Question

Find in the form of a determinant the condition that the expression
$u{a}^2+v{\beta }^2+{\gamma }^2+2u^{\prime }\beta \gamma +2v^{\prime }\gamma a+2u^{\prime }a\beta =0$
may be the product of two factors of the first degree in $a,\beta ,\gamma$.

Answer 1

$\alpha (u\alpha +w^{{}^{\prime }}\beta +{\upsilon }^{{}^{\prime }}\gamma )+\beta (w^{{}^{\prime }}\alpha +\upsilon \beta +u^{{}^{\prime }}\gamma )+\gamma ({\upsilon }^{{}^{\prime }}\alpha +u^{{}^{\prime }}\beta +w\gamma )$

For all values of $\alpha ,\beta ,\gamma$

$\frac{u\alpha +w^{{}^{\prime }}\beta +{\upsilon }^{{}^{\prime }}\gamma }{l}=\frac{w^{{}^{\prime }}\alpha +\upsilon \beta +u^{{}^{\prime }}\gamma }{m}=\frac{{\upsilon }^{{}^{\prime }}\alpha +u^{{}^{\prime }}\beta +w^{{}^{\prime }}\gamma }{n}$

(suppose)

Where $l,m,n$ are constants; then the given expression will be the product of two linear factors proportional to $(u\alpha +w^{{}^{\prime }}\beta +{\upsilon }^{{}^{\prime }}\gamma )(l\alpha +m\beta +n\gamma )$

The necessary condition is that abov equation should hold for all values of $\alpha ,\beta ,\gamma$ and therefore for such values as simultaneously satisfy;-

$u\alpha +w^{{}^{\prime }}\beta +{\upsilon }^{{}^{\prime }}\gamma =0$

$w^{{}^{\prime }}\alpha +\upsilon \beta +u^{{}^{\prime }}\gamma =0$

${\upsilon }^{{}^{\prime }}\alpha +u^{{}^{\prime }}\beta +w\gamma =0$

Eliminating $\alpha ,\beta ,\gamma$ we obtain

$\left|\begin{array}{ccc}u& w^{{}^{\prime }}& {\upsilon }^{{}^{\prime }}\\ w^{{}^{\prime }}& \upsilon & u^{{}^{\prime }}\\ {\upsilon }^{{}^{\prime }}& u^{{}^{\prime }}& w\end{array}\right|=0$