find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(\lambda+2)^2$.
By Cayley-Hamilton Theorem, we have $Ch_{A}(A)=0\implies (A -2)^3(A+2)^2=0\implies$ Either $(A -2)^3=0$ or $(A+2)^2=0$
Considering
$$(A+2)^2=0\implies A^2+4A+4I=0$$
$$\implies I=\left(-\frac{1}{4}\right)A^2+(-A)$$
$$\implies A^{-1}=\left(-\frac{1}{4}\right)A+(-I).$$
On comparing just recent equation with $A^{-1}=\alpha A+\beta I$,we get $\alpha =-\frac{1}{4}$ and $\beta =-1$
But the correct answer is $\alpha=\frac{1}{4}$ and $\beta =0$..
I wanted to know where I'm wrong?

Here's another way of solving your problem:
We're given that the characteristic polynomial of $A$ is $c_A(x)=(x-2)^3(x+2)^2$. Since $A$ is diagonalizable, that implies that the minimal polynomial must be $m_A(x)=(x-2)(x+2)$. Hence we have that
$$\begin{align*}
0 &= m_A(A)\\
&= (A-2I)(A+2I)\\
&= A^2-4I
\end{align*}$$
So $A^2-4I=0$, or equivalently $A^2=4I$.
Since $0$ is not an eigenvalue of $A$, we also know that $A$ is invertible. Multiplying by $A^{-1}$, we get that $A=4A^{-1}$. Thus $A^{-1}=\frac{1}{4}A$, $\alpha=\frac{1}{4}$, and $\beta=0$.
