Question

If $a, b, c$ are distinct positive real numbers, then the value of the determinant $\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}$ is

• A. < 0
• B. > 0
• C. 0
• D. $\geq 0$

Answer 1

Answer: (A)

Let $A=\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix}$
$\left[\right.$ apply $\left.C_{1} \rightarrow C_{1}+C_{2}+C_{3}\right]$
$=\begin{vmatrix}a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b\end{vmatrix}$
$=(a+b+C)\begin{vmatrix}1 & b & c \\ 1 & c & a \\ 1 & a & b\end{vmatrix}$
$\left[\right.$ apply $R_{2} \rightarrow R_{2}-R_{1}$ and $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$
$=(a+b+c)\begin{vmatrix}1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c\end{vmatrix}$
$=(a+b+c)[(c-b)(b-c)-(a-c)(a-b)]$
$=(a+b+c)\left[b c-b^{2}-c^{2}+b c-a^{2}\right.+a b+a c-b c]$
$=-(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$
Since, $a, b, c$ are distinct positive numbers.
$ \therefore $ The value of determinant $A$ is less than $0$ .