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If $\left(a_{1} / x\right)+\left(b_{1} / y\right)=c_{1},\left(a_{2} / x\right)+\left(b_{2} / y\right)=c_{2}$
$\Delta_{1}=\left|\begin{array}{ll}a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right|, \Delta_{2}=\left|\begin{array}{ll}b_{1} & c_{1} \\ b_{2} & c_{2}\end{array}\right|, \quad \Delta_{3}=\left|\begin{array}{ll}c_{1} & a_{1} \\ c_{2} & a_{2}\end{array}\right|$,
then $(x, y)$ is equal to which one of the following?
Options:
$\Delta_{1}=\left|\begin{array}{ll}a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right|, \Delta_{2}=\left|\begin{array}{ll}b_{1} & c_{1} \\ b_{2} & c_{2}\end{array}\right|, \quad \Delta_{3}=\left|\begin{array}{ll}c_{1} & a_{1} \\ c_{2} & a_{2}\end{array}\right|$,
then $(x, y)$ is equal to which one of the following?
Solution:
1725 Upvotes
Verified Answer
The correct answer is:
$\left(-\Delta_{1} / \Delta_{2},-\Delta_{1} / \Delta_{3}\right)$
Let $\frac{1}{x}=u, \frac{1}{y}=v$
$\therefore a_{1} u+b_{1} v=c_{1}$ and $a_{2} u+b_{2} v=c_{2}$
Using the method of cross multiplication $\frac{u}{b_{1} c_{2}-b_{2} c_{1}}=\frac{v}{c_{1} a_{2}-c_{2} a_{1}}=\frac{-1}{a_{1} b_{2}-a_{2} b_{1}}$
$\Rightarrow \frac{\frac{1}{x}}{\left|\begin{array}{ll}b_{1} & c_{1} \\ b_{2} & c_{2}\end{array}\right|}=\frac{\frac{1}{y}}{\left|\begin{array}{ll}c_{1} & a_{1} \\ c_{2} & a_{2}\end{array}\right|}$ $=\frac{-1}{\left|\begin{array}{ll}a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right|}$
$\begin{aligned} & \frac{\frac{1}{x}}{\Delta_{2}}=\frac{\frac{1}{y}}{\Delta_{3}}=-\frac{1}{\Delta_{1}} \\ \therefore & \frac{1}{x}=-\frac{\Delta_{2}}{\Delta_{1}} \\ \text { and } \frac{1}{y}=-\frac{\Delta_{3}}{\Delta_{1}} \\ \Rightarrow x &=-\frac{\Delta_{1}}{\Delta_{2}} \text { and } y=-\frac{\Delta_{1}}{\Delta_{3}} \end{aligned}$
$\therefore a_{1} u+b_{1} v=c_{1}$ and $a_{2} u+b_{2} v=c_{2}$
Using the method of cross multiplication $\frac{u}{b_{1} c_{2}-b_{2} c_{1}}=\frac{v}{c_{1} a_{2}-c_{2} a_{1}}=\frac{-1}{a_{1} b_{2}-a_{2} b_{1}}$
$\Rightarrow \frac{\frac{1}{x}}{\left|\begin{array}{ll}b_{1} & c_{1} \\ b_{2} & c_{2}\end{array}\right|}=\frac{\frac{1}{y}}{\left|\begin{array}{ll}c_{1} & a_{1} \\ c_{2} & a_{2}\end{array}\right|}$ $=\frac{-1}{\left|\begin{array}{ll}a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right|}$
$\begin{aligned} & \frac{\frac{1}{x}}{\Delta_{2}}=\frac{\frac{1}{y}}{\Delta_{3}}=-\frac{1}{\Delta_{1}} \\ \therefore & \frac{1}{x}=-\frac{\Delta_{2}}{\Delta_{1}} \\ \text { and } \frac{1}{y}=-\frac{\Delta_{3}}{\Delta_{1}} \\ \Rightarrow x &=-\frac{\Delta_{1}}{\Delta_{2}} \text { and } y=-\frac{\Delta_{1}}{\Delta_{3}} \end{aligned}$
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