Question

Prove that:

$\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|=4abc$

Answer 1

$\left|\begin{array}{ccc}b+c& a& a\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

$R_1\to R_1+R_2+R_3$

$=\left|\begin{array}{ccc}2\left(b+c\right)& 2\left(c+a\right)& 2\left(a+b\right)\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

Taking out $2$ common from $R_1$

$=2\left|\begin{array}{ccc}\left(b+c\right)& \left(c+a\right)& \left(a+b\right)\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

$R_1\to R_1-R_2{R}_3$

$=2\left|\begin{array}{ccc}0& -c& -b\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

$R_2\to R_2+R-1$

$R_3\to R_3+R_1$

$=2\left|\begin{array}{ccc}0& -c& -b\\ b& a& 0\\ c& 0& a\end{array}\right|$

Expanding along $R_1$

$=2\left[c\right(ab)-b(-ac\left)\right]=4abc$