If $a, b, c$ are distinct positive real numbers, then the value of the determinant…
If $a, b, c$ are distinct positive real numbers, then the value of the determinant $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is
$ < 0$
$>0$
$0$
$\geq 0$
Solution
$\begin{aligned}
& \text { Let } \mathrm{A}=\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right| \quad\left[\mathrm{C}_1 \rightarrow \mathrm{C}_1 \rightarrow \mathrm{C}_2 \rightarrow \mathrm{C}_3\right] \\
& =\left|\begin{array}{lll}
a+b+c & b & c \\
a+b+c & c & a \\
a+b+c & a & b
\end{array}\right| \\
& =(a+b+c)\left|\begin{array}{lll}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right| \\
& {\left[\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1 \text { and } \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1\right]} \\
& =(a+b+c)\left|\begin{array}{ccc}
1 & b & c \\
0 & c-b & a-c \\
0 & a-b & b-c
\end{array}\right| \\
& =(a+b+c)[(c-b)(b-c)-(a-c)(a-b)] \\
& =(a+b+c)\left[b c-b^2-c^2+b c-a^2+a b+a c-a b c\right] \\
& =-(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =-\frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \\
&
\end{aligned}$
Since it is given that $a, b, c$ are distinct positive numbers.
So, the value of determinant $A$ is less than 0 .