## could anyone help me to solve the issue

### jaah navi (view profile)

on 13 Sep 2019
Latest activity Commented on by jaah navi

on 17 Sep 2019

### John D'Errico (view profile)

AA=4
I am having matrix as below
A=[2 3;
1 4;
2 4;
3 4]
with respect to first row out of four i am having two numbers 2 and 3.
so i need to have an another variable B in the following manner
B={2 3} {1 4} for first row
{1 4} {2 3} for second row
{2 4} {1 3} third row
{3 4} {1 2} fourth row

#### 1 Comment

on 13 Sep 2019
Can you please try to give a more relevant title to your questions in general. Almost every question you ask you just give the same generic title. The question box should be for a short summary of what your question is about. Every question on the board could be titled 'Can you please help with this', but it isn't very useful for people scanning the questions for ones they may be able to contribute to.

### John D'Errico (view profile)

on 13 Sep 2019

Easy enough. Just use setdiff.
A=[2 3;
1 4;
2 4;
3 4];
That is, ...
B = A;
cind = [3 4];
for i = 1:4
B(i,cind(randperm(2))) = setdiff(1:4,B(i,1:2));
end
B is:
B
B =
2 3 4 1
1 4 2 3
2 4 1 3
3 4 1 2

jaah navi

### jaah navi (view profile)

on 17 Sep 2019
The above code works for the matrix
A=[2 3;
1 4;
2 4;
3 4];