To find the eigenvalues of the matrix \(A = \begin{pmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}\), we need to solve the characteristic equation, which is given by \(\det(A - \lambda I) = 0\), where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix.
The identity matrix \(I\) of order 3 is:
| \(\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) |
Therefore, the matrix \(A - \lambda I\) is:
| \(\begin{pmatrix} 0-\lambda & 0 & -1 \\ 0 & -1-\lambda & 0 \\ 0 & 0 & -1-\lambda \end{pmatrix}\) |
The determinant of this matrix is:
\(\det(A - \lambda I) = \begin{vmatrix} -\lambda & 0 & -1 \\ 0 & -1-\lambda & 0 \\ 0 & 0 & -1-\lambda \end{vmatrix}\)
Using the properties of determinants, we can calculate it as:
\(\det(A - \lambda I) = (-\lambda) \cdot \left[(-1-\lambda)(-1-\lambda)\right] = (-\lambda)(-1-\lambda)^2\)
Expanding the expression, we have:
\((-\lambda)(1 + 2\lambda + \lambda^2) = -\lambda - 2\lambda^2 - \lambda^3\)
Setting this expression equal to zero for the characteristic equation:
\(-\lambda - 2\lambda^2 - \lambda^3 = 0\)
This can be factored as:
\(\lambda (\lambda^2 + 2\lambda + 1) = 0\)
Factoring further, we get:
\(\lambda (\lambda + 1)^2 = 0\)
The solutions are found by setting each factor to zero:
Thus, the eigenvalues of the matrix are \(0, -1, -1\).
Hence, the correct answer is \(0, -1, -1\).