Suppose that $A$ is an invertible $5 \times 5$ matrix with characteristic polynomial $(\lambda-2)^3(\lambda+2)^2$. If $A$ is diagonalize find $\alpha$ and $\beta$ such that. $$A^{-1}=\alpha A+\beta I$$
Here is my plan. Suppose $A$ is diagonalizable. Then the diagonal matrix will be
$$D=\begin{bmatrix} 2& 0& 0 &0 &0\\ 0 &2 &0 &0& 0\\ 0 &0 &2 &0& 0\\ 0 &0 &0 &-2 &0\\ 0 &0 &0& 0 &-2 \end{bmatrix}$$
I think it is even easy. Namely, since you know that $A$ is diagonalizable with characteristic polynomial $(\lambda - 2)^3(\lambda + 2)^2$ it follows that the minimal polynomial of $A$ is $(\lambda - 2)(\lambda + 2) = \lambda^2 - 4$. So $$A^2 = 4 Id$$ hence $$A^{-1} = \frac{A}{4}$$ from which you get $\alpha = \frac{1}{4}$ and $\beta = 0$.