3} + 1}\\ c&{{c^2}}&{{c^3} + 1} \end{array}} \right|$ is 0?

$\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) = 0$

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a,b,c

are unequal what is the condition that the value of the determinant,

\mathrm{\Delta } \equiv \left| {\begin{array}{cccccccccccccccccccc} a&{{a^2}}&{{a^3} + 1}\\ b&{{b^2}}&{{b^3} + 1}\\ c&{{c^2}}&{{c^3} + 1} \end{array}} \right|

is 0?

View answer and explanation

Correct option: A

1 + abc = 0

Option A (

1 + abc = 0

) is correct because the condition for the determinant to be zero is

1 + abc = 0

when a, b, and c are unequal.

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\left|\begin{array}{ccc}a & b-c & b+c \\ c+a & b & c-a \\ a-b & a+b & c\end{array}\right|=0

a, b, c

a x+b y+c=0

{x_2} - {x_3} = 1,\; - {x_1} + 2{x_3} = - 2,\;{x_1} - 2,\;{x_1} - 2{x_2} = 3

2x + 3y + 1 = 0,3x + y - 2 = 0

ax + 2y - b = 0

\left|\begin{array}{ccc}1 & \left(\alpha^{2 x}-\alpha^{-2 x}\right)^2 & \left(\alpha^{2 x}+\alpha^{-2 x}\right)^2 \\ 1 & \left(\beta^{2 x}-\beta^{-2 x}\right)^2 & \left(\beta^{2 x}+\beta^{-2 x}\right)^2 \\ 1 & \left(\gamma^{2 x}-\gamma^{-2 x}\right)^2 & \left(\gamma^{2 x}+\gamma^{-2 x}\right)^2\end{array}\right|

\omega

\left| {\begin{array}{cccccccccccccccccccc} 1&\omega &{{\omega ^2}}\\ \omega &{{\omega ^2}}&1\\ {{\omega ^2}}&1&\omega \end{array}} \right|

A

2 \times 2

A(\operatorname{adj} A)=\left[\begin{array}{cc}10 & 0 \\ 0 & 10\end{array}\right]

|A|