Creating a loop for matrix multiplication

Hi. I am trying to create a loop to multiply two matricies 40 times such that A*B0=B1 and A*B1=B2 and so on. So far I have got this
%
a=1-0.0277;
b=(1/15)*(1-0.008);
c=0.0215*a;
d=(14/15)*(1-0.008);
A=[[a,b];[c,d]];
B0=[6820;2140];
B=[B0];
for j=1:40
B1=A*B0;
B0=B1;
B=[B;B1];
disp(round(B(j)))
end
plot(B)
It does work but the result comes up as a list of all the values together and I want it to be a in a 1x2 matrix format.
Can anyone help? Thank you

5 comentarios

dpb
dpb el 11 de Mayo de 2014
Editada: dpb el 11 de Mayo de 2014
How can you have a 1x2 matrix--you've got two of them which you've concatenated together vertically with the ';'. You can get a 2x2 if you simply horizontally concatenate as
B=[B B1];
but there are four elements. This is for the first pass -- after that you've got a difficulty in dimensions going forward. It's not at all clear what your end objective is, so not sure how to suggest a correction.
Star Strider
Star Strider el 11 de Mayo de 2014
Editada: Star Strider el 11 de Mayo de 2014
B actually is in an array. Changing that line to:
disp(round(B(:,j)))
will display the new columns of B as they are created.
Else, I agree.
dpb
dpb el 11 de Mayo de 2014
@Star== Whiff... :) Edited the bonehead comment out.
Wojtek
Wojtek el 12 de Mayo de 2014
Editada: Wojtek el 12 de Mayo de 2014
Basically, this loop creates an array B which has 82 entries in one column. What I would like to have is 81 separate arrays created using elements of B.
So for example C1=B(1:2,1), C2=B(3:4,1) and so on.
I have tried to write something like this:
for k=1:41
for a=1:2:81
for b=2:2:82
C(k)=B(a:b,1)
disp(C)
end
end
end
But that doesn't work.
John D'Errico
John D'Errico el 12 de Mayo de 2014
NO NO NO NO. Do not create 81 different named variables. Learn to use arrays of variables. Your code will be the better for it. Your mental health a bit better too, because your code will not drive you crazy.

Iniciar sesión para comentar.

Respuestas (1)

Wojtek
Wojtek el 13 de Mayo de 2014
Problem solved. I have used this
x=length(B);
y=x/2;
C=[];
for i=1:41
C=[C B(2*i-1:2*i)];
end
Thank you all for your comments.

1 comentario

for i=1:41,
C = [C B(2*i-1:2*i)]
end
is exactly the same as
C = B(1:82)

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.

Productos

Preguntada:

el 11 de Mayo de 2014

Comentada:

el 13 de Mayo de 2014

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by