Step-by-Step Solution
Key Concept: A nilpotent matrix satisfies $A^k = 0$ for some $k$, and this property greatly simplifies higher powers.
Computing successive powers of $A = \begin{pmatrix} 3 & 1 \\ -9 & -3 \end{pmatrix}$, we find that $A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, so $A$ is nilpotent. This gives $A^{100} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ and leads to $B^2 - 2B + I = 0$.
Correct Answer: 2