If a, b, c are distinct positive real numbers and \(\Delta = \begin{vmatrix}a & b & c\\ b & c & a\\ c & a & b\end{vmatrix}\), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\Delta \lt 0\)
Each row has the same sum. Applying \(C_1 \to C_1 + C_2 + C_3\) and taking \(a + b + c\) out:
\( \displaystyle \Delta = (a + b + c)\begin{vmatrix}1 & b & c\\ 1 & c & a\\ 1 & a & b\end{vmatrix} \)
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\), then expanding along \(C_1\):
\(\Delta = (a + b + c)\big[(c - b)(b - c) - (a - c)(a - b)\big]\)
\(\Delta = -(a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)\)
The second bracket can be written as a sum of squares:
\( \displaystyle a^{2} + b^{2} + c^{2} - ab - bc - ca = \frac{1}{2}\big[(a - b)^{2} + (b - c)^{2} + (c - a)^{2}\big] \)
For distinct a, b, c this is positive, and \(a + b + c \gt 0\). So \(\Delta \lt 0\).
Check with a = 1, b = 2, c = 3: \(\Delta = -6 \times 3 = -18\).