Answer: C
If the $ \lambda $ is the eigen value of the matrix $A$ then $\lambda^{k}$ is the eigen value of the matrix $A^{k}$, why?
We know that:
$Ax = \lambda x$ -----------(1), on left multiplying the matrix $A$ on both sides we get:
$AAx = A \lambda x$, since $\lambda$ is a scalar value it can commute
$AAx = \lambda Ax = \lambda \lambda x$, using equation $(1)$
$A^{2}x = \lambda^{2}x$
Since eigen values of $A$ are $\lambda_{1}, \lambda_{2}, \lambda_{3} = -3, -3, 5$.
The eigen values of $A^{2}$ are $\lambda_{a}, \lambda_{b}, \lambda_{c} = 9, 9, 25$, using the similar calculation as above we can conclude that,
eigen values of $3A^{2}$ are $\lambda_{m}, \lambda_{n}, \lambda_{o} = (3 \times 9), (3 \times 9), (3 \times 25) = 27, 27, 75.$
that of $A^{3}$ are $\lambda_{x}, \lambda_{y}, \lambda_{z} = -27, -27, 125.$
and finally that of $A^{3} – 3A^{2} = (\lambda_{x}-\lambda_{m}), (\lambda_{y}-\lambda_{n}), (\lambda_{z}-\lambda_{o}) = (-27-27), (-27-27), (125-75) = -54, -54, 50.$
The trace of a matrix $P = tr(P)$ is defined as the sum of elements of the principle diagonal. So trace of the given matrix is $-2+1+0 = -1$.
This is same as the sum of all the eigen values of the matrix, why? see: https://people.math.harvard.edu/~knill/teaching/math19b_2011/handouts/lecture28.pdf and https://math.stackexchange.com/a/546167
So the $tr(A^{3}-3A^{2}) = -54 + (-54) + 50 = -58$
