[THOUGHT] We need to count pairs (A, G) where A is a 6×6 matrix over F2, G is a simple labeled graph on vertices {1,...,6}, satisfying:

1. A has no zero rows.
2. |Z(A)| = 2, where Z(A) is the set of column indices whose columns are all-zero.
3. In G, no vertex in Z(A) is isolated.
4. G has no edges with one endpoint in Z(A) and the other in its complement.

Interpretation: Z(A) is a 2-element subset of {1,...,6} (choose which columns are all-zero). For the remaining 4 columns (non-zero columns), they must be non-zero (since Z(A) are exactly the all-zero columns). Also, A has no zero rows, so each of the 6 rows must have at least one 1 in the non-zero columns (since zero columns contribute nothing). But careful: The condition "no zero rows" means for each row, the row vector is not all-zero. Since columns in Z(A) are zero, the row must have at least one 1 in the columns outside Z(A). So each row must have at least one 1 among the 4 non-zero columns.

Now, A is over F2. For a fixed Z(A) (size 2), we have 4 columns that are not all-zero. Each of those 4 columns can be any non-zero column vector of length 6 over F2 (since they are not all-zero). But also we must ensure that overall, the matrix has no zero rows. So we need to count 6×4 matrices over F2 with no zero rows and no all-zero columns. Wait: Actually, the 4 columns are not all-zero, but they could be any non-zero column vector. However, they are independent choices? Not exactly: The condition "no zero rows" is a global condition across the 4 columns.

Let’s formalize: Let S = Z(A) (size 2). Let T = complement of S, size 4. Columns in S are all-zero. Columns in T are arbitrary but not all-zero. Additionally, the whole matrix A must have no zero rows. Since columns in S are zero, the row sum over T must be at least 1 for each row (i.e., each row must have at least one 1 in the columns in T). So we need to count the number of 6×4 matrices over F2 with:

• Each column is non-zero (since columns in T are not all-zero).
• Each row has at least one 1.

But careful: The condition "columns in T are not all-zero" is exactly that each column is non-zero. So we need to count 6×4 binary matrices with no zero columns and no zero rows.

That's a known combinatorial count: number of 6×4 binary matrices with no all-zero rows and no all-zero columns. Let’s compute.

Total 6×4 binary matrices: 2^(6*4) = 2^24. But we need to exclude matrices with at least one zero row or at least one zero column. Use inclusion-exclusion.

Let rows be r=6, columns be c=4. Let A_i be the set of matrices where row i is zero. Let B_j be the set where column j is zero.

We want matrices with no zero rows and no zero columns: N = total - |∪A_i| - |∪B_j| + |∪A_i ∩ ∪B_j|? Actually inclusion-exclusion over both row and column conditions.

Better: Use inclusion-exclusion over rows and columns together: Let U be all matrices. Let R be the set of matrices with at least one zero row. Let C be the set with at least one zero column. We want N = |U| - |R ∪ C| = |U| - |R| - |C| + |R ∩ C|.

Compute |R|: matrices with at least one zero row. Use inclusion-exclusion over rows: |R| = sum_{k=1}^6 (-1)^(k-1) * C(6,k) * 2^( (6-k)*4 )? Wait: If we fix a set of k rows to be zero, then those k rows are forced to be zero, the remaining 6-k rows can be anything (each of the 4 columns). So number = 2^((6-k)*4). So by inclusion-exclusion: |R| = ∑_{k=1}^6 (-1)^(k-1) C(6,k) 2^((6-k)*4).

Similarly, |C|: matrices with at least one zero column. Fix set of j columns to be zero: then those columns are zero, remaining 4-j columns can be anything (each of 6 rows). So number = 2^(6*(4-j)). So |C| = ∑_{j=1}^4 (-1)^(j-1) C(4,j) 2^(6*(4-j)).

Now |R ∩ C|: matrices with at least one zero row AND at least one zero column. Use inclusion-exclusion over both? Actually we can compute directly: number of matrices with some rows zero and some columns zero. Use double inclusion-exclusion: Let I be set of rows forced zero, J set of columns forced zero. Then rows in I are zero, columns in J are zero, the remaining entries (rows not in I, columns not in J) are free. So for fixed I of size a, J of size b, number = 2^((6-a)*(4-b)). Then by inclusion-exclusion over rows and columns separately:

|R ∩ C| = ∑{a=1}^6 ∑{b=1}^4 (-1)^(a+b-2) C(6,a) C(4,b) 2^((6-a)(4-b)). Wait careful: Actually we want matrices with at least one zero row and at least one zero column. The inclusion-exclusion formula for intersection of two properties: |R ∩ C| = total - |no zero rows| - |no zero columns| + |no zero rows and no zero columns|. But we already have formula for |no zero rows and no zero columns| = N. That's circular. Better to compute directly via inclusion-exclusion over rows and columns together: The number of matrices with no zero rows and no zero columns is N = ∑{a=0}^6 ∑{b=0}^4 (-1)^(a+b) C(6,a) C(4,b) 2^((6-a)(4-b)). That's a known formula. Let's verify: For a=0,b=0 we get +2^(24). For a=1,b=0 we subtract C(6,1)2^(54)=6*2^20, etc. So we can compute N directly using that double sum.

Thus N = ∑{a=0}^6 ∑{b=0}^4 (-1)^(a+b) C(6,a) C(4,b) 2^((6-a)*(4-b)).

We'll compute that in Python.

Now, that's for a fixed Z(A) of size 2. But Z(A) can be any 2-subset of {1,...,6}. Number of choices for Z(A): C(6,2) = 15.

So total number of matrices A satisfying conditions 1 and 2: 15 * N.

Now condition 3 and 4 involve graph G. G is a simple labeled graph on vertices {1,...,6}. Conditions depend on Z(A). For a given Z(A) (size 2), condition 3: no vertex in Z(A) is isolated. Condition 4: no edges between Z(A) and its complement. That means edges within Z(A) are allowed, edges within complement are allowed, but no edges across. So G is essentially the disjoint union of two subgraphs: one on vertex set Z(A) (size 2) and one on vertex set complement (size 4), with no edges between them. Additionally, in the subgraph on Z(A), no vertex is isolated. Since Z(A) has exactly 2 vertices, the only way to have no isolated vertices is that the two vertices must be adjacent (since with 2 vertices, if they are not adjacent, each is isolated). So the subgraph on Z(A) must be exactly the edge connecting them. That's 1 possibility.

Now the subgraph on the complement (size 4) can be any simple graph (no restrictions). Number of simple graphs on 4 labeled vertices: 2^{C(4,2)} = 2^6 = 64.

Thus for a given Z(A), the number of graphs G satisfying conditions 3 and 4 is: 1 * 64 = 64.

Therefore, for each Z(A), number of pairs (A,G) = N * 64.

Total pairs = 15 * N * 64 = 960 * N.

We need to compute N and then total.

Let's compute with Python. [/THOUGHT] [PYTHON] import math

def nCr(n, r): if r < 0 or r > n: return 0 return math.comb(n, r)

Compute N = number of 6x4 binary matrices with no zero rows and no zero columns

N = 0 for a in range(0, 7): for b in range(0, 5): term = ((-1)(a+b)) * nCr(6, a) * nCr(4, b) * (2((6-a)*(4-b))) N += term print("N =", N)

Total pairs

total = 15 * N * 64 print("Total pairs =", total) [/PYTHON]

[THOUGHT] The Python computation gave N = 10924399 and total pairs = 10487423040. This is the count of pairs (A, G) satisfying all conditions. The answer is an integer, so we output it as is. [/THOUGHT] \boxed{10487423040}