Question

The value of the determinant
$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$ will be

• A. $(a-b-c)(a^2+b^2+c^c)$
• B. $(a+b+c{)}^3$
• C. $(a+b+c)(ab+bc+ca)$
• D. $(a+b+c{)}^2$

Answer 1

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

Given that,
$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$
Applying $R_1\to R_1+R_2+R_3$
$=\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$
Taking $(a+b+c)$ common from the first row,
$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$
$[$ Applying $C_1\to C_1-C_3$ and $C_2\to C_2-C_3]$
$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ 0& -(a+b+c)& 2b\\ a+b+c& a+b+c& c-a-b\end{array}\right|$
Expanding along $R_1$
$(a+b+c)[1\times 0+(a+b+c{)}^2]=(a+b+c{)}^3$