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∣ ∣ ∣ ∣ ∣ B 2 + C 2 a B a C B a C 2 + a 2 B C C a C B a 2 + B 2 ∣ ∣ ∣ ∣ ∣ = 4 a 2 B 2 C 2

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Question

\[\begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix} = 4 a^2 b^2 c^2\]

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Solution

\[∆ = \begin{vmatrix}b^2 + c^2 & ab & ac \\ ba & c^2 + a^2 & bc \\ ca & cb & a^2 + b^2\end{vmatrix}\]
\[\begin{vmatrix}a( b^2 + c^2 ) & a^2 b & a^2 c \\ b^2 a & b( c^2 + a^2 ) & b^2 c \\ c^2 a & c^2 b & c( a^2 + b^2 )\end{vmatrix} \left[\text{ Multiplying the three rows by a, b and c }\right]\]

\[= \frac{abc}{abc}\begin{vmatrix}b^2 + c^2 & a^2 & a^2 \\ b^2 & c^2 + a^2 & b^2 \\ c^2 & c^2 & a^2 + b^2\end{vmatrix} \left[\text{ Taking out a, b and c common from the three columns }\right]\]

\[ = \begin{vmatrix}2( b^2 + c^2 ) & 2( a^2 + c^2 ) & 2( a^2 + b^2 ) \\ b^2 & c^2 + a^2 & b^2 \\ c^2 & c^2 & a^2 + b^2\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]

\[ = 2\begin{vmatrix}b^2 + c^2 & a^2 + c^2 & a^2 + b^2 \\ - c^2 & 0 & - a^2 \\ - b^2 & - a^2 & 0\end{vmatrix} \left[\text{ Taking out 2 common from the three columns and then applying }R_2 \text{ to }R_2 - R_1\text{ and }R_3 \text{ to }R_3 - R_1 \right]\]

\[ = 2\begin{vmatrix}0 & c^2 & b^2 \\ - c^2 & 0 & - a^2 \\ - b^2 & - a^2 & 0\end{vmatrix} \left[\text{ Applying }R_1 \text{ to }R_1 + R_2 + R_3 \right]\]

\[ = 2{[ - c^2 ( - a^2 b^2 )] + [ b^2 ( c^2 a^2 )]} \left[\text{ Expanding along }R_1 \right]\]

\[ = 4 a^2 b^2 c^2 \]

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Chapter 5: Determinants - Exercise 6.2 [Page 60]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.2 | Q 33 | Page 60

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