# Unique solution to a system of n equations

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

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

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.