For $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, if X and Y are square matrices of order 2 such that $X Y=X$ and $Y X=Y$, then $\left(Y^2+2 Y\right)$ equals to:

$3Y$

The correct answer is Option (4) → $3Y$

$XY=X,\;YX=Y.$

$Y^2=Y(XY)= (YX)Y = Y^2.$

$YX=Y \Rightarrow Y(X-I)=0.$

$XY=X \Rightarrow (Y-I)X=0.$

$YX=Y \Rightarrow Y^2=Y.$

$Y^2+2Y=Y+2Y=3Y.$