Question

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

The given determinant

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

$R_1\to R_1+R_2$
$=\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|=(a+b+c)\times \left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

$C_2\to C_2-C_1$ and $C_3\to C_3-C_1$
$=(a+b+c)\times \left|\begin{array}{ccc}1& 0& 0\\ 2b& b-c-a& 0\\ 2c& 0& c-a-b\end{array}\right|$
$=(a+b+c)\times \left|\begin{array}{cc}-b-c-a& 0\\ 0& -c-a-b\end{array}\right|=(a+b+c{)}^3$.