why not by cramers rule in determinant

why not by cramers rule in determinant

i am extremly sorry i am wrong

Ax = b for rectangular sys.

x = {(A'A)^(-1) A'b} provided inversion exists...!

another option is x = {A' (AA')^(-1)b} provided inversion exists...!

and i wonder which one will the CGA gives.........?!

thanks guys for the comments but i have not used CGA before,please give me some links to get a head start.

Ravi does that work for a multi variable system with more equations than the variables.

one of the two given solutions must work ...

since you have more equations .... my guess is

x = {(A'A)^(-1) A'b}

will work... an this is what ConjugateGradient Algo. gives also.

If you like i can teach you but for this way its difficult, mi email is [email protected] if you write me i send a papers with the methods to solve this problem!

Ravi,Ruiz thanks for your contributions.Hassan,they are algebraic equations.

Bartl thanks,it makes sense.if i used a maths software,would i still need to go through choosing the norm?

Generalized Banach algebra (a new Banach algebra )

http://www.researchgate.net/profile/Yang_Xiaojun/blog/23896_Generalized_Banach_algebra_a

In this case one of the equations must be dependent on one or several of the other equations because otherwise you have an overdetermined system. This is the same as saying that one of the equations is not independent of the others.

Jemimah: Actually you have a system of linear equations that can be expressed as

Y = XA, where Y is an (n x 1) vector, X is a (n x m) matrix, and A is an (m x 1) vector, with n is greater than m. The vector A is the vector of the unknowns.

As X is a rectangular matrix, it can not be inverted. However (X (transposed). X) is a square matrix. If this square matrix can be inverted, you would then get

A = (X (transposed) .X) (inverse).(X (transposed) Y).

This is very basic, if there are more equations then unknowns then just use as may equations as you have unknowns. My perferred method is addition.