The value of \( \begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix} \) is negative, if
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Values of \( a,b,c \) are distinct positive.
Expanding the determinant along the first row,
\( \displaystyle \begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix}=3abc-a^3-b^3-c^3 \)
\( \displaystyle =-(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \)
The second bracket is a sum of squares:
\( \displaystyle a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \)
For distinct a, b, c it is positive, so the determinant is negative exactly when a + b + c > 0. This is certainly true when a, b, c are distinct positive numbers.
Why the other options are incorrect
Option (2) Values of \( a,b,c \) are negative — Incorrect. Then a + b + c < 0, so the value is positive or 0, not negative.
Option (3) Values of \( a,b,c \) are equal — Incorrect. If a = b = c, every square in the bracket is 0, so the value is 0.
Option (4) Value of any 2 are positive and the third is negative — Incorrect. The sign of a + b + c is not fixed; for example, a = b = 1, c = −5 gives a + b + c < 0 and a positive value.