Answer
See below
Work Step by Step
We can write set $S$ as $S=\{A =\begin{bmatrix}
a & b \\ c &d
\end{bmatrix} \in M_2(R): a+b+c+d=0\}$.
We can notice that $\begin{bmatrix}
0 & 0 \\ 0 & 0
\end{bmatrix} \in S$, hence $S$ is nonempty set.
A1. Let $A=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix},B=\begin{bmatrix}
e & f \\ g & h
\end{bmatrix} \in S$
Obtain: $A+B=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}+\begin{bmatrix}
e & f \\ g & h
\end{bmatrix}=\begin{bmatrix}
a+e & b+f \\ c+g & d+h
\end{bmatrix}=0$
Since $A+B \in S$, set $S$ is closed under addition.
A2. Let $A=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix} \in S$ and $k \in R$
Obtain: $kA=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}=A=\begin{bmatrix}
ka & kb\\ kc & kd
\end{bmatrix}=k.0=0$
Since $kA \in S$, set $S$ is closed under scalar multiplication (2)
From (1),(2) and since $S$ is nonempty subset, $S$ is a subspace of $M_2(R)$
