Question

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

• A. $(a-b-c)(a^2+b^2+c^2)$
• B. ${(a+b+c)}^3$
• C. $(a+b+c)(ab+bc+ca)$
• D. none of the above

Answer 1

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

Applying $R_1\to R_1+R_2+R_3$ and taking common from $R_1$, we get

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

Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$

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

$={(a+b+c)}^3$