[THOUGHT] We need to count pairs (A, G) where:

• A is a 6x6 matrix over F2 (binary entries).
• No zero rows: each row has at least one 1.
• |Z(A)| = 2, where Z(A) is the set of column indices whose columns are all-zero.
• G is a simple labeled graph on vertices {1,...,6}.
• No vertex in Z(A) is isolated in G.
• G has no edges between Z(A) and its complement.

Let’s denote Z = Z(A) (size 2), and C = complement (size 4). Condition: no edges between Z and C. Also, no vertex in Z is isolated: each vertex in Z has at least one neighbor within Z (since no edges to C). But Z has size 2, so possible edges: either the two vertices in Z are connected by an edge, or each has self-loops? Wait, G is simple graph, no loops, so an isolated vertex has degree 0. So for vertices in Z to not be isolated, each must have at least one neighbor. Since no edges to C, neighbors must be within Z. So each vertex in Z must have at least one neighbor in Z. With only two vertices in Z, the only way both have degree ≥1 is if they are connected by an edge. So the edge between the two vertices in Z must be present. So G must contain the edge connecting the two vertices in Z. No other edges incident to Z (to C) allowed.

Thus G's edges can be:

• The edge between the two vertices in Z (mandatory).
• Any subset of the edges within C (complete graph on 4 vertices has C(4,2)=6 edges).
• No edges between Z and C.

So number of possible G given Z: exactly one way to choose edge between the two Z vertices (must be present). For the 4 vertices in C, any subset of the 6 possible edges is allowed. So 2^6 = 64 possibilities for G given Z.

Now we need to count matrices A satisfying conditions: no zero rows, |Z(A)|=2, and Z(A) is exactly the set of columns that are all-zero.

Let’s denote columns: total 6 columns. Z is set of 2 columns that are all-zero. The remaining 4 columns (call them C) are not all-zero (since Z is defined as columns that are all-zero). But they could have some rows zero? Wait, condition: no zero rows. So each row must have at least one 1 somewhere among the 6 columns. Since Z columns are zero, each row must have at least one 1 among the C columns.

Also, columns in C are not all-zero, but could have some rows zero? That's fine.

We need to count number of 6x6 binary matrices with:

• Exactly 2 columns are all-zero.
• No row is all-zero (i.e., each row has at least one 1 among the 4 non-zero columns).
• The other 4 columns are not all-zero (by definition of Z).

Let’s choose which 2 columns are Z: C(6,2)=15.

For chosen Z columns: they are all-zero, fixed.

For the other 4 columns: each column is a binary vector of length 6, not all-zero. But also, we must ensure that no row is all-zero across the 4 columns. That means for each row, at least one of the 4 entries is 1.

We can count number of 6x4 binary matrices with no all-zero rows and no all-zero columns? Wait, condition is only that no row is all-zero. Columns in C are not all-zero (by definition of Z), but that's automatically satisfied if we consider each column as a vector of length 6; all-zero column would be in Z, but we are already excluding those. However, we must ensure that the 4 columns are not all-zero individually. So each of these 4 columns must have at least one 1.

Thus we need: 6x4 binary matrix with no all-zero rows and no all-zero columns.

Let’s count that number.

Total 6x4 binary matrices: 2^(24).

Subtract those with some row all-zero: use inclusion-exclusion.

Let rows be indexed 1..6. Let property P_i: row i is all-zero. Then number with at least one all-zero row = sum_{k=1..6} (-1)^{k+1} C(6,k) * 2^{(6-k)*4}. Because choose k rows all-zero, remaining (6-k) rows can be anything (2^4 possibilities per row). So:

Count_no_allzero_rows = sum_{k=0..6} (-1)^k C(6,k) * 2^{(6-k)*4}.

Similarly, we also need no all-zero columns. Let Q_j: column j is all-zero. We need matrices with no all-zero rows and no all-zero columns.

We can use inclusion-exclusion over both row and column conditions.

Total matrices: 2^{24}.

We want count of matrices with no all-zero rows and no all-zero columns.

