Question

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

Answer 1

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

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

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

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

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

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

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