This is how I finally did:
if $a=d$ and $b=-c$, the set $SM_2(\mathbb{R})=\bigg\{\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \bigg| a, b \in \mathbb{R}\bigg\}$ turns into a field because:
$SM_2(\mathbb{R})$ is commutative, because:
\begin{equation*}
\begin{aligned}
\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} &= \begin{pmatrix} ac-bd & -ad-bc \\ bc+ad & -bd+ac\\ \end{pmatrix}
\\ &= \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}
\end{aligned}
\end{equation*}
The multiplicative inverse exists and is the inverse of matrix $A$, because $\det(A)=a^2-(-b)b=a^2+b^2\neq 0$, so $A$ is invertible and $A^{-1}=\frac{1}{a^2+b^2}\begin{pmatrix} a & b \\ -b & a\\ \end{pmatrix}$ and
\begin{equation*}
\begin{aligned}
\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \begin{pmatrix} \frac{a}{a^2+b^2} & \frac{b}{a^2+b^2} \\ \frac{-b}{a^2+b^2} & \frac{a}{a^2+b^2}\\ \end{pmatrix}
&= \begin{pmatrix} \frac{a^2+b^2}{a^2+b^2} & \frac{ab-ba}{a^2+b^2} \\ \frac{ba-ab}{a^2+b^2} & \frac{b^2+a^2}{a^2+b^2}\\ \end{pmatrix}
\\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}
\end{aligned}
\end{equation*}
Thus, $SM_2(\mathbb{R})$ is a field. Plus, it cannot be an ordered field. For this, consider $\varphi: SM_2(\mathbb{R}) \longrightarrow \mathbb{C}$ such that $\varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\right)=a+bi$. We have:
For any $A, B \in SM_2(\mathbb{R})$, we have:
\begin{equation*}
\begin{aligned}
\varphi(AB)&= \varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\begin{pmatrix}
c & -d \\ d & c\\
\end{pmatrix}\right)= \varphi\left( \begin{pmatrix} ac-bd & -ad-bc \\ bc+ad & -bd+ac\\ \end{pmatrix} \right) \\
&= (ac-bd)+(ad+bc)i= ac+adi+bci+bdi^2\\
&= a(c+di)+bi(c+di)= (a+bi)(c+di)\\
&= \varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\right) \varphi\left(\begin{pmatrix}
c & -d \\ d & c\\
\end{pmatrix}\right)=\varphi(A)\varphi(B)
\end{aligned}
\end{equation*}
So, $\varphi$ is a homomorphism.
For any $A, B \in SM_2(\mathbb{R})$, assume that $\varphi(A)=\varphi(B)$, then $a+bi=c+di$, so $(a-c)+(b-d)i=0$, and hence, $a=c$ and $b=d$ which means that $A=\begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}= \begin{pmatrix}
c & -d \\ d & c\\
\end{pmatrix} =B$. So, $\varphi$ is injective.
Now, let $a+bi\in \mathbb{C}$. Then, since we defined matrices in $SM_2(\mathbb{R})$ to be all the $2\times 2$ matrices with $a=d$, and $b=-c$, there exists $A=\begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix} \in SM_2(\mathbb{R})$ such that $\varphi\left( \begin{pmatrix}
a & -b \\ b & a\\
\end{pmatrix}\right)=a+bi$.
So $\varphi$ is an isomorphism. Hence, $SM_2(\mathbb{R})$ is isomorphic to $\mathbb{C}$ and since no order can be defined on $\mathbb{C}$, the aforementioned field cannot take any orders as well.
