Question

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

• A. $0$
• B. $a+b+c$
• C. ${(a+b+c)}^2$
• D. ${(a+b+c)}^3$

Answer 1

The correct option is D ${(a+b+c)}^3$

Let $\Delta =\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$
Applying $R_1\to R_1+R_2+R_3$ and taking common $(a+b+c)$ from $R_1$
$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$
Applying $C_2\to C_2\to C_1$ and $C_3\to C_3\to C_1$
$=(a+b+c)\left|\begin{array}{ccc}1& 0& 0\\ 2b& -b-c-a& 0\\ 2c& 0& -a-b-c\end{array}\right|$
$=(a+b+c)\left[(-b-c-a)(-a-b-c)\right]$
$=(a+b+c)$