# 2x2 matrix multiplied by a 2x1 column vector gives erratic results

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Timothy Goldsmith am 6 Jun. 2016
Kommentiert: Stephen23 am 7 Jun. 2016
For example: A=[3,-2;2,-2] times v=[1;-1] works, but fails if A=[1,2;3,4]. The problem seems to be that in Matlab matrix multiplication the elements in row A are multiplied by the corresponding columns in B. Here B has only one column, and needs that the column elements in A be multiplied by the corresponding row elements in B. I have circumvented this problem by writing a function that does the latter, but as the need is for applying a vector to a transformation matrix, I am surprised to discover that the standard matrix multiplication algorithm cannot be relied upon. When it fails it takes A=[a1,b1;a2,b2] and computes v1a1+v2b1;v1a2+v2b2] instead of v1(a1+a2);v2(b1+b2). Is there away around this problem other than my function?
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Moe_2015 am 6 Jun. 2016
Bearbeitet: Moe_2015 am 6 Jun. 2016
I think I figured out what he wants and commented on my answer. He wants the elements in the first column of A to each be multiplied by the first row in v (just one element) and be summed together. Then the elements in the second column of A to each be multiplied by the second row in v and be summed together. These two results then should be assembled in a column vector. I think when he says it "fails" he means it does not do what he wants (even though what he wants is not correct matrix multiplication and MATLAB is not failing).
Stephen23 am 7 Jun. 2016
Bearbeitet: Stephen23 am 7 Jun. 2016
"doing the correct calculation" is an interesting definition of "fail": MATLAB correctly returns the mathematically universal matrix product. How is this very basic mathematical operation either "erratic" or a "fail" ?

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Moe_2015 am 6 Jun. 2016
Bearbeitet: Moe_2015 am 6 Jun. 2016
That is not how matrix multiplication works at all. For a matrix and a vector:
A= [1 2 v= [1
3 4] -1]
A*v= [1*1+2*(-1)
3*1+4*(-1)]
MATLAB does not fail. It does it correctly.
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Timothy Goldsmith am 7 Jun. 2016
I like your solution, as I learned a couple of things new. I have also resolved the "erratic" behavior I encountered. The matrix I was interested in -- [3,2;-2,-2] multiplied by [1;-1] just happens to give the same result as your solution to my problem as does the function I wrote. Too much sameness in the second column. Many thanks for your help!
Stephen23 am 7 Jun. 2016
@Timothy Goldsmith: please give us complete examples of when "I find that A*v sometimes gives me the answer I want, and sometimes it gives the 'correct' answer". We need to see complete working examples of this, i.e. the input and output matrices, and the actual and expected output matrices.

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