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?
 A: Your error comes from the fact that you cannot say that if $M_1M_2=0$ that $M_1=0$ or $M_2=0$...
For example : 
$$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
A: 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 $\beta=0$.
A: Since $A$ is diagonalizable, and the characteristic polynomial is $A$ is $c_A(x)=(x-2)^3(x+2)^2$, it follows that $A=PDP^{-1}$, where $P$ is an invertible $5\times5$ matrix, and
$$D=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&-2&0\\0&0&0&0&-2\end{pmatrix}.$$
It follows that $A^{-1}=PD^{-1}P^{-1}$. Note that
$$D^{-1}=\begin{pmatrix}\frac{1}{2}&0&0&0&0\\0&\frac{1}{2}&0&0&0\\0&0&\frac{1}{2}&0&0\\0&0&0&-\frac{1}{2}&0\\0&0&0&0&-\frac{1}{2}\end{pmatrix}.$$
So we have that $4\cdot D^{-1}=D$. Hence
$$\begin{align*}
4\cdot A^{-1} &= 4\cdot PD^{-1}P^{-1}\\
&= P(4\cdot D^{-1})P^{-1}\\
&= PDP^{-1}\\
&= A\end{align*}$$
So $4\cdot A^{-1}=A$. It follows that $A^{-1}=\frac{1}{4}\cdot A$. Hence $\alpha=\frac{1}{4}$ and $\beta=0$.
As I pointed out in the comments, you made a mistake when you assumed that $(A-2I)^2(A+2I)^2=0$ implied that either $(A-2I)^2=0$ or $(A+2I)^2=0$. Also, note that the Cayley-Hamilton theorem turned out to not be necessary for this particular problem.
