If $A$ is invertible with $A\in Mat_3(\mathbb{C})$, prove for all $x\not = 0$ we have:
$P_{A^{-1}}(x)=\frac{x^{3}}{P_A(0)}P_A(\frac{1}{x}) $
My work
We know by Cayley-Hamilton theorem:
$P_{A}(x) = x^{3} - \mathrm{tr}(A)x^{2} + \mathrm{tr}(\mathrm{adj}(A))x - \det(A)$
Then,
$P_A(0)=-det(A)$
$P_A(1/x)=\frac{1}{x^3}-tr(A)\frac{1}{x^2}+tr(adj(A))\frac{1}{x}-det(A)$
Then,
$$\frac{x^3}{P_A(0)}P_A(\frac{1}{x})=(\frac{x^3}{-det(A)})(\frac{1}{x^3}-tr(A)\frac{1}{x^2}+tr(adj(A))\frac{1}{x}-det(A))=-\frac{1}{det(A)}+\frac{tr(A)x}{det(A)}-\frac{tr(adj(A)x^2}{det(A)}+x^3$$
Moreover,
$P_{A^{-1}}(x)=\frac{1}{x^3}-tr(A^{-1})\frac{1}{x^2}+tr(adj(A^{-1})\frac{1}{x}-det(A^{-1})$
Here i'm stuck. Can someone help me?
I'm sure you mean $$P_{A^{-1}}(x)=x^3-tr(A^{-1})x^2+tr(adj(A^{-1}))x-det(A^{-1}).$$
You showed that $$\frac{x^3}{P_A(0)}P_A\left(\frac{1}{x}\right)=-\frac{1}{det(A)}+\frac{tr(A)x}{det(A)}-\frac{tr(adj(A))x^2}{det(A)}+x^3.$$
To reach the desired conclusion, recall that:
$\det{A}\times\det{A^{-1}}=1;$
$A^{-1}=\dfrac{1}{\det A}\text{adj}(A)$.