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Question: \[\left| \begin{matrix} a - 1 & a & bc \\ b - 1 & b & ca \\ c - 1 & c & ab \end{matrix} \right| =\]...

a - 1 & a & bc \\ b - 1 & b & ca \\ c - 1 & c & ab \end{matrix} \right| =$$
A

0

B

(a−b)(b−c)(c−a)(a - b)(b - c)(c - a)

C

a3+b3+c3−3abca^{3} + b^{3} + c^{3} - 3abc

D

None of these

Answer

None of these

Explanation

Solution

∣a−1abcb−1bcac−1cab∣=∣aabcbbcaccab∣−∣1abc1bca1cab∣\left| \begin{matrix} a - 1 & a & bc \\ b - 1 & b & ca \\ c - 1 & c & ab \end{matrix} \right| = \left| \begin{matrix} a & a & bc \\ b & b & ca \\ c & c & ab \end{matrix} \right| - \left| \begin{matrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{matrix} \right|

= −∣aa21bb21cc21∣=−∣aa21b−ab2−a20c−ac2−a20∣- \left| \begin{matrix} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{matrix} \right| = - \left| \begin{matrix} a & a^{2} & 1 \\ b - a & b^{2} - a^{2} & 0 \\ c - a & c^{2} - a^{2} & 0 \end{matrix} \right|

[By R2→R2−R1;R3→R3−R1R_{2} \rightarrow R_{2} - R_{1};R_{3} \rightarrow R_{3} - R_{1}]

= −(a−b)(b−c)(c−a)- (a - b)(b - c)(c - a).