<s>[INST] Finding $\alpha$ and $\beta$ for the given matrix equation. Let $\textbf{A}$ be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda - 2)^3(\lambda + 2)^2$. If $\textbf{A}$ is diagonalizable, find $\alpha$ and $\beta$ such that :
$\textbf{A}^{-1} = \alpha \textbf{A} + \beta \textbf{I} $
Here is my approach so far:
Using Cayley-Hamilton theorem we get:
$\textbf{A}^{5} - 2\textbf{A}^{4} - 8 \textbf{A}^{3} + 16 \textbf{A}^{2} + 16\textbf{A} - 32 = 0 $
Now from $\textbf{A}^{-1} = \alpha \textbf{A} + \beta \textbf{I} $
we get : $\textbf{I} = \alpha \textbf{A}^{2} + \beta \textbf{A} $
Similarly I multiply the above equation by $\textbf{A}$ to get $\textbf{A} = \alpha \textbf{A}^{3} + \beta \textbf{A}^{2} $ Now we can substitute the value of $\textbf{A}^{2}$ from previous equation.
If I continue this way I obtain all the powers of $\textbf{A}$ in terms of $A$. 
Then I substitute these into Cayley equation but the whole thing becomes a mess involving powers of $\alpha$ and $\beta$.
Moreover, I am not able to figure out how to make use of diagonalizable property.
Please help.    
 [/INST] There is some invertible $B$ such that $A = B^{-1}DB$ where $D$ is the diagonal matrix with three times a $2$, two times a $-2$. $D^{-1}$ is just the matrix with the inverses of $2$ and $-2$, so $\frac{1}{2}$ and $-\frac{1}{2}$, which is just $\frac{1}{4}D$.
But then $A^{-1} = (B^{-1}DB)^{-1} = B^{-1}D^{-1}B = \frac{1}{4}B^{-1}DB = \frac{1}{4}A$.
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