Question

If $p+q+r=0$, then prove that

$\left|\begin{array}{ccc}pa& qb& rc\\ qc& ra& pb\\ rb& pc& qa\end{array}\right|=pqr\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$.

Answer 1

Given $p+q+r=0$

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

$\Rightarrow p^3+q^3+r^2-3pqr=0$ (i)

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

Applying $C_1\to C_1+C_2+C_3$

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

$=pqr(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& a& b\\ 1& c& a\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$=pqr(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 0& a-b& b-c\\ 0& c-b& a-c\end{array}\right|$

$=pqr(a+b+c)\left|\begin{array}{cc}a-b& b-c\\ c-b& a-c\end{array}\right|$

$=pqr(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$=pqr(a^3+b^2+c^2-2abc)$

$=pqr(a^3+a^3+c^3-3abc)-abc(p^3+q^3+r^3-3pqr)$ .......{from (i) }

$=pqr(a^3+b^3+c^3)-abc(p^3+q^3+r^3)$

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

$=\left(ap\right)\left\{\right(ra\left)\right(qa)-(pb\left)\right(pc\left)\right\}-\left(qb\right)\left\{\right(pc\left)\right(qa)-(pb\left)\right(rb\left)\right\}+\left(rc\right)\left\{\right(pc\left)\right(qc)-(ra\left)\right(rb\left)\right\}$

$=\left|\begin{array}{ccc}pa& qb& rc\\ qc& ra& pb\\ rb& pc& qa\end{array}\right|=L.H.S.$

Hence proved