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Matrix problem, iterative matrix calculation

Asked by Light on 6 Jun 2013

Hello my matrix problem?? I tried it in week. But no success!

Here is the matrix

A=[-1,-1,0,0,0;0,0,0,-1,0;0,1,-1,0,-1;0,0,1,0,0;1,0,0,1,1]

Every column has only one 1, one -1 and i have to add them with whole row. Then that column will be 0. It is the first step.. Result must be like that A matrix must be changed A=[-1,-1,0,0,0;0,0,0,-1,0;0,1,0,0,-1;0,0,0,0,0;1,0,0,1,1] 4th row added to the 3rd row. and 4th elements changed 0

Then that operation will be continued until all row changed 0. I think if i write iterate Code, it will be more easier. But failed, some help!

I couldn't write a CODE. How can i write CODE. Please tell me a CODE or structure...

1 Comment

David Sanchez on 6 Jun 2013

your explanation is not clear, could you please rewrite your question in a way it be clearer?

Light

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1 Answer

Answer by Iain on 6 Jun 2013
Edited by Iain on 6 Jun 2013

I think you are asking how to make your entire matrix zero, by adding columns of data that already exist.

 while any(A) & ~some error
  for i = 1:size(A,2) % each column
   col = A(:,i);
   [plus indp] = max(col);
   [nega indn] = min(col);
   if ~(nega => 0 || plus =< 0 )
    A(row_no,:) = A(indp,:) - A(indn,:)*plus/nega;
   end
 end
 some error = some check for infinite loops etc.
end

4 Comments

Iain on 6 Jun 2013
 >=, and <= instead of => and =<, I always get them mixed up. 
Light on 6 Jun 2013

A=[-1,-1,0,0,0;0,0,0,-1,0;0,1,-1,0,-1;0,0,1,0,0;1,0,0,1,1];

while any(A) & ~ some error

for i = 1:size(A,2) % each column
 col = A(:,i);
 [plus indp] = max(col);
 [nega indn] = min(col);
 if ~(nega >= 0 || plus <= 0 )
  A(row_no,:) = A(indp,:) - A(indn,:)*plus/nega;
 end
end
some error = some check for infinite loops etc.
end

Undefined function or variable 'some'.

:-(

Iain on 6 Jun 2013

"some error" is psuedocode - you should use it to prevent infinite loops, or any other errors that might arise.

If your sole aim is to set that matrix equal to zero by the simplest means possible, you can simply:

 A(:) = 0;
Iain

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