# Finding $A^{-3}$ using Cayley Hamilton Theorem

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.