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Question: If ∆<sub>1</sub> = \(\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matr...

If ∆1 = ∣111abca2b2c2∣,Δ2=∣1bca1cab1abc∣\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \right|,\Delta_{2} = \left| \begin{matrix} 1 & bc & a \\ 1 & ca & b \\ 1 & ab & c \end{matrix} \right| then

A

∆1+∆2=0

B

∆1+2∆2=0

C

∆1= ∆2

D

∆1= 2∆2

Answer

∆1+∆2=0

Explanation

Solution

∴\therefore Δ1=\Delta_{1} = ∣111abca2b2c2∣,=(a−b)(b−c)(c−a)\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \right|, = (a - b)(b - c)(c - a)& Δ2=∣1bca1cab1abc∣\Delta_{2} = \left| \begin{matrix} 1 & bc & a \\ 1 & ca & b \\ 1 & ab & c \end{matrix} \right|

= 1abc=∣aabca2babcb2cabcc2∣\frac{1}{abc} = \left| \begin{matrix} a & abc & a^{2} \\ b & abc & b^{2} \\ c & abc & c^{2} \end{matrix} \right| = ∣a1a2b1b2c1c2∣\left| \begin{matrix} a & 1 & a^{2} \\ b & 1 & b^{2} \\ c & 1 & c^{2} \end{matrix} \right|

=−∣1aa21bb21cc2∣- \left| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \right| =−∣111abca2b2c2∣= - \left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \right| =−Δ1= - \Delta_{1}

∴Δ1+Δ2=0\therefore\Delta_{1} + \Delta_{2} = 0