Proof related to matrix

Question

Let $A$ and $B$ be $n \times n$ real matrices such that $A^2 = I, B^2 = I$ and $(AB)^2 = I$. Prove that $AB = BA$.

Someone help me with this problem

You could help yourself by searching. – Jun 20 '15 at 02:17 — , Jun 20 '15 at 02:17

score 3 · Answer 1 · answered Jun 19 '15 at 23:43

3

$I = (AB)^2 = (AB)(AB) = A(BA)B \implies AB = A(A(BA)B)B = (A^2)(BA)(B^2) = BA$

answered Jun 19 '15 at 23:43

Marcus M

score 2 · Answer 2 · answered Jun 19 '15 at 23:44

2

$AB=(AB)^{-1} \iff AB=B^{-1}A^{-1} \iff AB=BA$

answered Jun 19 '15 at 23:44

Malcolm

how do AB becomes equal to its inverse – Ampatishan Arun Jun 19 '15 at 23:47
@AmpatishanArun: Because $(AB)^2=I$. That's the definition of inverse. – hmakholm left over Monica Jun 19 '15 at 23:48
$AB^2=I \iff (AB)^{-1}(AB)^2=AB^{-1}I \iff AB=AB^{-1}$ – Malcolm Jun 19 '15 at 23:48
okay i got it thanks – Ampatishan Arun Jun 19 '15 at 23:49

2 Answers2