The correct answer is $\boxed{\left[ {\begin{array}{{20}{c}} 2 \ 1 \end{array}} \right]\left[ {\begin{array}{{20}{c}} 1 \ { – 2} \end{array}} \right]}$.
To find the eigenvector pair of a matrix, we can use the following formula:
$$\mathbf{v} = \left[ {\begin{array}{*{20}{c}} x_1 \ x_2 \end{array}} \right]$$
where $\mathbf{v}$ is the eigenvector, $x_1$ and $x_2$ are the eigenvalues, and $A$ is the matrix.
In this case, we have the following matrix:
$$A = \left[ {\begin{array}{*{20}{c}} 3&4 \ 4&{ – 3} \end{array}} \right]$$
The eigenvalues of $A$ are found by solving the equation $|A – \lambda I| = 0$. In this case, we get the following eigenvalues:
$$\lambda_1 = 2 \text{ and } \lambda_2 = -3$$
Now that we have the eigenvalues, we can find the eigenvectors. To do this, we can use the following formula:
$$\mathbf{v} = \frac{1}{\lambda – A}\mathbf{u}$$
where $\mathbf{u}$ is a vector that is not a multiple of the null vector.
In this case, we can choose $\mathbf{u} = \left[ {\begin{array}{*{20}{c}} 1 \ 0 \end{array}} \right]$. Substituting this into the formula, we get the following eigenvector:
$$\mathbf{v} = \frac{1}{2 – \left[ {\begin{array}{{20}{c}} 3&4 \ 4&{ – 3} \end{array}} \right]}\left[ {\begin{array}{{20}{c}} 1 \ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2 \ 1 \end{array}} \right]$$
Therefore, the eigenvector pair of the matrix $\left[ {\begin{array}{{20}{c}} 3&4 \ 4&{ – 3} \end{array}} \right]$ is $\boxed{\left[ {\begin{array}{{20}{c}} 2 \ 1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \ { – 2} \end{array}} \right]}$.
The other options are incorrect because they do not correspond to the eigenvalues of the matrix.