Question:

<div>$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix} \Rightarrow |A| = 6$<br/>$\Rightarrow A^{-1} \Rightarrow \frac{adj A}{|A|} = \frac{1}{6} \begin{bmatrix} 6 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 2 & 1 \end{bmatrix}$<br/>$\Rightarrow A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 5 \\ 0 & -10 & 14 \end{bmatrix}$<br/>$\Rightarrow A^{-1} = \frac{1}{6} [A^2 + cA + dI]$<br/>Using characteristic equation: $|A - \lambda I| = 0$<br/>$(1-\lambda)((1-\lambda)(4-\lambda) + 2) = 0$<br/>$(1-\lambda)(\lambda^2 - 5\lambda + 6) = 0$<br/>$\lambda^2 - 5\lambda + 6 - \lambda^3 + 5\lambda^2 - 6\lambda = 0$<br/>$-\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0$<br/>$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0$<br/>By Cayley-Hamilton theorem: $A^3 - 6A^2 + 11A - 6I = 0$<br/>Multiply by $A^{-1}$: $A^2 - 6A + 11I - 6A^{-1} = 0$<br/>$6A^{-1} = A^2 - 6A + 11I$<br/>$A^{-1} = \frac{1}{6}(A^2 - 6A + 11I)$<br/>Comparing with $A^{-1} = \frac{1}{6}(A^2 + cA + dI)$, we get $c = -6$ and $d = 11$.</div>