Basic definitions
Given a real quadratic polynomial $$$q(x,y)=ax^2+2bxy+cy^2+2dx+2ey+f$$$ with the rectangular coordinates $$(x, y)$$, we will say that the equation $$q (x, y) = 0$$ defines the analytical conic, which we will denote with $$C_q$$.
Notice that one condition for using this definition is that the main part of $$q (x, y)$$, $$q_2(x,y)=ax^2+2bxy+cy^2$$, is not zero at all its points.
A point $$(m, n)$$ belongs to the analytical conic $$C_q$$ if and only if $$q (m, n) = 0$$. The point is called real if the two coordinates are real, and it's called complex if one or more of its two coordinates are complex. As it is an equation with real coefficients, if a complex point $$(m, n)$$ belongs to the conic, so does its conjugate.
If $$(x', y')$$ is another rectangular system of coordinates and $$$q'(x',y')=a'x'^2+2b'x'y'+c'y'^2+2d'x'+2e'y'+f'$$$ is a quadratic polynomial in $$(x', y')$$, and we say that the equations $$q (x, y) = 0$$ and $$q' (x', y') = 0$$ define the same analytical conic if and only if a real number $$K$$ other than zero exists in such a way that: $$q'(x',y')=K\cdot q(x,y)$$ where one considers $$q (x, y)$$ a polynomial in the $$(x', y')$$ coordinates system that is obtained when substituting $$(x, y)$$ for the values given by the formulas of the change of coordinates.
In particular, notice that two polynomials $$q (x, y)$$ and $$q' (x, y)$$ in the same rectangular coordinates define the same conic if, and only if, a real number $$K$$, other than zero exists so that $$q(x',y')=K \cdot q(x,y)$$.
Matrix and main matrix of the conics
Consider a symmetrical real matrix $$$\displaystyle \overline{A}=\begin{bmatrix} a & b & d \\ b & c & e \\ d & e & f \end{bmatrix}$$$ we can assign one polynomial to it $$$q^{\overline{A}}(x,y)=ax^2+2bxy+cy^2+2dx+2ey+f$$$
The main matrix of $$\bar{A}$$ is the non zero matrix $$$\displaystyle \begin{bmatrix} a & b \\ b & c \end{bmatrix}$$$ and the polynomial $$q^{\bar{A}}(x,y)$$ defines the analytical conic $$C_{q^{\bar{A}}}$$: we say that the analytical conic is determined by the matrix $$\bar{A}$$ regarding the coordinates $$(x,y)$$.
Notice that if $$K$$ is a real number other than zero, $$K\overline{A}$$ and $$KA$$ determine the same analytical conic with reference to the same coordinates system.
Notice also that the main part of the polynomial $$q^{\overline{A}}(x,y)$$ is the polynomial $$q^A(x,y)=ax^2+2bx+cy^2$$ corresponding to the main matrix $$A$$.
Reciprocally, given an analytical conic of the form: $$$q(x,y)=ax^2+2bxy+cy^2+2dx+2ey+f$$$ we can assign a real symmetrical matrix, where $$a, b, c, d, e, f$$ are the coefficients of the polynomial $$q (x, y)$$.
As the matrix $$\overline{A}$$ is defined, except from a scalar factor different from zero, we will write $$[\overline{A}]$$ to denote it and will say that it is the matrix of the conic regarding the coordinates $$(x, y)$$.
The main part of the matrix referring to the conic is $$A$$, where $$A$$ is given by the coefficients of the main part of $$q (x, y)$$. We can write: $$$\overline{A}= \begin{bmatrix} A & \omega^T \\ \omega & c \end{bmatrix}, \omega=(g,h)$$$
Then, the following results are true:

Considering a real symmetrical matrix , the conic that is determined is given by the equation $$ (x,y,1) \cdot \overline{A} \cdot (x,y,1)^T=0 $$ (with reference to the coordinates $$(x, y)$$).

Let $$\overline{A}$$ and $$\overline{A}'$$' be two symmetrical real matrixes of dimension 3. Then, the analytical conic defined by $$\overline{A}$$  with reference to the rectangular coordinates $$(x, y)$$  coincides with the one defined by $$\overline{A}'\overline{A}$$ with reference to the coordinates $$(x', y')$$ , if and only if a matrix exists $$\overline{M}=\begin{bmatrix} M & p \\ 0 & 1\end{bmatrix}$$, $$M \in O(2)$$, $$p=(r,s)^T$$ and $$r,s$$ reals and a number $$K$$ other than zero, such that: $$\overline{A}'=K\cdot \overline{M}^T \overline{A}\overline{M}$$.
Notice also that in this case we have an equality involving the main matrixes $$A$$ and $$A'$$,$$A' =K \cdot M^TAM$$
Consider the matrix $$\begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$$, and find the conic associated with $$A$$.
To compute the equation of the conic associated with matrix $$A$$, we must solve the following product: $$$\begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix}2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2\end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}=\begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix}2x+1 \\ y \\ x2\end{bmatrix} =$$$ $$$=(2x+1)x+y^2+x2=2x^2+y^2+2x2$$$