Asked by Brwa
on 25 May 2013

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

Answer by Azzi Abdelmalek
on 25 May 2013

Edited by Azzi Abdelmalek
on 25 May 2013

Accepted answer

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

Brwa
on 25 May 2013

yes, you are right about that, but my question is does loop can change the variavle C inside D for each value of j ? Or i need to write some thing like this C(j) or C(:,:,j) or any thing else ?

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

Brwa
on 26 May 2013

Thank you

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