I present to you the sandwich compliment of good faith. This is where I give the bad news in between two slices of good news. First, I would like to compliment you on your excellent lateral thinking for seeing a pattern I missed when I first posted this. I see you have taken each column and made it such so that there are at least 4 (0’s OR 1’s). You made a mistake though because you presented (what I assume) is a 5 x 6 matrix when you really should of given a 6 x 6 matrix to make it symmetrical. I fixed this mistake for you below although to give you credit, I never said the matrix had to be symmetrical. Below is the answer, please note the number of 0’s and 1’s on both columns and rows.

1 0 1 0 0 | 0
1 1 0 0 1 | 1
0 1 1 0 1 | 1
0 0 1 0 1 | 0
1 0 0 1 1 | 1
“1 0 1 1 0” | 1

Actually a pretty cool thing is that you can keep generating the matrix to be a maximum of 8 x 8 giving:

1 0 1 0 0 0 1 1
1 1 0 0 1 1 0 0
0 1 1 0 1 1 0 0
0 0 1 0 1 0 1 1
1 0 0 1 1 1 0 0
1 0 1 1 0 1 0 0
0 1 0 1 0 0 1 1
0 1 0 1 0 0 1 1

which means the 7 x 7 should look like this:

1 0 1 0 0 0 1
1 1 0 0 1 1 0
0 1 1 0 1 1 0
0 0 1 0 1 0 1
1 0 0 1 1 1 0
1 0 1 1 0 1 0
0 1 0 1 0 0 1

What about 9 x 9? well you can get any complimentary pair for row 9 and column 9. There are (by my calculations) 2380 possibilities so that at there are only 4 (0’s OR 1’s). For matrices 10 x 10 and up, you can repeat the adjacent term to preserve the 4 (0’s OR 1’s) thus creating a null matrix (example A). A cleaner answer that avoids the probability of combinatorics would be if you evolve the current rule to evolve with the matrix side for 10 x 10. (let there be n (0’s OR 1’s) where n = ((Matrix Dimension) / 2) IF (Matrix Dimension > 8) ELSE let n = 4). This generates the random pattern seen in Matrix C however a more regular pattern in Matrix B is also possible if we have use a double conditional (IF (Matrix Dimension) > 8) IF let there be n (0’s), let there be n (1’s) ELSE (initial condition).

MATRIX A

1 0 1 0 0 0 1 1 1 1 1 1 1 1 1
1 1 0 0 1 1 0 0 1 1 1 1 1 1 1
0 1 1 0 1 1 0 0 1 1 1 1 1 1 1
0 0 1 0 1 0 1 1 1 1 1 1 1 1 1
1 0 0 1 1 1 0 0 0 0 0 0 0 0 0
1 0 1 1 0 1 0 0 0 0 0 0 0 0 0
0 1 0 1 0 0 1 1 0 0 0 0 0 0 0
0 1 0 1 0 0 1 1 0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0
1 1 1 1 0 0 0 0 |0 0 0 0 0 0 0

MATRIX B

1 0 1 0 0 0 1 1 1 0 1 0 1 0 1
1 1 0 0 1 1 0 0 0 1 0 1 0 1 0
0 1 1 0 1 1 0 0 1 0 1 0 1 0 1
0 0 1 0 1 0 1 1 0 1 0 1 0 1 0
1 0 0 1 1 1 0 0 1 0 1 0 1 0 1
1 0 1 1 0 1 0 0 0 1 0 1 0 1 0
0 1 0 1 0 0 1 1 1 0 1 0 1 0 1
0 1 0 1 0 0 1 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

MATRIX C

1 0 1 0 0 0 1 1 1 1 0 0 1 0 1
1 1 0 0 1 1 0 0 0 0 1 1 0 1 0
0 1 1 0 1 1 0 0 1 0 1 0 1 1 0
0 0 1 0 1 0 1 1 0 1 0 1 0 0 1
1 0 0 1 1 1 0 0 1 0 1 0 1 0 1
1 0 1 1 0 1 0 0 0 0 1 1 1 0 0
0 1 0 1 0 0 1 1 1 1 0 0 0 1 1
0 1 0 1 0 0 1 1 0 1 0 1 0 1 0
1 0 0 1 1 0 1 0 1 0 1 0 1 0 1
0 1 1 0 0 0 1 1 0 0 1 1 0 0 1
0 1 1 0 1 1 0 0 1 1 0 0 1 1 0
1 0 0 1 0 1 0 1 0 1 1 0 0 1 0
1 0 1 0 1 0 1 0 1 0 0 1 1 0 1
0 1 1 0 0 1 0 1 0 1 1 0 0 1 0
1 0 0 1 1 0 1 0 1 0 0 1 1 0 1

I believe I have significantly addressed your answer and even provided evidence to support it. I’m sorry if I lost it a little though but that is what Mathematics is all about.

Now for the REALLY good news; THE REAL CHALLENGE! There is another answer out there which has nothing to do with Matrices. Think about it this way, try converting the matrix to one line then figure out what goes in either end. I’ll give you a clue, the number of digits HAS to be a power of 2. So it could be 2, 4, 64, 1024 or whatever but it has to be a power of 2. Can you give me this sequence to say 32 digits then?

Best of luck,
Vainstar