The correct answer is $\boxed{\text{(A)}}$.
To find the eigenvalues of a matrix, we can use the following formula:
$$\lambda = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
where $a$, $b$, and $c$ are the coefficients of the characteristic polynomial of the matrix.
The characteristic polynomial of the matrix $A$ is given by:
$$p(x) = |A – xI| = \begin{vmatrix} 1 – x & 2 & 3 \\ 7 & 8 – x & 6 \\ 4 & 5 & 9 – x \end{vmatrix}$$
Expanding the determinant, we get:
$$p(x) = x^3 – 20x^2 + 117x – 180$$
The roots of the characteristic polynomial are the eigenvalues of the matrix. To find the eigenvalues, we can use the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$
In this case, we have:
$$a = 1, \quad b = -20, \quad c = 117$$
Substituting these values into the quadratic formula, we get:
$$x = \frac{20 \pm \sqrt{(-20)^2 – 4 \cdot 1 \cdot 117}}{2 \cdot 1}$$
$$x = \frac{20 \pm \sqrt{-304}}{2}$$
$$x = \frac{20 \pm 2\sqrt{76}}{2}$$
$$x = 10 \pm \sqrt{76}$$
$$x = 10 \pm 2\sqrt{19}$$
Therefore, the eigenvalues of the matrix $A$ are $\boxed{1, -\sqrt{19}, \sqrt{19}}$.
Here is a brief explanation of each option:
- Option A: The eigenvalues of the matrix $A$ are $\boxed{1, -\sqrt{19}, \sqrt{19}}$. This is because the characteristic polynomial of the matrix $A$ is given by $p(x) = x^3 – 20x^2 + 117x – 180$, and the roots of the characteristic polynomial are the eigenvalues of the matrix. The roots of the characteristic polynomial $p(x)$ are $x = 10 \pm \sqrt{19}$, which are equal to $1, -\sqrt{19}, \sqrt{19}$.
- Option B: The eigenvalues of the matrix $A$ are $\boxed{1, 1, 0}$. This is not correct because the characteristic polynomial of the matrix $A$ is given by $p(x) = x^3 – 20x^2 + 117x – 180$, and the roots of the characteristic polynomial are $x = 10 \pm \sqrt{19}$, which are not equal to $1, 1, 0$.
- Option C: The eigenvalues of the matrix $A$ are $\boxed{1, 1, -1}$. This is not correct because the characteristic polynomial of the matrix $A$ is given by $p(x) = x^3 – 20x^2 + 117x – 180$, and the roots of the characteristic polynomial are $x = 10 \pm \sqrt{19}$, which are not equal to $1, 1, -1$.
- Option D: The eigenvalues of the matrix $A$ are $\boxed{1, 0, 0}$. This is not correct because the characteristic polynomial of the matrix $A$ is given by $p(x) = x^3 – 20x^2 + 117x – 180$, and the roots of the characteristic polynomial are $x = 10 \pm \sqrt{19}$, which are not equal to $1, 0, 0$.