Any one can help me to check this code ?
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A = [ 1 1 5; 2 1 6; 3 7 5 ];
B = ones(3,3)
S = eye(3,3)
O = [2 7 6; 3 7 0]
for j = 1: 1: 3
R = A(1:end,j) % get 3 column vectors. therefore i will have 3 (R)'s
C = B * R ; % 3x1 vector
D = [B*S ; O,B*C]
H = [zeros(3,3); S]
end
I have doubt about the value of C, does it change with j in the loop?
by the way C is a matrix, i can not write C(j) because i want to use the whole matrix (3x3) not just some specific values inside C.
If i need to change something, please let me know.
Thanks
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Accepted Answer
Azzi Abdelmalek
on 25 May 2013
Edited: Azzi Abdelmalek
on 25 May 2013
There is a small mistake, it should be O' instead of O,
A = [ 1 1 5; 2 1 6; 3 7 5 ];
B = ones(3,3)
S = eye(3,3)
O = [2 7 6; 3 7 0]
for j = 1: 3
R = A(1:end,j) % get 3 column vectors. therefore i will have 3 (R)'s
C = B * R ; % 3x1 vector
D = [B*S ; O' B*C]
H = [zeros(3,3); S]
end
3 Comments
Azzi Abdelmalek
on 25 May 2013
Yes C is changing every iteration, if you need to store each value of C , you can add in the loop
C1{j}=C
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