Question

By using properties of determination, show that
$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|={(a+b+c)}^3$

Answer 1

$R_1\Rightarrow R_1+R_2+R_3$

$\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

$c_2\Rightarrow c_2-c_1$; $c_3\Rightarrow c_3-c_1$

$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& -a-b-c& 0\\ 2c& 2c& -a-b-c\end{array}\right|$

$=(a+b+c)(a+b+c)(a+b+c)={(a+b+c)}^3$