What is the equation associated with the matrix of the following quadric? $$$\displaystyle \overline{A}= \begin{bmatrix} 1 & 0 & 1 & 3 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 3 & 0 & 2 & 5 \end{bmatrix}$$$
See development and solution
Development:
To find the equation of the conical curve, it is sufficient to develop the following product:
$$$\begin{bmatrix} x & y & z & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & 1 & 3 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 3 & 0 & 2 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix}=\begin{bmatrix} x & y & z & 1\end{bmatrix} \begin{bmatrix}x+z+3 \\ 2y \\ x+z+2 \\ 3x+2z+5\end{bmatrix}=$$$ $$$=x^2+xz+3z+2y^2+xz+z^2+2z+3x+2z+5=$$$ $$$=x^2+2y^2+z^2+2xz+6x+4z+5$$$
Solution:
The associate equation is $$x^2+2y^2+z^2+2xz+6x+4z+5$$