Question

Using properties of determinants, prove that :
$\left|\begin{array}{ccc}a& b-c& c+b\\ a+c& b& c-a\\ a-b& b+a& c\end{array}\right|=(a+b+c)(a^2+b^2+c^2)$

Answer 1

$\begin{array}{c}Wehavetoprovethat\\ \left|\begin{array}{ccc}a& b-c& c+b\\ a+c& b& c-a\\ a-b& b+a& c\end{array}\right|=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\\ Now,\\ \Delta =\frac{1}{a}\left|\begin{array}{ccc}{a}^2& b-c& c+b\\ a^2+ac& b& c-a\\ a^2-ab& b+a& c\end{array}\right|\left[multiplyingfirstcolumnbya\right]\\ \Delta =\frac{\left(a^2+b^2+c^2\right)}{a}\left|\begin{array}{ccc}1& b-c& c+b\\ 1& b& c-a\\ 1& b+a& c\end{array}\right|\left[C_1\to C_1+b{C}_2+c{C}_2\right]\\ R_2\to R_2-R_1\\ R_3\to R_3-R_1\\ \Delta =\frac{\left(a^2+b^2+c^2\right)}{a}\left|\begin{array}{ccc}1& b-c& c+b\\ 0& b& -a-b\\ 0& a+c& -b\end{array}\right|\\ =\frac{\left(a^2+b^2+c^2\right)}{a}\left(-bc+a^2+ac+ba+bc\right)\\ =\left(a^2+b^2+c^2\right)\left(a+b+c\right)\\ proved.\end{array}$