Question 1077157
<pre>
(AB)'(AB) = I

(B'A')(AB) = [(B'A')A]B = [B'(A'A)]B = [B'I]B = B'B = I = (AB)'(AB)

So

(B'A')(AB) = (AB)'(AB)

Right multiply both sides by (AB)'

[(B'A')(AB)](AB)' = [(AB)'(AB)](AB)'

(B'A')[(AB)](AB)'] = (AB)'[(AB)(AB)']

(B'A')I = (AB)'I

B'A' = (AB)'

That's what we were to prove.

Edwin</pre>