Question

Using the properties of determinants, show that:

$\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|=(x-y)(y-z)(z-x)(xy+yz+zx)$

Answer 1

Let $\Delta =\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|$
Applying $R_1=R_1-R_2,\text{\&}{R}_2=R_2-R_3$
$\Delta =\left|\begin{array}{ccc}x-y& x^2-y^2& z(y-x)\\ y-z& y^2-z^2& x(z-y)\\ z& z^2& xy\end{array}\right|$
Taking $(x-y)\text{\&}(y-z)$ common from first and second row respectively we get,
$\Delta =(x-y)(y-z)\left|\begin{array}{ccc}1& x+y& -z\\ 1& y+z& -x\\ z& z^2& xy\end{array}\right|$
Now applying $R_1=R_1-R_2$ we get,
$\Delta =(x-y)(y-z)\left|\begin{array}{ccc}0& x-z& x-z\\ 1& y+z& -x\\ z& z^2& xy\end{array}\right|$
Taking $(z-x)$ common from first row, we get,
$\Delta =(x-y)(y-z)(z-x)\left|\begin{array}{ccc}0& -1& -1\\ 1& y+z& -x\\ z& z^2& xy\end{array}\right|$
Expanding along $R_1$ we get,
$\Delta =(x-y)(y-z)(z-x)[xy+xz-1(z^2-yz-z^2\left)\right]$
$\therefore \Delta =(x-y)(y-z)(z-x)(xy+yz+zx)$
Hence proved.