Answer
See below
Work Step by Step
Let $A,B \in S$
then $A=\begin{bmatrix}
2 & 2\\
6 & 3
\end{bmatrix}\\
B=\begin{bmatrix}
4 & 8\\
1 & 2
\end{bmatrix}$
$\Rightarrow A+B=\begin{bmatrix}
2 & 2\\
6 & 3
\end{bmatrix}+\begin{bmatrix}
4 & 8\\
1 & 2
\end{bmatrix}=\begin{bmatrix}
6 & 9\\
6 & 6
\end{bmatrix}$
$\det (A+B)=A=\begin{vmatrix}
6 & 9\\
6 & 6
\end{vmatrix}=6.6-6.9=36-54=-18\ne 0$
Hence $A+B\notin S$.
$\Rightarrow A+B$ is not closed under addition.
Set of Rational numbers is closed under usual addition.
Because our scalars can come from set of real numbers
In particular, let $\lambda$ be a scalar.
$\Rightarrow \lambda A=\lambda \begin{bmatrix}
2 & 1\\
6 & 3
\end{bmatrix}=\begin{bmatrix}
2\lambda & \lambda\\
6\lambda & 3\lambda
\end{bmatrix}$
$\det (\lambda A)=\begin{vmatrix}
2\lambda & \lambda\\
6\lambda & 6\lambda
\end{vmatrix}=2\lambda.3\lambda-\lambda.6\lambda=6\lambda^2-6\lambda^2=0$
Hence $\lambda A \in S$
Therefore $S$ is closed scalar multiplication.
