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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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