Please anyone cross-check the explanation of my answer.
Question: Consider the $~5\times 5~$ matrix $A$ such that it's characteristic polynomial is $~(x-2)^3(x+2)^2~$. Now if $A$ is diagonalizable then find the value of $\alpha$ and $\beta$ such that $~A^{-1}=\alpha A+\beta I~.$
Answer: Since $A$ satisfies the condition $$~A^{-1}=\alpha A+\beta I\implies \alpha A^2+\beta A-I=0~,$$so the minimal polynomial of the matrix $A$ must be of degree $2$ and is of the above form.
Again it is given that $A$ is diagonalizable and also from the characteristic polynomial it is clear that it has no zero on the diagonal, so $A^{-1}$ exists.
Now from the characteristic polynomial, possible $2$-degree minimal polynomial is $~(x-2)(x+2)=x^2-4~.$
So Cayley–Hamilton theorem gives, $$~A^2-4I=0\implies A-4A^{-1}=0\implies A^{-1}=\dfrac 14 A~.$$ Hence $~\alpha=\frac 14~$ and $~\beta=0~.$
Note: I know one may say there be similar question (not exactly, I think) on here find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$.. But here I just want to cross-check my explaination nothnig else.