# Matrix multiplication.

• Nov 2nd 2010, 04:38 AM
stolla
Matrix multiplication.
to prove
Η * Η’ = Η’ * Η

Η= [a b]
[c d]

H'=[a' b']
[c' d']
• Nov 2nd 2010, 04:42 AM
Unknown008
Do the multiplication matrix and show us what you get! (Smile)
• Nov 2nd 2010, 04:49 AM
stolla
i Do the multiplication and i find a='a and d='d
but my teacher say a’ = a b’ = c c’ = b and d’ = d
• Nov 2nd 2010, 05:09 AM
Unknown008
$\displaystyle \left(\begin{array}{cc}a & b \\ c & d\end{array}\right) \cdot\left(\begin{array}{cc} a' & b' \\ c' & d'\end{array}\right) = \left(\begin{array}{cc}(aa'+bc') & (ab'+bd') \\(ca'+dc') & (cb' + dd')\end{array}\right)$

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.
• Nov 2nd 2010, 06:09 AM
stolla
we dont know that a = a', d = d', c = b' and b = c',
we are trying to find that h*h'=h'*h
and i think that the anser is a = a', d = d', c = b' and b = c',
but i am not completely sure
• Nov 2nd 2010, 06:11 AM
Unknown008
Well, for H.H' to be equal to H'.H, we must have a = a', d = d', c = b' and b = c'
• Nov 2nd 2010, 06:16 AM
stolla
thnks dude!
may the force be with you!!
• Nov 2nd 2010, 06:18 AM
Unknown008
Uh...okay... (Itwasntme)

Thanks!
• Nov 2nd 2010, 06:19 AM
stolla
sorry can you explain the answer plzzzzz??
• Nov 2nd 2010, 06:24 AM
Unknown008
You cannot prove that

H.H' = H'H

unless we know that a = a', d = d', c = b' and b = c'
• Nov 2nd 2010, 07:38 AM
HallsofIvy
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.