Given $A^2+cA+cI=0$, how to find inverse of $A+(c-1)I$? 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?
 A: Instead of all this, rewrite your starting equation as $A^2+(c-1)A+A+(c-1)I=-I$. Then factor the left-hand-side...
But if you favor a more systematic approach you can also proceed as you do until
$$d_1A^2+d_1cA-d_1A+d_2A+(c-1)d_2I=I$$
At this point you decide to set $(c-1)d_2=1$, but that is too early! First get rid of the $A^2$ using $A^2=-c(A+I)$, to get
$$-d_1c(A+I)+d_1cA - d_1A + d_2A + (c-1)d_2I = I$$
which simplifies to
$$(d_1c-d_1+d_2-d_1c)A + ((c-1)d_2-d_1c-1)I = 0$$
$$(d_2-d_1)A + ((c-1)d_2-cd_1-1)I$$
Now set both coefficients to 0. This gives immediately $d_1=d_2$, and we then need to solve
$$(c-1)d-cd-1=0$$
in which the $cd$'s cancel and give us $d=-1$. So the sought inverse is $-A-I$.
A: Perhaps a more transparent way: "change of variables".  Let $B = A + (c-1) I$, or $A = B - (c-1) I$.  Then $0 = A^2 + c A + c I = (B - (c-1) I)^2 + c (B - (c-1) I)  + c I= B^2 + (2-c) B + I$.  Multiply by $B^{-1}$ to get $B + (2-c) I + B^{-1} = 0$, i.e. $B^{-1} = -B + (c-2) I$
or $(A + (c-1) I)^{-1} = -A - (c-1) I + (c-2) I = -A - I$.  
You may find it a bit dodgy to multiply by $B^{-1}$ before you know that $B^{-1}$ exists, but once you have the result $-A-I$ it's easy to verify that this works by multiplying it by $A + (c-1) I$.
A: $$
\begin{eqnarray*}
x^2+cx+c = \left[x+(c-1)\right](x+1)+1\\
\therefore 0 = A^2+cA+cI = \left[A+(c-1)I\right](A+I)+I\\
\therefore \left[A+(c-1)I\right](A+I)=-I\\
\therefore \left[A+(c-1)I\right]^{-1} = -(A+I).
\end{eqnarray*}
$$
A: NEW ANSWER. In the "long" division 
$$X^2+cX+c=(X+c-1)(X+1)+1,$$ 
replace $X$ by $A$: 
$$0=A^2+cA+cI=\Big(A+(c-1)I\Big)\Big(A+I\Big)+I,$$ 
$$\Big(A+(c-1)I\Big)^{-1}=-A-I.$$ 
[EDIT. I'm realizing that this answer is the same as user1551's, who posted it before. Sorry...] 
OLD ANSWER. Let $A$ be an $n$ by $n$ matrix with coefficients in a field $K$, let $f\in K[X]$ be a polynomial annihilating $A$, and let $g\in K[X]$ be any polynomial. 
If $g$ is prime to $f$, then $g(A)$ is invertible, and the inverse of $g(A)$ is given by $h(A)$ where $h\in K[X]$ is an inverse of $g$ mod $f$. 
Moreover, there is a closed formula for such an $h$ (see this answer). 
