The value of the determinant $\begin{vmatrix}1&a&b+c\\1&b&c+a\\1&c&a+b\end{vmatrix}$ is
- A. 0
- B. 1
- C. $(a+b+c)$
- D. 3
Answer: A. 0
Explanation: Applying the column operation $C_3 \rightarrow C_2+C_3$, the determinant becomes $\Delta = \begin{vmatrix}1&a&a+b+c\\1&b&a+b+c\\1&c&a+b+c\end{vmatrix} = (a+b+c)\begin{vmatrix}1&a&1\\1&b&1\\1&c&1\end{vmatrix}$. Since $C_1=C_3$, $\Delta=0$.