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Question: \(\left| \begin{matrix} bc & bc^{'} + b^{'}c & b^{'}c^{'} \\ ca & ca^{'} + c^{'}a & c^{'}a^{'} \\ ab...

∣bcbc′+b′cb′c′caca′+c′ac′a′abab′+a′ba′b′∣\left| \begin{matrix} bc & bc^{'} + b^{'}c & b^{'}c^{'} \\ ca & ca^{'} + c^{'}a & c^{'}a^{'} \\ ab & ab^{'} + a^{'}b & a^{'}b^{'} \end{matrix} \right| is equal to.

A

(ab−a′b′)(bc−b′c′)(ca−c′a′)(ab - a^{'}b^{'})(bc - b^{'}c^{'})(ca - c^{'}a^{'})

B

(ab+a′b′)(bc+b′c′)(ca+c′a′)(ab + a^{'}b^{'})(bc + b^{'}c^{'})(ca + c^{'}a^{'})

C

(ab′−a′b)(bc′−b′c)(ca′−c′a)(ab^{'} - a^{'}b)(bc^{'} - b^{'}c)(ca^{'} - c^{'}a)

D

(ab′+a′b)(bc′+b′c)(ca′+c′a)(ab^{'} + a^{'}b)(bc^{'} + b^{'}c)(ca^{'} + c^{'}a)

Answer

(ab′−a′b)(bc′−b′c)(ca′−c′a)(ab^{'} - a^{'}b)(bc^{'} - b^{'}c)(ca^{'} - c^{'}a)

Explanation

Solution

Sol. Trick : Put a=1,b=−1,c=0a = 1,b = - 1,c = 0

a′=2,b′=2,c′=1\mathbf{a}^{\mathbf{'}}\mathbf{= 2,}\mathbf{b}^{\mathbf{'}}\mathbf{= 2,}\mathbf{c}^{\mathbf{'}}\mathbf{= 1}

Then the determinant is $\left| \begin{matrix} 0 & - 1 & 2 \ 0 & 1 & 2 \

  • 1 & 0 & 4 \end{matrix} \right| = 4$

Option (3) also gives the same value.