If $A$ is a matrix, then express it in the form $A^{-1} = A^2 + pA + qI$

Question

I am given a matrix $ A = \begin{bmatrix}5 & 7 & 3\\ 1 & 5 & 2\\ 3 & 2 & 1\end{bmatrix}$ and I have to express it in the form $A^{-1} = A^2 + pA + qI$
I thought for a while but cannot think of anything else other than computing $A^{-1}$ and $A^2$ and then actually solving it. Or a better method to multiply $A$ on both sides and then compute the result requiring me to compute $A^3$ in that case.
My question is, that is it possible to find out this expansion using a better and more efficient method because the one above seems quite tedious.

Answer 1

Since the characteristic polynomial of $A$ is $-\lambda ^3+11 \lambda ^2-15 \lambda +1$, the Cayley-Hamilton theorem tells you that$$-A^3+11A^2-15A+\operatorname{Id}=0,$$and therefore$$-A^2+11A-15\operatorname{Id}+A^{-1}=0.$$So,$$A^{-1}=A^2-11A+15\operatorname{Id}.$$

Answer 2

1. Evaluate the characteristic polynomial of $\;A\;$ (up to multiplication by $\;-1\;$ ) :

$$p_A(t)=\det (xI-A)=x^3-11x^2+15x-1$$

2. Deduce the matrix is invertible

3. Apply Cayley-Hamilton Theorem:

$$A^3-11A^2+15A-I=0\; (=\text{ the zero matrix, of course})$$

4. Multiply the above by $\;A^{-1}\;$

5. Solve your problem