R2→R2-R1 =xpqp-xx-p0pqx =(x-p) xpq-110pqx =(x-p) xx+pq-100pp+qx ∴ C2→C2+C1 =(x-p) (x-p) (-x-0)+q(-p-q) =(x-p) x2+px-pq-q2 =(x-p) (x-q) (x+q)+p(x-q) =(x-p) (x-q) (x+p+q)

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$|\begin{matrix} x & p & q \\ p & x & q \\ p & q & x \end{matrix}| =$

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(x–p) (x–q)

(x–p) (x–q) (x+p+q)

(x–p) (x+p+q)

(x–q) (x+p+q)

answer is B.

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Detailed Solution

$\begin{array}{rcl} R_{2} & \to & R_{2} - R_{1} \\ = & |\begin{matrix} x & p & q \\ p - x & x - p & 0 \\ p & q & x \end{matrix}| \\ = & (x - p) |\begin{matrix} x & p & q \\ - 1 & 1 & 0 \\ p & q & x \end{matrix}| \\ = & (x - p) |\begin{matrix} x & x + p & q \\ - 1 & 0 & 0 \\ p & p + q & x \end{matrix}| (∴ C_{2} \to C_{2} + C_{1}) \\ = & (x - p) ((x - p) (- x - 0) + q (- p - q)) \\ = & (x - p) (x^{2} + px - pq - q^{2}) \\ = & (x - p) ((x - q) (x + q) + p (x - q)) \\ = & (x - p) (x - q) (x + p + q) \end{array}$