Q92 / 170 · Manipal (MET) 2018 · Mathematics
If $a, b$ and $c$ are negative and different real numbers, then $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is
- A$ \lt 0$
- B0
- Correct answerC$\gt0$
- D$\leq 0$
Solution
Given, $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$
Applying, $C_1 \rightarrow C_1+C_2+C_3$ and taking common from $C_1$ $\begin{aligned} \Delta & =(a+b+c)\left|\begin{array}{lll} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right| \\ & =-(a+b+c)\left[a^2+b^2+c^2-a b-b c-c a\right] \\ & =-\frac{(a+b+c)}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\ & \gt0 \text { as } a, b, c \lt 0 \end{aligned}$
Applying, $C_1 \rightarrow C_1+C_2+C_3$ and taking common from $C_1$ $\begin{aligned} \Delta & =(a+b+c)\left|\begin{array}{lll} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array}\right| \\ & =-(a+b+c)\left[a^2+b^2+c^2-a b-b c-c a\right] \\ & =-\frac{(a+b+c)}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\ & \gt0 \text { as } a, b, c \lt 0 \end{aligned}$
Hence, the correct answer is option: 'C'.