Step-by-step Solution:

Given: \[ 6 \mathrm{A}^{-1} = \mathrm{A}^2 + \mathrm{cA} + \mathrm{dI} \] Multiplying both sides by \(\mathrm{A}\): \[ 6 \mathrm{I} = \mathrm{A}^3 + \mathrm{cA}^2 + \mathrm{dA} \] Rearranging: \[ \mathrm{A}^3 + \mathrm{cA}^2 + \mathrm{dA} - 6 \mathrm{I} = 0 \tag{1} \] From the Cayley-Hamilton theorem, the characteristic equation of \(\mathrm{A}\) is: \[ \det(\mathrm{A} - \lambda \mathrm{I}) = 0 \] \[ \mathrm{A} - \lambda \mathrm{I} = \begin{bmatrix} 1 - \lambda & 0 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & -2 & 4 - \lambda \end{bmatrix} \] \[ \det(\mathrm{A} - \lambda \mathrm{I}) = (1-\lambda) \cdot \det\begin{bmatrix} 1 - \lambda & 1 \\ -2 & 4 - \lambda \end{bmatrix} \] \[ \det\begin{bmatrix} 1 - \lambda & 1 \\ -2 & 4 - \lambda \end{bmatrix} = (1-\lambda)(4-\lambda) + 2 \] \[ = (1-\lambda)(4-\lambda) + 2 = \lambda^2 - 5\lambda + 6 \] \[ \det(\mathrm{A} - \lambda \mathrm{I}) = (1-\lambda)(\lambda^2 - 5\lambda + 6) \] \[ = \lambda^3 - 6\lambda^2 + 11\lambda - 6 \tag{2} \] Comparing Equation (1) and Equation (2), we get: \[ \mathrm{c} = -6, \, \mathrm{d} = 11 \] --- **Final Answer:** \[ \boxed{c = -6, \, d = 11} \]