Invertible matrix properties of a matrix I have here the following question:

Let $X$ be the $5 \times 5$ matrix "full of ones":
$X = \begin{pmatrix}1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 1\end{pmatrix}$
$(a)$ Is $X$ invertible? Explain.
$(b)$ Find a number $c$ such that $X^2=cX$.
$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $a\neq 0$, and $a\neq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)

I already did part $(a)$ and $(b)$. 
$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.
$(b)$ $X^2 = \begin{pmatrix}5 & 5 & 5 & 5 & 5 \\ 5 & 5 & 5 & 5 & 5 \\ 5 & 5 & 5 & 5 & 5 \\ 5 & 5 & 5 & 5 & 5\\ 5 & 5 & 5 & 5 & 5 \end{pmatrix}$
Thus, $c=5$.
$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.
$=(X-aI_5)(X+(a-c)I_5)$
$=-a(a-c)I_5$
If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:
$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$
I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?
For reference, the answer should be:
$M^{-1}=\frac{X+(a-c)I_5}{-a(a-c)}$

Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.
