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Question: \[\left| \begin{matrix} b^{2} + c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^...

b^{2} + c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2} + b^{2} \end{matrix} \right| =$$
A

abcabc

B

4abc4abc

C

4a2b2c24a^{2}b^{2}c^{2}

D

a2b2c2a^{2}b^{2}c^{2}

Answer

4a2b2c24a^{2}b^{2}c^{2}

Explanation

Solution

Δ=∣b2+c2a2a2b2c2+a2b2c2c2a2+b2∣\Delta = \left| \begin{matrix} b^{2} + c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2} + b^{2} \end{matrix} \right|

=−2∣0c2b2b2c2+a2b2c2c2a2+b2∣- 2\left| \begin{matrix} 0 & c^{2} & b^{2} \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2} + b^{2} \end{matrix} \right|,by R1→R1−(R2+R3)R_{1} \rightarrow R_{1} - (R_{2} + R_{3})

= −2∣0c2b2b2a20c20a2∣- 2\left| \begin{matrix} 0 & c^{2} & b^{2} \\ b^{2} & a^{2} & 0 \\ c^{2} & 0 & a^{2} \end{matrix} \right|, by R2→R2−R1R_{2} \rightarrow R_{2} - R_{1} $$R_{3} \rightarrow R_{3} - R_{1}

= −2{−c2(b2a2)+b2(−c2a2)}=4a2b2c2- 2\{ - c^{2}(b^{2}a^{2}) + b^{2}( - c^{2}a^{2})\} = 4a^{2}b^{2}c^{2}.

Trick: Put α\alpha so that the option give different values.