Question

Prove that $\left|\begin{array}{ccc}\frac{a^2+b^2}{c}& c& c\\ a& \frac{b^2+c^2}{a}& a\\ b& b& \frac{c^2+a^2}{b}\end{array}\right|=4abc$

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}\frac{a^2+b^2}{c}& c& c\\ a& \frac{b^2+c^2}{a}& a\\ b& b& \frac{c^2+a^2}{b}\end{array}\right| \\ =\frac{1}{abc}\left|\begin{array}{ccc}{a}^2+b^2& c^2& c^2\\ a^2& b^2+c^2& a^2\\ b^2& b^2& c^2+a^2\end{array}\right|\left[\mathrm{Multiplying}{R}_1,R_2\mathrm{and}{R}_3\mathrm{by}c,a\mathrm{and}b\mathrm{and}\mathrm{then}\mathrm{dividing}\mathrm{by}abc\right] \\ =\frac{1}{abc}\left|\begin{array}{ccc}{a}^2+b^2& c^2-a^2-b^2& c^2-a^2-b^2\\ a^2& b^2+c^2-a^2& 0\\ b^2& 0& c^2+a^2-b^2\end{array}\right|\left[\mathrm{Applying}{C}_2\to C_2-C_1\mathrm{and}{C}_3\to C_3-C_1\right] \\ =\frac{1}{abc}\left|\begin{array}{ccc}0& -2b^2& -2a^2\\ a^2& b^2+c^2-a^2& 0\\ b^2& 0& c^2+a^2-b^2\end{array}\right|\left[\mathrm{Applying}{R}_1\to R_1-R_2-R_3\right] \\ =\frac{1}{abc}[-a^2\left|\begin{array}{cc}-2b^2& -2a^2\\ 0& c^2+a^2-b^2\end{array}\right|+b^2\left|\begin{array}{cc}-2b^2& -2a^2\\ b^2+c^2-a^2& 0\end{array}\right|\left[\mathrm{Expanding}\mathrm{along}{C}_1\right] \\ =\frac{1}{abc}\left[-a^2\left\{-2b^2(c^2+a^2-b^2)\right\}+b^2\left\{0+2a^2\left(b^2+c^2-a^2\right)\right\}\right] \\ =\frac{1}{abc}\left[-a^2\left\{-2b^2{c}^2-2b^2{a}^2+2b^4\right\}+b^2\left\{2a^2{b}^2+2a^2{c}^2-2a^4\right\}\right] \\ =\frac{1}{abc}\left[2a^2{b}^2{c}^2+2a^4{b}^2-2a^2{b}^4+2a^2{b}^4+2a^2{b}^2{c}^2-2a^4{b}^2\right] \\ =\frac{1}{abc}4a^2{b}^2{c}^2=4abc \end{gathered}$$

Hence proved.