<s>[INST] 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?
 [/INST] 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.
</s>