Question

If $x+y+z=0,$ then the value of $\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|$ is equal to

• A. $0$
• B. $\left(xyz\right)\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$
• C. $\left(x^2{y}^2{z}^2\right)\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$
• D. $\frac{1}{\left(xyz\right)}\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$

Answer 1

The correct option is B $\left(xyz\right)\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|$

Using 'Bagula' method :$\left|\begin{array}{ccc}xa& yb& zc\\ yc& za& xb\\ zb& xc& ya\end{array}\right|=[xyz{a}^3+xyz{b}^3+xyz{c}^3]-[abc{z}^3+abc{x}^3+abc{y}^3]=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)=xyz(a^3+b^3+c^3-3abc)[\because x+y+z=0;x^3+y^3+z^3=3xyz]$
We know that,
$\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|=a^3+b^3+c^3-3abc$
$\Rightarrow \left(xyz\right)\left|\begin{array}{ccc}a& b& c\\ c& a& b\\ b& c& a\end{array}\right|=xyz(a^3+b^3+c^3-3abc)$