Question

If $p+q+r=0$ then $\left|\begin{array}{ccc}pa& qb& rc\\ qc& ra& pb\\ rb& pc& qa\end{array}\right|$ =

• A. $\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$
• B. $\left|\begin{array}{ccc}p& q& r\\ q& r& p\\ r& p& q\end{array}\right|$
• C. $pqr\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$
• D. $abc\left|\begin{array}{ccc}p& q& r\\ q& r& p\\ r& p& q\end{array}\right|$

Answer 1

Answer: (C)

$pqr\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$

$\left|\begin{array}{ccc}pa& qb& rc\\ qc& ra& pb\\ rb& pc& qc\end{array}\right|=?$

$p+q+r=0$

$p^3+q^3+r^3=3pqr$

$pa(qr{a}^2-p^{2}bc)-qb(q^{2}ac-pr{b}^2)+rc(pq{c}^2-r^{2}ab)$

$pqr{a}^3-p^{3}abc-q^{3}abc+pqr{b}^3+pqr{c}^3-abc{r}^3$

$pqr(a^3+b^3+c^3)-abc\frac{p^3+q^3+r^3}{3pqr}$

$pqr(a^3+b^3+c^3-3abc)$

$=pqr\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$