TITLE: Given $A^2+cA+cI=0$, how to find inverse of $A+(c-1)I$?
QUESTION [6 upvotes]: Suppose a square matrix $A$ such that $A^2+cA+cI=0$ for all $c \in \mathbb{Z}$. How can I show that $A+(c-1)I$ is invertible and find its inverse?
I started off this way:
$A+(c-1)I = A+cI-I$
Then $(A+cI-I)(d_1A+d_2I)=I$, where $d_1, d_2 \in \mathbb{Z}$.
Expand it and it becomes:
$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$
$\Rightarrow 

(c-1)d_2=1 \;\; and \; \; d_1A^2+d_1cA-d_1A+d_2A=0$
$\Rightarrow 
d_2=\frac{1}{(c-1)}$
Continue from $d_1A^2+d_1cA-d_1A+d_2A=0$, after some manipulation, I got $d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$.
Since given that $A^2+cA+cI=0$,
$d_1(A^2+cA+cI)-d_1cI-d_1A+d_2A=0$
$\Rightarrow 
-d_1cI-d_1A+d_2A=0$
$\Rightarrow 
-d_1cI-d_1A+\frac{1}{c-1}A=0$
$\Rightarrow 
-d_1cI-d_1A=-\frac{1}{c-1}A$
$\Rightarrow 
d_1(cI+A)=\frac{1}{c-1}A$
At this point, since I don't know if $A$ is invertible yet, I cannot do it as $d_1=\frac{1}{c-1}\frac{A}{(cI+A)}$.
Even if I did this, I still cannot find a value for $d_1$ and $d_2$ to find the inverse of $A+(c-1)I$. How should I continue from here?

REPLY [5 votes]: $$
\begin{eqnarray*}
x^2+cx+c = \left[x+(c-1)\right](x+1)+1\\
\therefore 0 = A^2+cA+cI = \left[A+(c-1)I\right](A+I)+I\\
\therefore \left[A+(c-1)I\right](A+I)=-I\\
\therefore \left[A+(c-1)I\right]^{-1} = -(A+I).
\end{eqnarray*}
$$