Answer
See below
Work Step by Step
Let $S=\{A \in M_2(R):tr(A)=0\}$
Assume $A_1=\begin{bmatrix}
1 & 0 \\0 & -1
\end{bmatrix}\\
A_2=\begin{bmatrix}
0 & 1 \\0 & 0
\end{bmatrix}\\
A_3=\begin{bmatrix}
0 & 0 \\1 & 0
\end{bmatrix} \in S$
Let $a,b,c$ be scalars and matrix $A=\begin{bmatrix}
a & b \\c & -a
\end{bmatrix}$
then obtain $a\begin{bmatrix}
1 & 0 \\0 & -1
\end{bmatrix}+b\begin{bmatrix}
0 & 1 \\0 & 0
\end{bmatrix}+c\begin{bmatrix}
0 & 0 \\1 & 0
\end{bmatrix}\\
=\begin{bmatrix}
a & 0 \\0 & -a
\end{bmatrix}+\begin{bmatrix}
0 & b \\0 & 0
\end{bmatrix}+\begin{bmatrix}
0 & 0 \\c & 0
\end{bmatrix}\\
=\begin{bmatrix}
a & b \\c & -a
\end{bmatrix}\\
=A$
Thus, any matrix $A \in S$ can be written as a linear combination of $A_1,A_2, A_3$ then $\{A_1,A_2,A_3\}$ spans $S$.
Obtain $aA_1+bA_2+cA_3=\begin{bmatrix}
0 & 0 \\0 & 0
\end{bmatrix}\\
a\begin{bmatrix}
1 & 0 \\0 & -1
\end{bmatrix}+b\begin{bmatrix}
0 & 1 \\0 & 0
\end{bmatrix}+c\begin{bmatrix}
0 & 0 \\1 & 0
\end{bmatrix}\\=\begin{bmatrix}
a & 0 \\0 & -a
\end{bmatrix}+\begin{bmatrix}
0 & b \\0 & 0
\end{bmatrix}+\begin{bmatrix}
0& 0 \\c & 0
\end{bmatrix}\\
=\begin{bmatrix}
a & b \\c & -a
\end{bmatrix}$
From that we have $a=b=c=0$
Hence, the set $\{A_1,A_2,A_3\}$ is linearly independent in $M_2(R)$
Consequently, $\{A_1,A_2,A_3\}$ is a basic for $M_2(R)$ and $\dim[S]=3$
