Prove that $T$ is surjective 
Let $T:S_2 \to \mathbb{R}_3[x]$ such that $$T\left(\begin{bmatrix} a &
 b\\  b & c \end{bmatrix}\right)=(a+b+c) + (-a+2c)x+(2a+3b+6c)x^2.$$
Prove that $T$ is surjective.

My attempt:
Let $w \in \mathbb{R}_3[x]$ such that $w=(a+b+c) + (-a+2c)x+(2a+3b+6c)x^2$.
So $w=a(1-x+2x^2)+b(1+3x^2)+c(1+2x+6x^2)$.
Is that proving that $T$ is surjective? I think that I missed something here.
Thanks!

A function $f:X\rightarrow Y$ is said to be surjective if:
For every $y\in Y$ there exists and $x\in X$ such that $f(x)=y$.
We need to show that for every $a+bx+cx^2 \in \mathbb{R}_3[x]$ for arbitrary $a,b,c \in \mathbb{R}$ that there exists some symmetric matrix $S\in S_2$ such that $T(S)=a+bx+cx^2$.
The short answer is yes because given $\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}\in S_2$ then
$T\left(\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}\right)=a+bx+cx^2\in \mathbb{R}_3[x]$

The calculations below show you how I got the matrix $\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}$:
$T\left(\begin{bmatrix} A& B\\B&C \end{bmatrix}\right) = (A+B+C)+(-A+2C)x+(2A+3B+6C)x^2=a+bx+c^2$
and equating the coefficients, we can translate into matrix notation:
$$\begin{align}A+B+C & =a\\
-A+2C &=b\\
2A+3B+6C &=c\end{align} \Longrightarrow \begin{bmatrix}1&1&1\\-1&0&2\\2&3&6\end{bmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$$
so if $M=\begin{bmatrix}1&1&1\\-1&0&2\\2&3&6\end{bmatrix}$ is invertible, then we could solve for $\begin{pmatrix}A\\B\\C\end{pmatrix}$.
You can find the inverse by creating the matrix $[M|I]$ and bring that matrix into RREF to get $[I|M^{-1}]$ through a series of row operations, you can see this here where the step-by-step process is provided as well. It turns out that $M$ is invertible and its inverse is
$$M^{-1}=\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}$$
(Note that the column vectors of $M$ are exactly the vectors in your linear combination $a(\color{aqua}{1-x+2x^2})+b(\color{aqua}{1+3x^2})+c(\color{aqua}{1+2x+6x^2})$ and the fact that $M$ is invertible means that $[M|I]\cong[I|M^{-1}]$ are row equivalent, thus the RREF of $M$ is $I$ and the column vectors of $M$ are linearly independent, and so, are the vectors in your sum. This would suffice as a proof of surjectivity as well.)
thus $\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}\begin{bmatrix}1&1&1\\-1&0&2\\2&3&6\end{bmatrix}\begin{pmatrix}A\\B\\C\end{pmatrix}=I\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$
and $\begin{pmatrix}A\\B\\C\end{pmatrix}=\begin{bmatrix}-6&-3&2\\10&4&-3\\-3&-1&1\end{bmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$, only left to multiply and equate:
$$\begin{align}A&= -6a-3b+2c\\
B &=10a+4b-3c\\
C &=-3a-b+c\end{align}$$
$\require{cancel} T\left(\begin{bmatrix}-6a-3b+2c & 10a+4b-3c \\ 10a+4b-3c & -3a-b+c \end{bmatrix}\right)=a+bx+cx^2\\ \text{Check:}\\ 
\begin{align}a &=-6a-\cancel{3b}+\cancel{2c}+10a+\cancel{4b}-\cancel{3c}-3a-\cancel{b}+\cancel{c}  \\
b &= -(-6a-3b+2c)+2(-3a-b+c)=\cancel{6a}+3b-\cancel{2c}-\cancel{6a}-2b+\cancel{2c} \\
c &=2(-6a-3b+2c)+3(10a+4b-3c)+6(-3a-b+c)=-\cancel{12a}-\cancel{6b}+4c+\cancel{30a}+\cancel{12b}-9c-\cancel{18a}-\cancel{6b}+6c \end{align} $
And we've found a matrix $S$ s.t. $T(S)=a+bx+cx^2$, for every $a+bx+cx^2 \in \mathbb{R}_3[x]$.
