Consider a homogeneous system $\textbf{Ax}=\textbf{0}$ whose coefficient matrix is
$$\textbf{A}= \begin{pmatrix} x_{1}^{2}& x_{1}y_{1}& y_{1}^{2}& x_{1}& y_{1}& 1\\ x_{2}^{2}& x_{2}y_{2}& y_{2}^{2}& x_{2}& y_{2}& 1 \\ x_{3}^{2}& x_{3}y_{3}& y_{3}^{2}& x_{3}& y_{3}& 1 \\ x_{4}^{2}& x_{4}y_{4}& y_{4}^{2}& x_{4}& y_{4}& 1\\ x_{5}^{2}& x_{5}y_{5}& y_{5}^{2}& x_{5}& y_{5}& 1 \end{pmatrix}\in Mat(\mathbf{R} ;5\times 6 ) .$$ $$ \textbf{x}= \begin{pmatrix} A\\ B\\ C\\ D\\ E\\ F \end{pmatrix}\in\mathbf{R^6};\qquad\textbf{0}=\begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix}.$$
Show that
$rank(\textbf{A})=5;$ there is an $\mathbf{x} $ such that $\textbf{Ax}=\textbf{0}$ and $\det(\begin{pmatrix} 2A& B& D \\ B& 2C& E \\ D& E& 2F \ \end{pmatrix})\ne 0$.$\Longleftrightarrow $ For any $ k_{1},k_{2},k_{3}\in \{1,2,3,4,5\} $ and $k_{1}<k_{2}<k_{3}$, $ rank\begin{pmatrix} x_{k_{1}}& y_{k_{1}}& 1\\ x_{k_{2}}& y_{k_{2}}& 1 \\ x_{k_{3}}& y_{k_{3}}& 1 \end{pmatrix}=3.$
I want to apply only the knowledge of linear algebra to proof " There is a unique conic through any five points in a real plane, and it is nondegenerate if and only if no three of the points are collinear ."