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Please help me to solve the issue for the following code

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Prabha Kumaresan
Prabha Kumaresan am 26 Jan. 2018
Kommentiert: Prabha Kumaresan am 29 Jan. 2018
I have asked this question already but I couldnt get the answers what i actually need.
A=[0.01 0.02 0.03 0 0 0 0.02;
0.02 0.03 0.04 0 0 0 0.03;
0.03 0.02 0.02 0 0 0 0.04]
s=sum(sum(A))
if i run the code it gives the value 0.3100.
But how to replace the numbers such that the sum of A should be 0.35. could anyone help me on this
  1 Kommentar
Rik
Rik am 26 Jan. 2018
If you don't get an answer to work, you should comment there, NOT open a new question.

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Akzeptierte Antwort

Rik
Rik am 26 Jan. 2018
You have a working answer here, you just didn't run the code correctly.
A=[0.01 0.02 0.03 0 0 0 0.02;
0.02 0.03 0.04 0 0 0 0.03;
0.03 0.02 0.02 0 0 0 0.04];
A = 0.35*A/sum(A(:))
s=sum(A(:))
  2 Kommentare
Prabha Kumaresan
Prabha Kumaresan am 29 Jan. 2018
Bearbeitet: Walter Roberson am 29 Jan. 2018
the command line A = 0.35*A/sum(A(:)) should not be included. On omitting the line s=0.31.
Could you please tell me how to change or replace the numbers in A such that the value of s=0.35.
Walter Roberson
Walter Roberson am 29 Jan. 2018
Why should that line not be included? If it gets you the answer you want, then why not use it?
If you want to generate A so that it already has the right sum, then use Roger's randfixedsum() from the File Exchange.

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Weitere Antworten (1)

Prabha Kumaresan
Prabha Kumaresan am 29 Jan. 2018
if i use the command line it gives the result what i need but the addition of the numbers of the matrix was not changed.
  2 Kommentare
Walter Roberson
Walter Roberson am 29 Jan. 2018
Bearbeitet: Walter Roberson am 29 Jan. 2018
I do not understand that remark.
Perhaps you need
new_A = 0.35*A./sum(A(:));
additions_to_A = new_A - A;
and now the additions of numbers to A that you need to reach the desired total is in the matrix additions_to_A, and you can get the desired matrix by using
desired_matrix = A + additions_to_A;
Prabha Kumaresan
Prabha Kumaresan am 29 Jan. 2018
Thanks a lot.It works.

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