The correct answer is $\boxed{\left\{ {\begin{array}{*{20}{c}} 2 \ { – 1} \end{array}} \right\}}$.
To find the eigenvalues and eigenvectors of a matrix, we can use the following formula:
$$\lambda v = A v$$
where $\lambda$ is the eigenvalue, $v$ is the eigenvector, and $A$ is the matrix.
In this case, we have:
$$\begin{align}
\lambda v &= A v \
\lambda \left( \begin{array}{c} 2 \ -1 \end{array} \right) &= \left( \begin{array}{cc} 2 & 2 \ 1 & 3 \end{array} \right) \left( \begin{array}{c} 2 \ -1 \end{array} \right) \
\lambda \left( \begin{array}{c} 2 \ -1 \end{array} \right) &= \left( \begin{array}{c} 4 \ 3 \end{array} \right) \
\lambda &= 4 \
\end{align}$$
Therefore, one of the eigenvalues of the matrix is $\lambda = 4$.
To find the corresponding eigenvector, we can solve the following equation:
$$\begin{align}
Av &= \lambda v \
\left( \begin{array}{cc} 2 & 2 \ 1 & 3 \end{array} \right) \left( \begin{array}{c} x \ y \end{array} \right) &= \left( \begin{array}{c} 4 \ 3 \end{array} \right) \
2x + 2y &= 4 \
x + 3y &= 3 \
\end{align}$$
Solving this system of equations, we get:
$$x = 2, \quad y = -1$$
Therefore, one of the eigenvectors of the matrix is $\boxed{\left\{ {\begin{array}{*{20}{c}} 2 \ { – 1} \end{array}} \right\}}$.
The other eigenvalue and eigenvector of the matrix can be found similarly.