Need to Produce a different size matrix on each iteration

Good Morning,

I need your urgent help. I have a presentation on this in a few hours. I am trying to create one matrix concatenated to the previous per loop.

the concept is that the abc first matrix is [1 51]. the second matrix is [1 51;2 52]; the third matrix is [1 51;2 52;3 53]; and so on.

wqe=[1;2;3;4;5;6;7;8;9;10;11;12;13;14;15];
kli=[51;52;53;54;55;56;57;58;59;60;61;62;63;64;65];
hsd=size(kli,1);
kas=zeros(hsd,1);
sak=zeros(hsd,1);
abc=repmat({zeros(hsd,1)},hsd,1);
for i=1:hsd
    kas(i)=wqe(i);   
    sak(i)=kli(i);
end
for h=2:1:hsd    
    abc{i}=[kas(i-1) sak(i-1);kas(i) sak(i)];
end

Answer 1

[Alan Stevens]

Try the following

wqe=[1;2;3;4;5;6;7;8;9;10;11;12;13;14;15];
kli=[51;52;53;54;55;56;57;58;59;60;61;62;63;64;65];
hsd=size(kli,1);
kas=zeros(hsd,1);
sak=zeros(hsd,1);
abc=zeros(hsd,2);
for i=1:hsd
    kas(i)=wqe(i);   
    sak(i)=kli(i);
end
abc(1,:) = [kas(1) sak(1)];
for h=2:1:hsd    
    abc(h,:)=[kas(h) sak(h)];
end
disp(abc)

     1    51
     2    52
     3    53
     4    54
     5    55
     6    56
     7    57
     8    58
     9    59
    10    60
    11    61
    12    62
    13    63
    14    64
    15    65

[DARLINGTON ETAJE]

Thanks for the attempt Alan. The final answer has to have 15 third dimensions...meaning 15 matrices.

the abc first matrix is [1 51]. the second matrix is [1 51;2 52]; the third matrix is [1 51;2 52;3 53]; and so on.

[Alan Stevens]

More like this?

wqe=[1;2;3;4;5;6;7;8;9;10;11;12;13;14;15];
kli=[51;52;53;54;55;56;57;58;59;60;61;62;63;64;65];
hsd=size(kli,1);
kas=zeros(hsd,1);
sak=zeros(hsd,1);
abc=zeros(hsd,2);
for i=1:hsd
    kas(i)=wqe(i);   
    sak(i)=kli(i);
end
abc(1,:) = [kas(1) sak(1)];
for h=2:1:hsd    
    abc(h,:)=[kas(h) sak(h)];
end
matrix{1}=abc(1,:);
for i = 2:hsd
    matrix{i}=abc(1:i,:);
end

[DARLINGTON ETAJE]

this response is correct. Thank you so much.