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If $$A = \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ \end{bmatrix}, $$ then use the Cayley-Hamilton Theorem to find $A^{-3}$.
This is how far I have gotten: \begin{align} p(\lambda) &= \lambda^2 -3\lambda -2 \\ p(A) &= 0 = A^2-3A-2I \\ A^2 &= 3A + 2I \\ A^3 &= 3A^2 + 2A = 11A+6I \end{align}
How do I proceed from here?
The characteristic polynomial is $(2-\lambda)(1-\lambda)-4=2-\lambda -2\lambda +\lambda^2 -4=\lambda^2-3\lambda -2$.
CH ensures you that $$ A^2 - 3A = +2I $$ hence $$ A(A-3I)=2I \qquad (A-3I)A = 2I $$ so that $A^{-1}=\frac{1}{2}(A-3I)$. From the equation above we also have, as you found, $$ A^3 = 3A^2 +2A = 3(3A+2I) + 2A= 11A+6I. $$ All in all, $$ A^{-3} = (A^{-1})^3 = \frac{1}{8}(A-3I)^3 = 8^{-1}(A^3 - 9A^2 + 27A - 27I) $$ which can be further simplified into $$ = 8^{-1}(11A+6I - 9(3A+2I)+27A-27I) = $$ $$ = \frac{1}{8}(11A+6I-27A -18 I +27A - 27I) = \frac{1}{8}(11A -39I) $$ thus $$ \boxed{A^{-3} = \frac{1}{8}\begin{bmatrix} -17 & 44 \\ 11 & -28 \end{bmatrix}}. $$
The Cayley-Hamilton theorem tells you that $A^2-3A-2I=0$; multiply by $A^{-1}$ to find $$ 2A^{-1}=A-3I $$ Multiply again by $2A^{-1}$: $$ 4A^{-2}=2I-6A^{-1}=2I-3(A-3I)=11I-3A $$ Multiply again by $2A^{-1}$: $$ 8A^{-3}=22A^{-1}-6I=11(A-3I)-6I=11A-39I $$ In a different way, you know from CH that $A^{-3}=\alpha A+\beta I$; then $$ I=\alpha A^4+\beta A^3 $$ and you can use CH for reducing the expression on the right.
Now go the other way around: \begin{align} p(\lambda) &= \lambda^2 -3\lambda -2 \\ p(A) &= O = A^2-3A-2I \\ I &= A \color{blue}{A^{-1}} = \tfrac12A^2 -\tfrac32 A = A \color{blue}{(\tfrac12 A -\tfrac32 I)} \\ A^{-1} &= \tfrac12 A -\tfrac32 I \end{align} Then $A^{-3}$ is just $A^{-1}$ times $A^{-1}$ times $A^{-1}$. You can reduce powers of $A$ using $p(A)$ if needed.
