Question

The value of determinant $\begin{vmatrix}a-b&b+c&a\\ b-a&c+a&b\\ c-a&a+b&c \end{vmatrix}$ is

• A. 1
• B. 0
• C. 2
• D. 3

Answer 1

Answer: (B)

Operating $R_1\rightarrow R_1-R_2$
$R_2\rightarrow R_2-R_3$
$\begin{vmatrix}0 & a-b & (a-b)(a+b+c)\\0&b-c&(b-c)(a+b+c)\\1 &c&c^2-ab\end{vmatrix}$
$\Rightarrow (a-b)(b-c)\begin{vmatrix}0 &1 &a+b+c\\0 &1&a+b+c\\1&c&c^2-ab\end{vmatrix}=0$
Because $R_1$ and $R_2$ are identical.
Hence (b) is the correct answer.