Question

If $x+y+z=0$, then prove that $\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$

Answer 1

Given $x+y+z=0$
To prove:
$\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$
$\therefore$ LHS $=\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|$
$=xa(za.ya-xb.xc)-yb(yc.ya-xb.zb)+zc(yc.xc-za.zb)=xa(a^{2}yz-x^{2}bc)-yb(y^{2}ac-b^{2}xz)+zc(c^{2}xy-z^{2}ab)=xyz{a}^3-x^{3}abc-y^{3}abc+b^{3}xyz+c^{3}xyz-z^{3}abc=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)=xyz(a^3+b^3+c^3)-abc\left(3xyz\right)$
$[\because x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz]$
$=xyz(a^3+b^3+c^3-3abc)$ .....(i)
Now, RHS $=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|=xyz\left|\begin{array}{ccc}a+b+c& b& c\\ a+b+c& a& b\\ a+b+c& c& a\end{array}\right|$ $[\because C_1\to C_1+C_2+C_3]$
$=xyz(a+b+c)\left|\begin{array}{ccc}1& b& c\\ 1& a& b\\ 1& c& a\end{array}\right|$ [taking (a+b+c) common from $C_1$]
$=xyz(a+b+c)\left|\begin{array}{ccc}0& b-c& c-a\\ 0& a-c& b-a\\ 1& c& a\end{array}\right|$
$[\because R_1\to R_1-R_{3}and{R}_2\to R_2-R_3]$
Expanding along $C_1$
$=xyz(a+b+c)\left[1\right(b-c\left)\right(b-a)-(a-c\left)\right(c-a\left)\right]=xyz(a+b+c)(b^2-ab-bc+ac+a^2+c^2-2ac)=xyz(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=xyz(a^3+b^3+c^3-3abc)$ .....(ii)
From Eqs. (i) and (ii),
LHS=RHS
$\Rightarrow \left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$
Hence proved.