Let $A$ and $B$ are square matrices of same order satisfying $AB=A$ and $BA=B$ ,Then ${(A^{2021}+B^{2021})}^{2022}$ is equal to

 $A^2=A\cdot A$
$=\left(AB\right)\left(AB\right)$
$=A\left(BA\right)B$
$=\left(AB\right)B$
$=AB$
$=A$
Similarly  $B^2=B$
$\therefore A=A^2=A^3=\cdots$
$B=B^2=B^3=\dots$
${(A^{2021}+B^{2021})}^{2022}={(A+B)}^{2022}$
${(A+B)}^2=(A+B)(A+B)$
$=A^2+AB+BA+B^2$
$=A^2+A+B+B^2=2(A+B)$
${(A+B)}^3=2^2(A+B)$
$\therefore {(A+B)}^{2022}=2^{2021}(A+B)$