# Calculate specific result by numeric series multiplication and their later sum

firstly I need to apologize for my non mathematical language. Secondly I'll try to explain what I would like to calculate:

I have 3 series/rows of numbers and every series has 5 numbers(for example):

1.series: 2,4,9,1,7

2.series: 3,1,8,6,2

3.series: 8,0,6,0,9

sum of their columns should be as close as possible to this result:

result: 90,60,70,21,45

any series may/may not be multiplied by any multiplier

Could someone advice me what method shall be used for the calculation please? Any help is greatly appreciated

Thank you

nickzde

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Put the system in the form $$Xw = y$$ i.e. (note each series is in a column of $X$) $$\begin{pmatrix} 2 & 3 & 8 \\ 4 & 1 & 0 \\ 9 & 8 & 6 \\ 1 & 6 & 0 \\ 7 & 2 & 9 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} = \begin{pmatrix} 90 \\ 60 \\ 70 \\ 21 \\ 45 \end{pmatrix}$$ Now to solve for $w,$ you can approximate it using least-squares as follows: $$Aw = y \\ A^{T} A w = A^{T} y \\ w = (A^{T} A)^{-1} A^{T} y \\$$ This will work as long as $A^{T} A$ is invertible.

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Hello, thank you very much for your answer. And how to resolve it in the case when A^{T}A is not invertible? Thank you –  hugo Apr 1 '12 at 15:10
When $A^{T}A$ is not invertible this means that one series is an exact multiple of some other series. Remove the redundant one are proceed. –  user2468 Apr 1 '12 at 16:55
Now I see. Could I have an additional question please? As least-squares gives me an approximation, it can happen that result of multiplication of resulting w value with X will be smaller than intended value y. Is there a way how to make every w big enough so that Xw >= y? And is it possible, at the same time, to make sure every Xw <= some other number? Many Thanks again! I am marking your reply as answer. –  hugo Apr 1 '12 at 17:34
@hugo I am not quite sure. AFAIK, least squares minimizes the square of the error. In other words it minimizes $(Aw - y)^2.$ So it can give you $Aw$ bigger or smaller than $y.$ Think $(2.1-2)^2$ is small. Also $(1.9-2)^2$ is small. What you need is minimize the error (not squared). I have to go through my course notes to recall which method. But please go ahead and post a new question; maybe someone else know readily. –  user2468 Apr 1 '12 at 17:49
Thank you very much for your help. I have posted new question link. We can meet there again :) Have a nice day.hugo –  hugo Apr 1 '12 at 19:34

Let say you multiply the rows by $a,b$ and $c$, respectively. Then if you add them up you get the following system of equations $2a+3b+8c=90$, $4a+1b+0=60$, $9a+8b+6c=70$, $1a+6b+0c=21$ and $7a+2b+9c=45$. You can solve the system of equations as usual.

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Hello, thank you very much for your answer. Is it possible to alter these equations> 2a+3b+8c=90 ...etc. in such a way they would implement not only a,b,c but also d,e,f.....in the case there was more series than just 3? Again many thanks. –  hugo Apr 1 '12 at 15:11