CallMeIshmael
06-09-2007, 03:26 PM
Im pretty rusty when it comes to matrices, and I was wondering if someone could help me with the following problem... I need it for some code Im writing
given that
A*B = C
where, A, B and C are all matrices of dimensions in the form:
A: (m X n)
B: (n X 1)
C: (m X 1) (with m, n > 1)
is there a QUICK method of determining B, given A and C?
Also, it might be important to note that the sum of each of the columns of A is 1, which also means that sum(B) == sum(C).
So, take, for example:
[0.8 0.6 0.4] X [0.4]..=..[0.48]
[0.2 0.4 0.6] .. [0.2]......[0.22]
.....................[0.1]
Is there a quick way to go from (0.48; 0.22), to (0.4, 0.2, 0.1), given that you also know the first matrix
Im not really sure how to define 'quick', but assume that what I want to do means that the above operation is carried out thousands of times. Bascially, I dont want to have to row reduce / solve the equation each time.
thanks
EDIT: m != n
given that
A*B = C
where, A, B and C are all matrices of dimensions in the form:
A: (m X n)
B: (n X 1)
C: (m X 1) (with m, n > 1)
is there a QUICK method of determining B, given A and C?
Also, it might be important to note that the sum of each of the columns of A is 1, which also means that sum(B) == sum(C).
So, take, for example:
[0.8 0.6 0.4] X [0.4]..=..[0.48]
[0.2 0.4 0.6] .. [0.2]......[0.22]
.....................[0.1]
Is there a quick way to go from (0.48; 0.22), to (0.4, 0.2, 0.1), given that you also know the first matrix
Im not really sure how to define 'quick', but assume that what I want to do means that the above operation is carried out thousands of times. Bascially, I dont want to have to row reduce / solve the equation each time.
thanks
EDIT: m != n