If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of the given matrix in $a, b, c$? 
If $a,b,c$ are the roots of $x^3-px+q=0$, then what is the determinant of 
  $$
    \begin{pmatrix}
    a & b & c \\
    b & c & a \\
    c & a & b \\
    \end{pmatrix} \,\,?
$$
  (A) $p^2+6q \quad$
  (B) $1 \quad$
  (C) $p \quad$
  (D) $0 \quad$

In this equation given we have product of eigenvalues given as $-q$ and we know product of eigenvalues is determinant then why isn't the determinant is $-q$?  
 A: Hint
Using Sarrus, one gets 
$$\det(A)=3abc-a^3-b^3-c^3.$$
Using that $a,b,c$ solve $x^3-px+q=0$ you can conclude.
A: $$\det\begin{pmatrix}a&b&c\\ b&c&a \\ c&a&b\end{pmatrix}=(a+b+c)\det\begin{pmatrix}1&1&1\\ b&c&a \\ c&a&b\end{pmatrix}=0$$
since $a+b+c=0$.
A: Determinant = $3abc-(a^3+b^3+c^3)$
By Vieta's theorem,
$$a+b+c=0\\ab+bc+ca=-p\\abc=-q\text.$$
A: 
In this equation given we have product of eigenvalues given as $−q$ and we know product of eigenvalues is determinant then why isn't the determinant is $−q$?

The statement you're thinking of is that the product of eigenvalues of a matrix $A$ is equal (up to sign, depending on parity) to the constant term of its characteristic polynomial $p_A$.
In this case, however, the given matrix, though constructed from the roots $a, b, c$ of the characteristic polynomial, doesn't have the property that its eigenvalues are $a, b, c$, and therefore the above general fact does not apply.
On the other hand, using, e.g., cofactor expansion gives that the determinant of the given matrix is
$$\det \pmatrix{a&b&c\\b&c&a\\c&a&b} = 3 a b c - (a^3 + b^3 + c^3).$$
On the other hand, writing the cubic polynomial as $$x^3 - p x + q = (x - a)(x - b)(x - c),$$ expanding, and comparing like terms recovers a special case of Vieta's Formulas:
$$\begin{array}{rcl}
 0 &=& - (a + b + c) \\
-p &=& bc + ca + ab \\
 q &=& -a b c 
\end{array} .$$
Manipulating these lets us write both terms of the expression $3 a b c - (a^3 + b^3 + c^3)$ for the determinant as polynomials in $p, q$. These manipulations give special cases of Newton's Identities.
