Question

If $(f^2-bc)x+(ch-fg)y+(bg+hf)z=0$,
$(ch-fg)x+(g^2-ca)y+(af-gh)z=0$,
$(bg-hf)x+(af-gh)y+(h^2-ab)z=0$,
show that $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$.

Answer 1

Eliminating $x,y,z$ we get

$\left|\begin{array}{ccc}{f}^2-bc& ch-fg& bg-hf\\ ch-fg& g^2-ca& af-gh\\ bg-hf& af-gh& h^2-ab\end{array}\right|$

The determinants formed by the minors of the elements of the determinant is equal square to the original determinant

$\therefore \left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=\left|\begin{array}{ccc}bc-f^2& fg-ch& hf-bg\\ fg-ch& ca-g^2& gh-af\\ hf-bg& gh-af& ab-h^2\end{array}\right|$

$\therefore$ Second determinant

$={(abc+2fgh-a{f}^2-b{g}^2-c{h}^2)}^2$

For the three equations to hold, the second determinant vanishes

$\therefore abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$