If a matrix $M$ has eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$, then any matrix defined by a polynomial $Q = f(M)$ will have eigenvalues $f(\lambda_1), f(\lambda_2), \dots, f(\lambda_n)$.
In this question:
Matrix $M$ eigenvalues: $\lambda_1 = 5$ and $\lambda_2 = -2$.
Polynomial $Q$: $Q = M^3 - 4M^2 - 2M$.
Eigenvalues of $Q$ ($\lambda_Q$): Substitute $\lambda_M$ into the polynomial $f(\lambda) = \lambda^3 - 4\lambda^2 - 2\lambda$.
Step-by-Step Calculation
1. for $\lambda_1 = 5$:
Substitute $5$ into the equation:
$$\lambda_{Q1} = (5)^3 - 4(5)^2 - 2(5)$$
$$\lambda_{Q1} = 125 - 4(25) - 10$$
$$\lambda_{Q1} = 125 - 100 - 10 = \mathbf{15}$$
2. For $\lambda_2 = -2$:
Substitute $-2$ into the equation:
$$\lambda_{Q2} = (-2)^3 - 4(-2)^2 - 2(-2)$$
$$\lambda_{Q2} = -8 - 4(4) + 4$$
$$\lambda_{Q2} = -8 - 16 + 4 = \mathbf{-20}$$
SO option $a,c$ is correct.