Question

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

Answer 1

Given $\left|\begin{array}{ccc}x& x^2& yz\\ y& y^2& zx\\ z& z^2& xy\end{array}\right|$

$R_1\to R_1-R_2{R}_2\to R_2-R_3$

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

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

$=(x-y)(y-z)\left|\begin{array}{ccc}1& \left(x+y\right)& -z\\ 1& (y+z)& -x\\ z& z^2& xy\end{array}\right|$

$R_1\to R_1-R_2$

$=(x-y)(y-z)\left|\begin{array}{ccc}0& \left(x-z\right)& (x-z)\\ 1& (y+z)& -x\\ z& z^2& xy\end{array}\right|$

$=(x-y)(y-z)(z-x)\left|\begin{array}{ccc}0& 1& 1\\ 1& (y+z)& -x\\ z& z^2& xy\end{array}\right|$

$C_2\to C_2-C_3$

$=(x-y)(y-z)(z-x)\left|\begin{array}{ccc}0& 1& 1\\ 1& y+z+x& -x\\ z& z^2-xy& xy\end{array}\right|$

$Expandingalong{R}_1$

$=(x-y)(y-z)(x-z)[z^2-xy-z(x+y+z\left)\right]$
$=(x-y)(y-z)(x-z)[z^2-xy-zx-yz-z^2]$
$=(x-y)(y-z)(z-x)(xy+yz+zx)$
=R.H.S

Hence proved.