Topic: Engineering Mathematics
The value of the determinant $|\begin{matrix}1&a&b+c\\ 1&b&c+a\\ 1&c&a+b\end{matrix}|$ is
- A. 0
- B. 1
- C. $(a+b+c)$
- D. 3
Answer: A. 0
Explanation: Applying column operation $C_{3} \rightarrow C_{3} + C_{2}$, the third column becomes $(a+b+c)$ for all its elements. We can take out the common factor $(a+b+c)$ from the third column, making the third column identically $(1, 1, 1)^{T}$, which makes it identical to the first column. Since two columns are identical, the determinant's value is 0.