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Question: \(2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} - bc & b^{2} - ac & c^{2} - ab \end{matrix}...

2∣111abca2−bcb2−acc2−ab∣=2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} - bc & b^{2} - ac & c^{2} - ab \end{matrix} \right| =

A

0

B

1

C

2

D

3abc3abc

Answer

0

Explanation

Solution

We have 2∣111abca2−bcb2−acc2−ab∣2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} - bc & b^{2} - ac & c^{2} - ab \end{matrix} \right|

= 2∣111abca2b2c2∣−2∣111abcbcacab∣2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \right| - 2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ bc & ac & ab \end{matrix} \right|

= 2∣111abca2b2c2∣−2abc∣abca2b2c2abcabcabc∣2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \right| - \frac{2}{abc}\left| \begin{matrix} a & b & c \\ a^{2} & b^{2} & c^{2} \\ abc & abc & abc \end{matrix} \right|

Applying C1(a),C2(b),C3(c)C_{1}(a),C_{2}(b),C_{3}(c)

=2∣111abca2b2c2∣−2abc(abc)∣abca2b2c2111∣=0= 2\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix} \right| - \frac{2}{abc}(abc)\left| \begin{matrix} a & b & c \\ a^{2} & b^{2} & c^{2} \\ 1 & 1 & 1 \end{matrix} \right| = 0.