The correct answer is $\boxed{\text{B}}$.
To find the eigenvalues and eigenvectors of a matrix, we can use the following formula:
$$\lambda v = A v$$
where $\lambda$ is an eigenvalue, $v$ is an eigenvector, and $A$ is the matrix.
In this case, we have the matrix $A = \left[ {\begin{array}{*{20}{c}} 5&{ – 1} \ 4&1 \end{array}} \right]$.
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:
$$| \left[ {\begin{array}{{20}{c}} 5&{ – 1} \ 4&1 \end{array}} \right] – \lambda \left[ {\begin{array}{{20}{c}} 1&0 \ 0&1 \end{array}} \right] | = \left[ {\begin{array}{*{20}{c}} 5 – \lambda &{ – 1} \ 4&1 – \lambda \end{array}} \right]$$
We can then solve the equation $| A – \lambda I | = 0$ to find the eigenvalues.
In this case, we get the eigenvalues $\lambda = 3$ and $\lambda = -3$.
To find the eigenvectors, we can substitute each eigenvalue into the equation $Av = \lambda v$ and solve for $v$.
In this case, we get the following eigenvectors:
$$v_1 = \left[ {\begin{array}{*{20}{c}} 1 \ 1 \end{array}} \right]$$
and
$$v_2 = \left[ {\begin{array}{*{20}{c}} -1 \ 1 \end{array}} \right]$$
We can see that the eigenvalue $\lambda = 3$ has multiplicity 2, since there are two linearly independent eigenvectors associated with it.
Therefore, the correct answer is $\boxed{\text{B}}$.