Problem with an Invertible matrix Good afternoon,

I need your help please in this question, and I'm sorry if my question is evident;

Let $X, Y\in M_n(\mathbb{R})$ be two real matrices and $I$ be the identity matrix. It is clear that
$$(X-I)(Y-I)=XY-X-Y+I$$ 

The question is: show that if $XY=X+Y$, then $I-X$ is invertible?

What I have done: if $XY=X+Y$ then one has $(X-I)(Y-I)=I$, how I continue?

PS: * I know that a matrix $X\in M_n(\mathbb{R})$ is invertible if and only if there exist $X'\in M_n(\mathbb{R})$ such that 
$$XX'=X'X=I$$
** Also we know that a matrix $X\in M_n(\mathbb{R})$ is invertible if and only if $|X|\neq0$.

Thank you
If $ XY = X+Y, $ then $$ (X-I)(Y-I) = XY - X - Y + I = X + Y - X - Y + I = I $$
So $$(X-I)^{-1} = (Y-I)$$ and clearly if $(X-I)$ is invertible so is $(I-X) = - (X - I)$. We also have to show that $(Y-I)$ is a left inverse of $(X-I)$, but this follows quite easily.
If $PQ=I$ then $\det(PQ)=\det(I)$, and hence $\det(P)\cdot \det(Q) = 1$. It follows that $\det(P) \neq 0$ and $\det(Q) \neq 0$, i.e. both $P$ and $Q$ are invertible. Moreover, $P^{-1}=Q$ and $Q^{-1}=P$.

If $R$ is an $n$-by-$n$ matrix then $\det(-R)=(-1)^n\cdot \det(R)$ and hence $\det(R) \neq 0 \iff \det(-R) \neq 0$, i.e. $R$ is invertible if and only if $-R$ is invertible.

You have shown that $(X-I)(Y-I)=I$. If $P=X-I$ and $Q=Y-I$ then we can conclude that both $X-I$ and $Y-I$ are invertible. Moreover, it follows that both $I-X$ and $I-Y$ are invertible.