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 the matrix $A = \left[ {\begin{array}{*{20}{c}} { – 4}&2 \ 4&3 \end{array}} \right]$.
To find the eigenvalues, we can solve the equation $Av = \lambda v$.
We can do this by first finding the characteristic polynomial of $A$, which is given by:
$$| A – \lambda I |$$
In this case, the characteristic polynomial is:
$$| \left[ {\begin{array}{{20}{c}} { – 4}&2 \ 4&3 \end{array}} \right] – \lambda \left[ {\begin{array}{{20}{c}} 1&0 \ 0&1 \end{array}} \right] | = \left[ {\begin{array}{*{20}{c}} { – 4}-\lambda &2 \ 4&3-\lambda \end{array}} \right]$$
We can then solve the equation $| A – \lambda I | = 0$ to find the eigenvalues.
In this case, the eigenvalues are $\lambda = 2$ and $\lambda = 5$.
Once we have found the eigenvalues, we can find the eigenvectors by solving the equation $Av = \lambda v$ for each eigenvalue.
In this case, we have the following equations:
$$\left[ {\begin{array}{{20}{c}} { – 4}&2 \ 4&3 \end{array}} \right] \left[ {\begin{array}{{20}{c}} x_1 \ x_2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \ 5 \end{array}} \right]$$
and
$$\left[ {\begin{array}{{20}{c}} { – 4}&2 \ 4&3 \end{array}} \right] \left[ {\begin{array}{{20}{c}} y_1 \ y_2 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \ 2 \end{array}} \right]$$
We can solve these equations for $x_1$ and $x_2$, and $y_1$ and $y_2$, respectively.
In this case, we find that the eigenvectors are $\left[ {\begin{array}{{20}{c}} 2 \ { – 1} \end{array}} \right]$ and $\left[ {\begin{array}{{20}{c}} 1 \ 2 \end{array}} \right]$.
Therefore, the correct answer is $\boxed{\left[ {\begin{array}{*{20}{c}} 2 \ { – 1} \end{array}} \right]}$.