Question

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

$=(a+b+c)(x+a+b+c{)}^2,x\ne 0$ and $a+b+c\ne 0$, then $x$ is equal to :

• A. $abc$
• B. $2(a+b+c)$
• C. $-(a+b+c)$
• D. $-2(a+b+c)$

Answer 1

The correct option is D $-2(a+b+c)$

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

$R_1\to R_1+R_2+R_3\Delta =\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

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

$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\Delta =(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ b+c+a& -b-c-a& 2b\\ 0& c+a+b& c-a-b\end{array}\right|$

$=(a+b+c)(a+b+c{)}^2$

Hence, $x=-2(a+b+c)$