Question

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

Answer 1

We have, $\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|=\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$ $[\because R_1\to R_1+R_2+R_3]$
$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$
[taking (a+b+c) common from the first row]
$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ 0& -(a+b+c)& 2b\\ (a+b+c)& (a+b+c)& (c-a-b)\end{array}\right|$
$[\because C_1\to C_1-C_{3}and{C}_2\to C_2-C_3]$
Expanding along $R_1$,
$=(a+b+c)\left[1\right\{0+(a+b+c{)}^2\}]=(a+b+c\left)\right[(a+b+c{)}^2]=(a+b+c{)}^3$