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element-by-element multiplication
Dear Mathwork users,
If i have two matrices,
A=magic(3)
B=magic(3)
a =[8 1 6; 3 5 7; 4 9 2]
a =
8 1 6
3 5 7
4 9 2
b =[8 1 6; 3 5 7; 4 9 2]
b =
8 1 6
3 5 7
4 9 2
What multiplication method do i use (e.g. cross/dot product) if i want to multiply each element by each element to result in a 3x3 matrix to get:
c= [54, 1 36; 9 25 49; 16 81 4]
c =
54 1 36
9 25 49
16 81 4
Thank you ^^
Natalie
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Reply
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 Natalie
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6/19/2010 6:19:05 PM |
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magic(3).*magic(3)
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Matt
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6/19/2010 6:27:05 PM
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Also, you really should get used to using MATLAB's help facilities. For example, if you had done:
docsearch('element-by-element')
you would have figured it out!
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Matt
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6/19/2010 6:30:22 PM
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"Natalie Sin Hwee " <sin.ng09@imperial.ac.uk> wrote in message <hvj1mp$p54$1@fred.mathworks.com>...
> Dear Mathwork users,
>
> If i have two matrices,
>
> A=magic(3)
> B=magic(3)
>
>
> a =[8 1 6; 3 5 7; 4 9 2]
> a =
> 8 1 6
> 3 5 7
> 4 9 2
>
> b =[8 1 6; 3 5 7; 4 9 2]
> b =
> 8 1 6
> 3 5 7
> 4 9 2
>
> What multiplication method do i use (e.g. cross/dot product) if i want to multiply each element by each element to result in a 3x3 matrix to get:
>
> c= [54, 1 36; 9 25 49; 16 81 4]
> c =
> 54 1 36
> 9 25 49
> 16 81 4
>
>
> Thank you ^^
> Natalie
Well, I don't know of ANY operator that will give you the
result you show, since 8*8 = 64, not 54.
If perhaps you really meant 64 there, then you might find
the .* operator of value.
help times
John
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John
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6/19/2010 6:30:22 PM
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a.*b
and 8 x 8 is 64
"Natalie Sin Hwee " <sin.ng09@imperial.ac.uk> skrev i meddelelsen
news:hvj1mp$p54$1@fred.mathworks.com...
> Dear Mathwork users,
>
> If i have two matrices,
>
> A=magic(3)
> B=magic(3)
>
>
> a =[8 1 6; 3 5 7; 4 9 2]
> a =
> 8 1 6
> 3 5 7
> 4 9 2
>
> b =[8 1 6; 3 5 7; 4 9 2]
> b =
> 8 1 6
> 3 5 7
> 4 9 2
>
> What multiplication method do i use (e.g. cross/dot product) if i want to
> multiply each element by each element to result in a 3x3 matrix to get:
>
> c= [54, 1 36; 9 25 49; 16 81 4]
> c =
> 54 1 36
> 9 25 49
> 16 81 4
>
>
> Thank you ^^
> Natalie
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0
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Reply
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John
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6/19/2010 6:59:46 PM
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4 Replies
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