2x2 matrix multiplied by a 2x1 column vector gives erratic results
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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?
3 Commenti
Steven Lord
il 6 Giu 2016
What EXACTLY do you mean by "fails if A=[1,2;3,4]"? Give a complete self-contained example of what you try to do (something we can copy and paste into MATLAB and run), show what MATLAB computes, and explain exactly why you consider that to be an incorrect answer.
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).
Risposte (1)
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.
Please review matrix multiplication:
4 Commenti
Timothy Goldsmith
il 6 Giu 2016
Modificato: Timothy Goldsmith
il 6 Giu 2016
Moe_2015
il 6 Giu 2016
ok, for what you want (which is not matrix multiplication) you could do the following:
A=[1,2;3,4];
v=[1;-1];
result=sum(A,1)'.*v
which gives:
result =
4
-6
Timothy Goldsmith
il 7 Giu 2016
Stephen23
il 7 Giu 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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