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The question is:

If $a,b,c$ are negative distinct real numbers,then the determinant $$ \begin{vmatrix} a & b & c \\ b & c & a\\ c & a & b \end{vmatrix} $$ is $$(a) \le 0 \quad (b) \gt 0 \quad (c) \lt 0 \quad (d) \ge 0 $$

My strategy: I identifed that the matrix is a circulant hence the determinant can be expressed in the form of $-(a^3 + b^3 + c^3 - 3abc)$ which implies that $-(a+b+c)\frac{1}{2}[(a-b)^2 + (b-c)^2 + (c-a)^2]$ hence $(-)(-)(+) \gt 0$ but the answers says it is $\ge 0$, so can we have three $a,b,c$ such that the answer is $0$ ?

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What happens if a=b=c? – Martin Sleziak Apr 11 '11 at 14:21
@Martin:The questions says that $a,b$ and $c$ are distinct negative numbers,so can a=b=c ? – Quixotic Apr 11 '11 at 14:23
Sorry, I should have read the question more carefully. – Martin Sleziak Apr 11 '11 at 14:25
@Martin:It's alright,happens to me all the time :) – Quixotic Apr 11 '11 at 14:30
up vote 4 down vote accepted

If you require $a, b, c$ distinct, no. The term $a+b+c$ can't be $0$ as they are all the same sign and the other term is a sum of squares which can only be $0$ if $a=b=c$.

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Exactly, so the correct answer should be (b) right? – Quixotic Apr 11 '11 at 14:25
I agree with you. Maybe the author forgot about "distinct". – Ross Millikan Apr 11 '11 at 14:28
+1,thanks :-) – Quixotic Apr 11 '11 at 14:29

Consider the equation:

$$(a^3 + b^3 + c^3)/3 = abc$$

as expressing the equality of the arithmetic and geometric means of $a^3,b^3,c^3$. By a well known result this is only possible if the three cubes are equal. So, no, we cannot get them distinct and have the determinant be zero.

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I knew this result but it never really occur to me till now to use it here,thanks and +1 :-) – Quixotic Apr 11 '11 at 14:33
@Debanjan: Yes, just going a different route! Your solution (embedded in the question) already shows that something less than all three distinct suffices to give a nonzero result, i.e. all three must be equal to reach zero. – hardmath Apr 11 '11 at 14:53

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