Q. Determinant $\left| \begin{matrix} b^{2} + c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2} + a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2} + b^{2} \end{matrix} \right|$ is equal to

A) $abc$
B) $4\ abc$
C) $4a^{2}b^{2}c^{2}$
D) $a^{2}b^{2}c^{2}$
Explanation

(c)
$\left| \begin{matrix}
b^{2} + c^{2} & a^{2} & a^{2} \\
b^{2} & c^{2} + a^{2} & b^{2} \\
c^{2} & c^{2} & a^{2} + b^{2}
\end{matrix} \right|$
$= - 2\left| \begin{matrix}
0 & c^{2} & b^{2} \\
b^{2} & c^{2} + a^{2} & b^{2} \\
c^{2} & c^{2} & a^{2} + b^{2}
\end{matrix} \right|$ $\ \lbrack R_{1} \rightarrow R_{1} - \left( R_{2} + R_{3} \right)\rbrack$
$= - 2\left| \begin{matrix}
0 & c^{2} & b^{2} \\
b^{2} & a^{2} & 0 \\
c^{2} & 0 & a^{2}
\end{matrix} \right|$ $\ \binom{R_{2} \rightarrow R_{2} - R_{1}}{R_{3} \rightarrow R_{3} - R_{1}}$
$= - 2\left\lbrack - c^{2}\left( b^{2}a^{2} - 0 \right) + b^{2}\left( 0 - a^{2}c^{2} \right) \right\rbrack$
$= - 2\left\lbrack - {2a}^{2}b^{2}c^{2} \right\rbrack = 4a^{2}b^{2}c^{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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