Question

Prove the following :
$\left|\begin{array}{cccc}0& x& y& z\\ -x& 0& c& b\\ -y& -c& 0& a\\ -z& -b& -a& 0\end{array}\right|={(ax-by+cz)}^2.$

Answer 1

Keeping in view the answer involving $ax-by+cz$ we multiply $C_2,C_3$ and $C_4$ by $a,b,c$ respectively and divide $\Delta$ by $abc$
$\therefore \Delta =\frac{1}{abc}\left|\begin{array}{cccc}0& ax& by& cz\\ -x& 0& bc& bc\\ -y& -ac& 0& ac\\ -z& -ab& -ab& 0\end{array}\right|$
Apply $C_2-C_3+C_4$
$\Delta =\frac{1}{abc}\left|\begin{array}{cccc}0& ax-by+cz& by& cz\\ -x& 0& bc& bz\\ -y& 0& 0& ac\\ -z& 0& -ab& 0\end{array}\right|$
Expand with 2nd column hence $-$ive sign.
$\therefore \Delta =\frac{-(ax-by+cz)}{abc}\left|\begin{array}{ccc}-x& bc& bc\\ -y& 0& ac\\ -z& -ab& 0\end{array}\right|$
Take $-1$, $b$ and $c$ common from $C_1,C_2$ and $C_3$ respectively.
$\therefore \Delta =\frac{ax-by+cz}{a}\left|\begin{array}{ccc}x& c& b\\ y& 0& ac\\ z& -a& 0\end{array}\right|$
$=\frac{ax-by+cz}{a}[x(0+a^2)-y(0+ab)+z(ac+0)]$
$=(ax-by+cz{)}^2$
Note : A skew-symmetric $\Delta$ of even order is perfect square but of odd order is zero.