The correct answer is $\boxed{\left\{ {\begin{array}{*{20}{c}} { – 1} \ 1 \end{array}} \right\}}$.
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 following matrix:
$$A = \left[ {\begin{array}{*{20}{c}} { – 5}&2 \ { – 9}&6 \end{array}} \right]$$
We can solve the equation $\lambda v = A v$ for $v$ to find the eigenvectors.
First, we can try to solve the equation for $v_1$:
$$\lambda v_1 = \left[ {\begin{array}{*{20}{c}} { – 5}&2 \ { – 9}&6 \end{array}} \right] v_1$$
We can reduce this equation to the following form:
$$\left[ {\begin{array}{*{20}{c}} { – \lambda – 5}&2 \ { – 9}&{6 – \lambda} \end{array}} \right] v_1 = 0$$
We can find the eigenvalues by solving the equation $|A – \lambda I| = 0$.
In this case, we have the following equation:
$$\left| \begin{array}{*{20}{c}} { – \lambda – 5}&2 \ { – 9}&{6 – \lambda} \end{array} \right| = 0$$
We can solve this equation to find the eigenvalues $\lambda = -2$ and $\lambda = 9$.
Now that we know the eigenvalues, we can find the eigenvectors.
To find the eigenvector corresponding to the eigenvalue $\lambda = -2$, we can solve the equation $v_1 \left[ {\begin{array}{*{20}{c}} { – 2 – 5}&2 \ { – 9}&{6 – 2} \end{array}} \right] = 0$.
We can reduce this equation to the following form:
$$v_1 \left[ {\begin{array}{*{20}{c}} { – 7}&2 \ { – 9}&{4} \end{array}} \right] = 0$$
We can solve this equation to find the eigenvector $v_1 = \left\{ {\begin{array}{*{20}{c}} { – 1} \ 1 \end{array}} \right\}$.
Therefore, one of the eigenvectors of the matrix $\left[ {\begin{array}{{20}{c}} { – 5}&2 \ { – 9}&6 \end{array}} \right]$ is $\left\{ {\begin{array}{{20}{c}} { – 1} \ 1 \end{array}} \right\}$.