Question

$\left|\begin{array}{ccc}0& xyz& x-z\\ y-x& 0& y-z\\ z-x& z-y& 0\end{array}\right|=$

Answer 1

$\left|\begin{array}{ccc}0& xyz& x-z\\ y-x& 0& y-z\\ z-x& z-y& 0\end{array}\right|=$
Evaluating the given determinant by using the row & column operation of determinant.

Let, $\Delta =\left|\begin{array}{ccc}0& xyz& x-z\\ y-x& 0& y-z\\ z-x& z-y& 0\end{array}\right|$

Applying $C_1\to C_1-C_3$

$\Rightarrow \Delta =\left|\begin{array}{ccc}z-x& xyz& x-z\\ z-x& 0& y-z\\ z-x& z-y& 0\end{array}\right|$

Taking $(z-x)$ common from $C_1$

$\Rightarrow \Delta =(z-x)\left|\begin{array}{ccc}1& xyz& x-z\\ 1& 0& y-z\\ 1& z-y& 0\end{array}\right|$

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

$\Rightarrow \Delta =(z-x)\left|\begin{array}{ccc}0& xyz& x-y\\ 0& y-z& y-z\\ 1& z-y& 0\end{array}\right|$

Taking $(y-z)$ common from $R_2$

$\Rightarrow \Delta =(z-x)(y-z)\left|\begin{array}{ccc}0& xyz& x-y\\ 0& 1& 1\\ 1& z-y& 0\end{array}\right|$

Expanding along $C_1$

$\Rightarrow \Delta =(z-x)(y-z)(xyz-x+y)$

Hence, the value of determinant is

$(z-x)(y-z)(y-x+xyz)$