Q.If $\frac{a_1}{x}+\frac{b_1}{y}=c_1,\ \frac{a_2}{x}+\frac{b_2}{y}=c_2,\ \Delta_1=\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix},\ \Delta_2=\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix},\ \Delta_3=\begin{vmatrix} c_1 & a_1 \\ c_2 & a_2 \end{vmatrix}$ then the value of $(x, y)$ is :

Using $A(\operatorname{adj}A)=|A|\,I_n$ and taking determinants gives $|\operatorname{adj}A|=|A|^{\,n-1}$, option (a).

Key identity. For any $n\times n$ matrix,

$A\,(\operatorname{adj}A)=|A|\,I_n.$

Take determinants of both sides.

$|A|\cdot|\operatorname{adj}A|=\big|\,|A|\,I_n\big|=|A|^{\,n}.$

…

Substitute $u=1/x,\ v=1/y$; Cramer's rule gives $(x,y)=\left(-\frac{\Delta_1}{\Delta_2},-\frac{\Delta_1}{\Delta_3}\right)$.

Let $u=\dfrac1x,\ v=\dfrac1y$. The system becomes linear in $u,v$:

$a_1u+b_1v=c_1,\qquad a_2u+b_2v=c_2.$

With $\Delta_1=\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}$, Cramer's rule gives

$u=\frac{\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}}{\Delta_1}=\frac{c_1b_2-c_2b_1}{\Delta_1}=\frac{-\Delta_2}{\Delta_1},\qquad v=\frac{\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}}{\Delta_1}=\frac{a_1c_2-a_2c_1}{\Delta_1}=\frac{-\Delta_3}{\Delta_1},$

…