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?
Best Answer
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 $$ \boxed{A^{-3} = \frac{1}{8}\begin{bmatrix} -17 & 44 \\ 11 & -28 \end{bmatrix}}. $$