If Δ1=∣∣ ∣∣111abca2b2c2∣∣ Ȣ

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If

Δ_{1} = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣, Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣∣ ∣

, then

Options:

A . Δ1+Δ2=0

B . Δ1+2Δ2=0

C . Δ1=Δ2

D . Δ1=2Δ2

Answer: Option A
:
A

Δ_{1} = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣,

&

Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣∣ ∣

In

Δ_{2}

,Transposing the determinant

Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 b c & c a & a b a & a & c \end{matrix} ∣ ∣∣ ∣

C_{1} \to a C_{1}, C_{2} \to b C_{2}, C_{3} \to C C_{3}

Δ_{2} = \frac{1}{a b c} ∣ ∣∣ ∣ \begin{matrix} a & b & c a b c & a b c & a b c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣

Taking abc common from

R_{2}

Δ_{2} = ∣ ∣∣ ∣ \begin{matrix} a & b & c 1 & 1 & 1 a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣

R_{1} \leftrightarrow R_{2}

Δ_{2} = - ∣ ∣∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣∣ ∣

\Rightarrow Δ_{2} = - Δ_{1} \Rightarrow Δ_{2} + Δ_{1} = 0

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