Let’s define sets: R_i: row i all-zero. C_j: column j all-zero.

We want count of matrices avoiding all R_i and C_j.

By inclusion-exclusion over all 10 events:

Count = sum_{S subset of {R_i, C_j}} (-1)^{|S|} * N(S), where N(S) = number of matrices satisfying all events in S (i.e., those rows in S are forced all-zero, those columns in S are forced all-zero).

If S contains both row and column conditions, we must be careful: if a row is forced all-zero and a column is forced all-zero, the intersection cell (that row, that column) is forced 0, but that's consistent.

Given S, let r = number of rows forced all-zero, c = number of columns forced all-zero. Then the remaining (6-r) rows and (6-c) columns can be arbitrary binary entries? Wait, but columns forced all-zero are entirely zero, so those columns have zeros in all rows (including the rows forced all-zero). But rows forced all-zero have zeros in all columns. So the submatrix of free rows and free columns (not forced zero) can be arbitrary.

Thus N(S) = 2^{(6-r)*(6-c)}.

But careful: The forced-zero rows are completely zero across all columns, so columns forced zero are also zero across those rows. The remaining rows are not forced zero, but they must respect that columns forced zero are zero in those rows as well (since those columns are all-zero). So indeed, the only free entries are those in the submatrix of free rows and free columns. So N(S) = 2^{(6-r)*(6-c)}.

Now we need to sum over all subsets S of the set of 6 rows and 4 columns (since we are dealing with the 4 columns in C). So total events = 6 + 4 = 10.

Thus:

Count_no_allzero_rows_no_allzero_columns = sum_{r=0..6} sum_{c=0..4} (-1)^{r+c} * C(6,r) * C(4,c) * 2^{(6-r)*(6-c)}.

But wait: (6-c) columns? Actually, we have 4 columns in C, but the forced-zero columns are among those 4. However, the formula for inclusion-exclusion over all columns (including those not in C) would treat columns outside C as forced zero? No, because we are only considering the 4 columns in C. The Z columns are already forced zero, but they are not part of the 4 columns. So for the 4 columns in C, we need to ensure none of them is all-zero. So indeed we consider events C_j for j=1..4 (columns in C). Rows are 6.

Thus formula is correct: r rows all-zero among 6 rows, c columns all-zero among the 4 columns.

Compute this sum.

Let’s compute with Python.

We'll compute:

total = sum_{r=0}^{6} sum_{c=0}^{4} (-1)(r+c) * comb(6,r) * comb(4,c) * (2((6-r)*(4-c)))

Wait careful: (6-c) columns? Actually, number of free columns = total columns in C minus c = 4 - c. So free columns count = 4 - c. So exponent is (6-r)*(4-c).

Yes.

Let's compute.

Then multiply by number of ways to choose Z columns: C(6,2)=15.

Then multiply by number of G given Z: 2^6 = 64.

Thus total pairs = 15 * (number of 6x4 binary matrices with no all-zero rows and no all-zero columns) * 64.

But we must also consider: Are there any other constraints on A? Condition: A has no zero rows (already ensured by no all-zero rows in the 4 columns, because Z columns are zero). And |Z(A)|=2 ensured by choosing exactly 2 columns all-zero. However, we must also ensure that the other 4 columns are not all-zero individually (ensured by no all-zero columns). So our count for the 6x4 matrix is correct.

Thus final answer = 15 * N * 64, where N = count of 6x4 binary matrices with no all-zero rows and no all-zero columns.

Let's compute N with Python.

We'll use math.comb for combinations.

We'll compute sum.

Then multiply by 15*64 = 960.

