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$$ \begin{cases} \sigma_1\left(\alpha x_1+\beta y_1\right)+\sigma_2\left(\gamma x_1+\delta y_1\right)=0&\\ \sigma_1\left(\alpha x_2+\beta y_2\right)+\sigma_2\left(\gamma x_2+\delta y_2\right)=0&\\ \vdots&\\ \sigma_1\left(\alpha x_n+\beta y_n\right)+\sigma_2\left(\gamma x_n+\delta y_n\right)=0& \end{cases} , \ n \in \mathbb{Z}_+$$ How does one have to choose coefficients$$ \alpha, \ \beta, \gamma \ \mathrm{and} \ \delta$$ for this system of equations to have a unique solution $$ \begin{cases} \sigma_1=0&\\ \sigma_2=0& \end{cases}$$?

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    $\begingroup$ Some condition on the integer $n$? Can it take any value $n\ge 1$? $\endgroup$
    – Piquito
    Nov 23, 2020 at 0:35
  • $\begingroup$ @Piquito Yes, any integer ≥ 1. $\endgroup$
    – mathslover
    Nov 23, 2020 at 7:35

1 Answer 1

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COMMENT.-Put $A_i=ax_i+by_i$ and $B_i=cx_i+dy_i$ with $i=1,2,\cdots,n$. Since $\sigma_1=-\dfrac{B_i}{A_i}\sigma_2$ and $\sigma_2=-\dfrac{A_i}{B_i}\sigma_1$ because of $\sigma_1=\sigma_2=0$ as only solution we need $A_i\ne0$ and $B_i\ne0$ for all $i$.

It follows for all $i$ the system $$\begin{cases}ax_i+by_i=h_i\ne0 \\ cx_i+dy_i=k_i\ne0\end{cases}$$ then

$$x_i=\frac{\left|\begin{matrix}h_i&b\\k_i&d\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|}\hspace3cm y_i=\frac{\left|\begin{matrix}a&h_i\\c&k_i\end{matrix}\right|}{\left|\begin{matrix}a&b\\c&d\end{matrix}\right|}$$ Since the point $P_i=(x_i,y_i)$ is unique we should have $\left|\begin{matrix}a&b\\c&d\end{matrix}\right|\ne0$ which is $$\color{red}{ad-bc\ne0}$$ as a first condition.

Another condition could be that no point $P_i$ is on the coordinate axes from which more restrictions could be deduced for the coefficients $a, b, c, d$. I leave this as an exercise for the $O.P.$ and for those who want to solve it.

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