Question

$\left|\begin{array}{ccc}x& x^2& 1+x^2\\ y& y^2& 1+y^2\\ z& z^2& 1+z^3\end{array}\right|$
Show that $1+xyz=0$

Answer 1

$\left|\begin{array}{ccc}x& x^2& 1+x^2\\ y& y^2& 1+y^2\\ z& z^2& 1+z^2\end{array}\right|$

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

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

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

$R_2\to R_2-R_1$

$R_3\to R_3-R_1$

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

solving above matrix, we get

$=(y-z)(z-x)(z-y)(1+xyz)$

$\Rightarrow (y-z)(z-x)(z-y)(1+xyz)=0$

$\Rightarrow 1+xyz=0$