How to get back the original matrix?

A=[1 2 3 2;4 1 2 3;3 4 3 2;2 4 1 1];

>> R=[1 3 4 2];

>> B=A(R,:);

>> C=B(:)';

>> D=[1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 1];

>> E=C(D==1);

Can we get back A?

[KSSV]

Already you have A...

[Ammy]

Yes But by the reverse process?

Answer 1

[Torsten]

A = [1 2 3 2;...
    4 1 2 3;...
    3 4 3 2;...
    2 4 1 1]
R = [1 3 4 2];
B = A(R,:);
Rinv(R) = 1:numel(R);
A_recovered = B(Rinv,:)

Answer 2

[Chunru]

A=[1 2 3 2;4 1 2 3;3 4 3 2;2 4 1 1];
% Reordering the rows of A is reversible
R=[1 3 4 2];
B=A(R,:);

% Straightening up the matrix into vector is also reversible
C=B(:)';

% Picking up part of the data is not reversible
D=[1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 1];
E=C(D==1);

[Ammy]

@Chunru thank you

How the following can be

reversible ,

A=[1 2 3 2;4 1 2 3;3 4 3 2;2 4 1 1];

% Reordering the rows of A is reversible

R=[1 3 4 2];

B=A(R,:);