Correct answer: (C) If M is invertible, M⁻¹ is a polynomial in M of degree 4.
Explanation
By Cayley–Hamilton, $M^5+a_4M^4+\cdots+a_1M+a_0I=0$. If M is invertible, a₀ ≠ 0 and multiplication by M⁻¹ expresses $M^{-1}=-(M^4+a_4M^3+\cdots+a_1I)/a_0$, proving C. Since dimension is odd, p(0) = −det M. The matrix 2I contradicts D. The two scalar relations in B do not force all characteristic roots to be real.
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