matrix multiplication

[Seb]

On Tue, 27 Feb 2018 12:25:30 +1300,
Gregory Ewing <greg.ewing at canterbury.ac.nz> wrote:

> Seb wrote:
>> I was wondering is whether there's a faster way of multiplying each
>> row (1x3) of a matrix by another matrix (3x3), compared to looping
>> through the matrix row by row as shown in the code.

> Just multiply the two matrices together.

> If A is an nx3 matrix and B is a 3x3 matrix, then C = A @ B is an nx3
> matrix where C[i] = A[i] @ B.

> (This is a property of matrix multiplication in general, nothing
> special about numpy.)

I think that's only true if B is the same for every row in A.  In the
code I posted, B varies by row of A.

-- 
Seb