If $A$ is diagonalizable, find $\alpha$ and $\beta$ 
Let $A$ be a $5 \times 5$ matrix whose characteristic polynomial is given by
  $$p_A(\lambda)=(λ + 2)^2 (λ − 2)^3$$
      If $A$ is diagonalizable, find $\alpha$ and $\beta$ such that
  $$A^{-1} = \alpha A + \beta I$$

I am unable to find the inverse of $5\times 5$ matrix, I only know how to invert $3\times 3$ matrices. I don't know how to find the values of $α$ and $β$.
If anybody can help me I would be very thankful to them.
 A: Since $A$ is diagonalizable and $p_A(\lambda)=(\lambda + 2)^2 (\lambda − 2)^3$,
we have that there exists an invertible matrix $M$ such that
$$A=M\mbox{diag}(-2,-2,2,2,2)M^{-1}.$$
Hence
\begin{align*}A^{-1}&=(M\mbox{diag}(-2,-2,2,2,2)M^{-1})^{-1}=
M\mbox{diag}\left(-\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)M^{-1}\\&=\frac{1}{4}M\mbox{diag}(-2,-2,2,2,2)M^{-1}=\frac{1}{4}A.
\end{align*}
A: The fact that A is diagonalizable means that there exist an invertible matrix, P, such that $PAP^{1}= \begin{bmatrix}2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 & -2 \end{bmatrix}$.  So $(PAP^{1})^{-1}= PA^{-1}P^{-1}= \begin{bmatrix}\frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & -\frac{1}{2} \end{bmatrix}$.  
So $A^{-1}= P^{-1}\begin{bmatrix}\frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & -\frac{1}{2} \end{bmatrix}P$.
