Matrix help needed

[CallMeIshmael]

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

[danny]

In general, if n > m as in your example, the problem is underdetermined. You have m equations and n unknowns. In your example, the vector (0.3, 0.4, 0) is also a solution.

On the other hand, if n < m the problem is overdetermined. It may not have a solution. In either case, you must have some criteria for choosing the "best" solution. The methods for solving problems of this type are somewhat complex. Try looking up singular value decomposition.

[m_the0ry]

Row reduction is the only way to do this kind of operation. I recommend using a TI89 because you can enter the matrix in standard form:

[A|C]

and just tell it to perform row reduction.

[Fly]

if A*B = C, then

A^-1 * A * B = A^-1 * C

so

B = A^-1 * C

[m_the0ry]

inverting a matrix is certainly not any easier than row reduction.

[CallMeIshmael]

is it even possible to invert a non square matrix?

I didnt think it was.. but, like I said, Im not that knowledgeable of matrices


(also... yeah, it does seem like the only way is to RR it)