How to multiply a scalar with matrix within loop and for each iteration store the values in new matrix?

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Rabia Zulfiqar
Rabia Zulfiqar 2020-5-18
评论： Rabia Zulfiqar 2020-5-19
I have a matrix of 24x365 but just to explain what I want to do I am taking here an example matrix of 2x2.
A=[1 2
3 4]
b=0.1
n=3 (no of years)
The new matrix for each year must be formed in a way that for first year it should be simply C(1)=A but for second year the scalar b is multiplied with the C so C(2)=b*C(1).For third year it should be C(3)=b*C(2) and so on...
The answer matrix should be like
C(1)=[1 2
3 4]
C(2)= [0.1 0.2
0.3 0.4]
C(3)=[0.01 0.02
0.03 0.04]
I have formed this code but I am not getting the desired results.
clc
clear all
A=[1 2;
3 4];
b = 0.1
n=3;
c=b*a;
d=c;
for i=1:n
c(1)=d
c(i+1)=b*c(i);
end
The error appears as "Unable to perform assignment because the left and right sides have a different number of elements
Error in Untitled (line 11)
c(1)=d"
Can anyone please help me???

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采纳的回答

Matt J
Matt J 2020-5-18
编辑：Matt J 2020-5-18
There is no need for a loop:
C=A.*reshape(b.^(0:n-1),1,1,[]);

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