Question

If $x+y+z=0$, then $\left|\begin{array}{ccc}ax& by& cz\\ cy& az& bx\\ bz& cx& ay\end{array}\right|=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$.

• A. True
• B. False

Answer 1

The correct option is A True

Given,

$x+y+z=0-\left(1\right)$

Taking LHS

$\left|\begin{array}{ccc}ax& by& cz\\ cy& az& bx\\ bz& cx& ay\end{array}\right|$

$=xa(yz{a}^2-x^{2}bc)-yb(y^{2}ac-b^{2}xz)+zc(c^{2}xy-z^{2}ab)=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)=xyz(a^3+b^3+c^3)-3abc\left(xyz\right)$ .......[since $x+y+z=0,t$ therefore,]

$=xyz(a^3+b^3+c^3-3abc)=xyz(a+b+c)(a^2+b^2+c^2-ab-ac-bc)-\left(1\right)$

Taking RHS

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

Applying $R_1\to R_1+R_2+R_3$

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

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

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$

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

$=xyz(a+b+c)\left[\right(a-c\left)\right(a-b)-(c-b\left)\right(b-c\left)\right]=xyz(a+b+c)(a^2+b^2+c^2-ab-bc-ac)-\left(2\right)$

from eqs.(1) and (2)

LHS=RHS

Hence,

$\left|\begin{array}{ccc}ax& by& cz\\ cy& az& bx\\ bz& cx& ay\end{array}\right|=xyz\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$