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 is equal to:-

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

Answer 1

Answer: (D)

$-2(a+b+c)$

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

$R_1\to R_1+R_2+R_3$

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

Take $a+b+c$ common from first row

$C_2\to C_2-C_1,$ $C_3\to C_3-C_2$

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

$(a+b+c)(a+b+c{)}^2$
$⟹x=-2(a+b+c)$