## Duplicate Each Element in a Matrix without using Repelem or Repmat

### Benjamin Ong (view profile)

on 31 Oct 2019
Latest activity Commented on by Benjamin Ong

on 2 Nov 2019

### Fangjun Jiang (view profile)

Is there a way I can make each element in a matrix duplicate into a 3 by 3? if
A=[1 0;0 1]
Is there a way i can turn it into
Aew=[1 1 1 0 0 0;1 1 1 0 0 0;1 1 1 0 0 0;0 0 0 1 1 1;0 0 0 1 1 1;0 0 0 1 1 1]
so basically
Anew=repelem(A,3,3)
but without repelem, repmat or any special functions?
Thanks

#### 1 Comment

on 31 Oct 2019
Yes, but the only reason I can see for wanting to do so is as a homework question. Like most things you can do it with a for loop.

### Fangjun Jiang (view profile)

on 31 Oct 2019
Edited by Fangjun Jiang

### Fangjun Jiang (view profile)

on 31 Oct 2019

a=eye(2);
k=3;
d=blkdiag(ones(k,1),ones(k,1));
b=d*a*d'
%%Or a more generic case
a=[1 2 3;4 5 6];
k=3;
[m,n]=size(a);
d1=ones(k,1)*ones(1,m);
d1=mat2cell(d1,k,ones(1,m));
d1=blkdiag(d1{:});
d2=ones(n,1)*ones(1,k);
d2=mat2cell(d2,ones(n,1),k);
d2=blkdiag(d2{:});
b=d1*a*d2

Benjamin Ong

### Benjamin Ong (view profile)

on 1 Nov 2019
Thanks this helped alot :D

### Stephen Cobeldick (view profile)

on 31 Oct 2019
Edited by Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 31 Oct 2019

>> A = [1,0;0,1]
A =
1 0
0 1
>> V = ceil((1:6)/3);
>> B = A(V,V)
B =
1 1 1 0 0 0
1 1 1 0 0 0
1 1 1 0 0 0
0 0 0 1 1 1
0 0 0 1 1 1
0 0 0 1 1 1

Benjamin Ong

### Benjamin Ong (view profile)

on 1 Nov 2019
This way only seems to produce a 2 by 2 everytime though no matter the size of the initial matrix
Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 2 Nov 2019
"This way only seems to produce a 2 by 2 everytime though no matter the size of the initial matrix"
Clearly it works, as the example in my answer shows.
Of course it can be trivially adjusted to replicate the elements of any sized array any number of times, to do so requires two very small changes to the V definition:
k = 3;
V = ceil((1:k*size(A,1))/k);
This is certainly a much simpler and more efficient solution than the one that you accepted.
Benjamin Ong

### Benjamin Ong (view profile)

on 2 Nov 2019
Ahh ya I did realise your method is simpler but i was having big brainfart at the moment and the solution i accepted I seem to understand how it works better? But thanks for the help anyways :D