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MathGroup Archive 2000

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Re: A question of matrix multiply, who can solve it?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg25719] Re: [mg25671] A question of matrix multiply, who can solve it?
  • From: BobHanlon at aol.com
  • Date: Thu, 19 Oct 2000 04:35:42 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 10/18/2000 4:02:59 AM, chenjs at iopp.ccnu.edu.cn writes:

>As a beginning user, I find a mistake of Mathematic 4.0(/3.0 for student).
>
>That is about the multipy method of matrix.
>
>For example, as I input the two matrices:
>
>A={a[1, 1], a[1, 2], a[1, 3], a[1, 4], a[1, 5]}
>{a[2, 1], a[2, 2], a[2, 3], a[2, 4], a[2, 5]}
>{a[3, 1], a[3, 2], a[3, 3], a[3, 4], a[3, 5]}
>{a[4, 1], a[4, 2], a[4, 3], a[4, 4], a[4, 5]}
>{a[5, 1], a[5, 2], a[5, 3], a[5, 4], a[5, 5]},
>B={{b[1, 1], b[1, 2], b[1, 3], b[1, 4], b[1, 5]}, 
> {b[2, 1], b[2, 2], b[2, 3], b[2, 4], b[2, 5]}, 
> {b[3, 1], b[3, 2], b[3, 3], b[3, 4], b[3, 5]}, 
> {b[4, 1], b[4, 2], b[4, 3], b[4, 4], b[4, 5]}, 
> {b[5, 1], b[5, 2], b[5, 3], b[5, 4], b[5, 5]}},
>
>then calculate the result of A B. The mathematics gives the following
>result:
>
>A B={{a[1, 1] b[1, 1], a[1, 2] b[1, 2], a[1, 3] b[1, 3], 
>  a[1, 4] b[1, 4], a[1, 5] b[1, 5]}, 
> {a[2, 1] b[2, 1], a[2, 2] b[2, 2], a[2, 3] b[2, 3], 
>  a[2, 4] b[2, 4], a[2, 5] b[2, 5]}, 
> {a[3, 1] b[3, 1], a[3, 2] b[3, 2], a[3, 3] b[3, 3], 
>  a[3, 4] b[3, 4], a[3, 5] b[3, 5]}, 
> {a[4, 1] b[4, 1], a[4, 2] b[4, 2], a[4, 3] b[4, 3], 
>  a[4, 4] b[4, 4], a[4, 5] b[4, 5]}, 
> {a[5, 1] b[5, 1], a[5, 2] b[5, 2], a[5, 3] b[5, 3], 
>  a[5, 4] b[5, 4], a[5, 5] b[5, 5]}}.
>
>As all knows, this is not correct. I think it is terrible.
>
>Do you think so? How can improve it?
>
>Looking forward to receiving the reply.

n = 2;

A = Array[a, {n, n}];

B = Array[b, {n, n}];

A*B

{{a[1, 1]*b[1, 1], a[1, 2]*b[1, 2]}, {a[2, 1]*b[2, 1], a[2, 2]*b[2, 2]}}

A*B does not mean what you think. It means that each of the corresponding 
elements are multiplied, that is, the result that you got. You need the dot 
product, A.B

A.B

{{a[1, 1]*b[1, 1] + a[1, 2]*b[2, 1], a[1, 1]*b[1, 2] + a[1, 2]*b[2, 2]}, 
  {a[2, 1]*b[1, 1] + a[2, 2]*b[2, 1], a[2, 1]*b[1, 2] + a[2, 2]*b[2, 2]}}

Dot[A, B] == A.B

True


Bob Hanlon


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