# Invertible matrices multiplication

I have two square matrices $X$ and $Y$ of the order of $n \times n$. It is given that $X = I - XY$. I have to show that $X$ is invertible and that $XY = YX$.

The first part is fairly easy, moving stuff around gets me to the point that $X(I + Y) = I$ and therefore $X$ multiplied by another matrix yields the identity matrix and therefore $X$ is Invertible.

I am stuck at the second part which requires showing that $XY = YX$.

I tried multiplying $(I-XY)B$ but it got me nowhere.

Remark: Suppose that $AB=I$; then $BA=I=AB$.

As you have mentioned $X(I+Y)=I$.

Considering the above remark we can write

$$X(I+Y)=I=(I+Y)X;$$

now expanding both sides, one gets that:

$$X+XY=X+YX \qquad \Longrightarrow \qquad XY=YX.$$

Since\begin{align}X&=\operatorname{Id}-XY\\&=XX^{-1}-XY\\&=X(X^{-1}-Y)\end{align}and since $X$ is invertible, $X^{-1}-Y=\operatorname{Id}$, and this means that $Y=X^{-1}-\operatorname{Id}$. Therefore, $XY=YX$, since obviously $X$ and $X^{-1}-\operatorname{Id}$ commute (both $X(X^{-1}-\operatorname{Id})$ and $(X^{-1}-\operatorname{Id})X$ are equal to $\operatorname{Id}-X$).