linear algebra - Given a Characteristic Polynomial of a Matrix...

This question contains three parts. I have already answered the first two. The last part is confusing me.

Suppose $A$ is a $4 \times 4$ matrix whose characteristic polynomial is $p(x) = (x - 1)(x + 2)^2(x - 3)$.

Part (a): Show that $A$ is invertible. Find the characteristic polynomial of $A^{-1}$.

We have that the roots of a characteristic polynomial are the eigenvalues of $A$. That is, $\lambda = -2, -2, 1, 3$ are our eigenvalues. The determinant of an $n \times n$ matrix is the product of its eigenvalues. Hence, det$A = 12$. An $n \times n$ matrix is invertible if and only if its determinant is nonzero. Therefore, $A$ is invertible.

Since none of the eigenvalues are zero, we have that $\lambda$ is an eigenvalue of $A$ if and only if $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$. Then, the characteristic polynomial for $A^{-1}$ is $q(x) = (x - 1)(x + 1/2)^2(x - 1/3)$.

Part (b): Find the determinant and trace of $A$ and $A^{-1}$.

This is easy since the determinant of an $n \times n$ matrix is the product of its eigenvalues and the trace of an $n \times n$ matrix is the sum of its eigenvalues.

Part (c): Express $A^{-1}$ as a polynomial in $A$. Explain your answer.

Not really sure what part (c) is getting at.

Answer

By the Cayley-Hamilton theorem, we have $(A-1)(A+2)^2(A-3)=0$, that is,
$A^4-9A^2-4A+12I=0$.

Multiply both sides by $A^{-1}$, and be amazed!