If AB = A and BA = B, where A c. · B are square matrices, then
B2=B and A2=A
B2≠B and A2=A
A2≠A,B2=B
A2≠A,B2≠B
We have,
A2=AA
⇒ A2=(AB)A [∵AB=A]⇒ A2=A(BA) ⇒ A2=AB [∵BA=B]⇒ A2=A [∵AB=A]
and,
B2=BB⇒B2=(BA)B [∵BA=B]⇒B2=B(AB)⇒B2=BA [∵AB=A]⇒B2=B [∵BA=B]