Mathematics - Determinant Question with Solution | TestHub

Let $\Delta =\left|\begin{array}{ccc}-2a& a+b& a+c\\ b+a& -2b& b+c\\ c+a& b+c& -2c\end{array}\right|$ Applying $C_1+C_3$ and $C_2+C_3$ $$\Delta =\left|\begin{array}{ccc}-a+c& 2a+b+c& a+c\\ 2b+a+c& -b+c& b+c\\ a-c& b-c& -2c\end{array}\right|$$ Now, applying $R_1+R_3$ and $R_2+R_3$ $$\Delta =\left|\begin{array}{ccc}0& 2(a+b& )a-c\\ 2(a+b& 0& b-c\\ a-c& b-c& -2c\end{array}\right|$$ On expanding, we get $$\begin{array}{rl}\Delta =& -2(a+b)\{-2c[2(a+b)]\\ & -(a-c)(b-c)\}\\ & +(a-c)[2(a+b)(b-c)]\\ \Delta =& 8\mathrm{c}(a+b)(a+b)\\ & +4(a+b)(a-c)(b-c)\end{array}$$ $$\begin{array}{rl}& =4(a+b)\left[2ac+2bc+ab-bc-ac+c^2\right]\\ & =4(a+b)\left[ac+bc+ab+c^2\right]\\ & =4(a+b)[c(a+c)+b(a+c)]\\ & =4(a+b)(b+c)(c+a)\\ & =\alpha (a+b)(b+c)(c+a)\end{array}$$ Hence, $\alpha =4$