Question

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

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

Answer 1

Answer: (B)

$ab-h^2$

Given $ax+hy+g=0$
$hx+by+f=0$
$gx+fy+c=k$
are consitent. then
$\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c-k\end{array}\right|=0$
$a(bc-kb)-a{f}^2-h(hc-kh)+hgf-g\left(hf\right)+g^{2}b=0$
$abc+fgh-a{f}^2-g^{2}b-c{h}^2=abk-h^{2}k$
$k=\frac{abc+fgh-a{f}^2-g^{2}b-c{h}^2}{ab-h^2}$

$k=\frac{1}{ab-h^2}\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|$

$\Rightarrow \lambda =ab-h^2$