Problem with matrix manipulation

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Assume user input data as below. I define my matrix is cost. The matrix i created is 3 by 3 matrix. So the matrix should form like this:
cost = [c11 c12 c13
c21 c22 c23
c31 c32 c33]
Since I want to display set of row, I do it like this:
c1 = cost(1,:); % it will become c1 = c11 c12 c13
c2 = cost(2,:); % it will become c2 = c21 c22 c23
c3 = cost(3,:); % it will become c3 = c31 c32 c33
Then I want the value in the matrix. I do it like this.
c11 = cost(1,1);
c12 = cost(1,2);
c13 = cost(1,3);
c21 = cost(2,1);
c22 = cost(2,2);
c23 = cost(2,3);
c31 = cost(3,1);
c32 = cost(3,2);
c33 = cost(3,3);
So this is the equation that I want to use for this type of matrix.
lambda = ((8*c13*c23*c33*Pdt)+(4*c12*c23*c33)+(4*c13*c22*c33)+(4*c13*c23*c32)) ./ (4*c23*c33)+(4*c13*c33)+(4*c13*c23));
So my problem is, if I want to make 4 by 3 matrix, and it would generate a matrix like this:
cost = [c11 c12 c13
c21 c22 c23
c31 c32 c33
c41 c42 c43]
The equation that I want to use for this matrix(4 by 3) is quite different. How can I do that? Do I need to use if else statement? Or do while? Can anyone help me solve this? Can anyone create the code?

Accepted Answer

Rob Graessle
Rob Graessle on 10 Feb 2011
You can use the SIZE function to check how many rows are in the cost matrix, then use IF/ELSE to change the equation used for lambda.
if size(cost, 1) == 3
% Use lambda equation for 3 rows
lambda = ((8*cost(1,3)*cost(2,3)*cost(3,3)*Pdt)+ (4*cost(1,2)*cost(2,3)*cost(3,3)+(4*cost(1,3)*cost(2,2)*cost(3,3)+(4*cost(1,3)*cost(2,3)*cost(3,2)) ./ (4*cost(2,3)*cost(3,3)+(4*cost(1,3)*cost(3,3)+(4*cost(1,3)*cost(2,3));
elseif size(cost, 1) == 4
% Use lambda equation for 4 rows
lambda = ...;
end
Also note that you don't need to create new variables c11, c12, etc. - it's not very memory-efficient.

More Answers (2)

Walter Roberson
Walter Roberson on 10 Feb 2011
It is not recommended that you use this approach at all. Please see the FAQ for alternatives.

slumberk
slumberk on 10 Feb 2011
@rob
what is the meaning of size(cost,1) == 3?? Does it have 3 row n column??
@walter
thx for the info =)
  2 Comments
slumberk
slumberk on 11 Feb 2011
@rob
thx rob =)

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