to prove
Η * Η’ = Η’ * Η
Η= [a b]
[c d]
H'=[a' b']
[c' d']
$\displaystyle \left(\begin{array}{cc}a & b \\ c & d\end{array}\right) \cdot\left(\begin{array}{cc} a' & b' \\ c' & d'\end{array}\right) = \left(\begin{array}{cc}(aa'+bc') & (ab'+bd') \\(ca'+dc') & (cb' + dd')\end{array}\right)$
And
$\displaystyle \left(\begin{array}{cc} a' & b' \\ c' & d'\end{array}\right) \cdot \left(\begin{array}{cc}a & b \\ c & d\end{array}\right) = \left(\begin{array}{cc}(a'a+b'c) & (a'b+b'd) \\(c'a+d'c) & (c'b + d'd)\end{array}\right)$
Well, I'm not sure what you are trying to do, but your original post says prove that H.H' = H'.H
Since we know that a = a', d = d', c = b' and b = c', then H.H' = H'.H is proved.
All you appear to have here is two unrelated matrices, H and H'.
You cannot then prove that "H*H'= H'*H" because it is not true: matrix multiplication is not commutative.
Or did you mean that H' is the transpose of H? No, that doesn't work either. So I don't know what you are asking.