Question

If the system of equations $ax+hy+g=0,hx+by+f=0$ and $gx+fy+c=k$
is consistent and
$k=\frac{1}{\lambda }\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|$, then $\lambda$ is equal to

• A. $h^2-ab$
• B. $ab-h^2$
• C. $h^2+ab$
• D. $g^2-ac$

Answer 1

The correct option is B $ab-h^2$

Given $ax+hy+g=0$,$hx+by+f=0$ and $gx+fy+c=k$ are consistent.
$\Rightarrow \Delta =0$
$\Rightarrow \Delta =\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c-k\end{array}\right|=0$

$\Rightarrow a(bc-kb)-a{f}^2-h(hc-kh)+hgf+g\left(hf\right)-g^{2}b=0$
$\Rightarrow abc+2fgh-a{f}^2-g^{2}b-c{h}^2=abk-h^{2}k$
$k=\frac{abc+2fgh-a{f}^2-g^{2}b-c{h}^2}{ab-h^2}$
$\therefore k=\frac{1}{ab-h^2}\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|$
$\Rightarrow \lambda =ab-h^2$