Question

If the equations
$ax+hy+g=0,hx+by+f=0,gx+fy+c=\lambda$ are consistent, show that $\lambda =\frac{abc+2fgh-a{f}^2-b{g}^2-c{h}^2}{ab-h^2}$

Answer 1

For consistency, we must have $D=0$ i.e
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c-\lambda \end{array}\right|=0$
Expand
$a(bc-f^2)-h(ch-gf)+g(hf-bg)-\lambda (ab-h^2)=0$
$\therefore \lambda =\frac{abc+2fgh-a{f}^2-b{g}^2-c{h}^2}{ab-h^2}$
Another form:
The last equation is
$x(ax+hy+g)+y(hx+by+f)+(gx+by+c-\lambda )=0orgx+fy+c=\lambda$
Hence the three equations are same as of part (a). Hence same value of $\lambda$.