Question

Prove that

\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]

Answer 1

\[∆ = \begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix}\]

\[ = \frac{1}{abc}\begin{vmatrix}a^2 + b^2 & c^2 & c^2 \\ a^2 & b^2 + c^2 & a^2 \\ b^2 & b^2 & c^2 + a^2\end{vmatrix} \left[\text{ Multiplying }R_1 , R_2 \text{ and }R_3\text{ by c, a and b and then dividing by abc }\right]\]

\[ = \frac{1}{abc}\begin{vmatrix}a^2 + b^2 & c^2 - a^2 - b^2 & c^2 - a^2 - b^2 \\ a^2 & b^2 + c^2 - a^2 & 0 \\ b^2 & 0 & c^2 + a^2 - b^2\end{vmatrix} \left[\text{ Applying }C_2\text{ to }C_2 - C_1\text{ and }C_3\text{ to }C_3 - C_1 \right]\]

\[ = \frac{1}{abc}\begin{vmatrix}0 & - 2 b^2 & - 2 a^2 \\ a^2 & b^2 + c^2 - a^2 & 0 \\ b^2 & 0 & c^2 + a^2 - b^2\end{vmatrix} \left[\text{ Applying }R_1\text{ to }R_1 - R_2 - R_3 \right]\]

\[ = \frac{1}{abc}[ - a^2 \begin{vmatrix}- 2 b^2 & - 2 a^2 \\ 0 & c^2 + a^2 - b^2\end{vmatrix} + b^2 \begin{vmatrix}- 2 b^2 & - 2 a^2 \\ b^2 + c^2 - a^2 & 0\end{vmatrix} \left[\text{ Expanding along }C_1 \right]\]

\[ = \frac{1}{abc}\left[ - a^2 \left\{ - 2 b^2 ( c^2 + a^2 - b^2 ) \right\} + b^2 \left\{ 0 + 2 a^2 \left( b^2 + c^2 - a^2 \right) \right\} \right]\]

\[ = \frac{1}{abc}\left[ - a^2 \left\{ - 2 b^2 c^2 - 2 b^2 a^2 + 2 b^4 \right\} + b^2 \left\{ 2 a^2 b^2 + 2 a^2 c^2 - 2 a^4 \right\} \right]\]

\[ = \frac{1}{abc}\left[ 2 a^2 b^2 c^2 + 2 a^4 b^2 - 2 a^2 b^4 + 2 a^2 b^4 + 2 a^2 b^2 c^2 - 2 a^4 b^2 \right]\]

\[ = \frac{1}{abc}4 a^2 b^2 c^2 = 4abc\]

Hence proved.