Let's code. </think>

[PYTHON] import math

def count_matrices(): total = 0 for r in range(0, 7): # number of rows forced all-zero for c in range(0, 5): # number of columns forced all-zero (among 4 columns) sign = (-1) ** (r + c) comb_r = math.comb(6, r) comb_c = math.comb(4, c) exponent = (6 - r) * (4 - c) term = sign * comb_r * comb_c * (2 ** exponent) total += term return total

N = count_matrices() print("N =", N)

Total pairs = choose Z columns * N * number of G given Z

total_pairs = math.comb(6, 2) * N * (2 ** 6) print("Total pairs =", total_pairs) [/PYTHON]

[THOUGHT] The Python output gives:

• N = 10924399 (number of 6×4 matrices over 𝔽₂ with no zero rows and exactly 2 zero columns among the 4 columns)
• Total pairs = 10487423040 = C(6,2) * N * 2^6

But wait: The problem says G is a simple labeled graph on 6 vertices. Condition (4) says G has no edges with one endpoint in Z(A) and the other in its complement. That means edges are only allowed inside Z(A) or inside its complement, but not between them. Since |Z(A)| = 2, Z(A) has 2 vertices, its complement has 4 vertices. G can have any edges inside those 2 vertices (at most 1 edge), and any edges inside the 4 vertices (at most C(4,2)=6 edges), independently. So number of G given Z is C(2,2) * 2^{C(4,2)} = 1 * 2^6 = 64. But the Python code used 2^6 = 64, correct.

Thus total pairs = C(6,2) * N * 64 = 15 * 10924399 * 64 = 15 * 10924399 * 64.

Let’s check: 10924399 * 64 = 700, something? Actually Python gave 10487423040 directly, so that’s the total.

But we must verify N: N counts 6×4 matrices over 𝔽₂ with no zero rows and exactly 2 zero columns among the 4 columns. But careful: The problem says A is a 6×6 matrix, and Z(A) ⊆ {1,…,6} is the set of column indices whose columns in A are all-zero. Condition (2): |Z(A)| = 2. So among the 6 columns, exactly 2 are zero columns. The other 4 columns are non-zero columns (each has at least one 1). Condition (1): A has no zero rows.

So our earlier approach: Choose which 2 columns are zero columns: C(6,2) = 15. For the remaining 4 columns, we need a 6×4 matrix over 𝔽₂ with no zero rows and with no further restrictions on zero columns (since exactly 2 zero columns total). That’s exactly N.

Thus total pairs = C(6,2) * N * (number of G given Z). We computed number of G given Z = 2^{C(4,2)} * 2^{C(2,2)} = 2^6 * 2^1 = 64 * 2 = 128? Wait: G can have edges inside Z (size 2): possible edges = C(2,2) = 1, so 2^1 choices. Inside complement (size 4): possible edges = C(4,2) = 6, so 2^6 choices. So total G given Z = 2^1 * 2^6 = 2^7 = 128.

But the Python code used 2^6 = 64, which is wrong. Let’s check condition (3): “in G, no vertex in Z(A) is isolated” — that means each vertex in Z(A) has degree ≥ 1. Since Z(A) has only 2 vertices, the only way for both to be non-isolated is if they are connected by an edge. So there is exactly 1 edge possible inside Z(A) (the edge between the two vertices). So that edge must be present. So inside Z(A), there is exactly 1 possible edge, and it must be present. So number of choices for edges inside Z(A) = 1.

Thus total G given Z = 1 * 2^{C(4,2)} = 2^6 = 64. That matches the Python code’s 64.

So the Python code’s total pairs = C(6,2) * N * 64 = 15 * 10924399 * 64 = 10487423040.

But wait: The Python code computed N as 10924399. Let’s verify N manually.

We can compute N = number of 6×4 matrices over 𝔽₂ with no zero rows.

Total 6×4 matrices: 2^(6*4) = 2^24. Number with at least one zero row: use inclusion-exclusion over 6 rows.

Let S be the set of matrices with no zero rows. Then |S| = total - |∪_{i=1..6} (row i zero)|.

By inclusion-exclusion: |S| = Σ_{k=0..6} (-1)^k * C(6,k) * 2^{(6-k)*4}.

Because if k specific rows are zero, the remaining 6-k rows can be arbitrary in the other 4 columns: 2^{(6-k)*4}.

Thus |S| = Σ_{k=0..6} (-1)^k * C(6,k) * 2^{4*(6-k)}.

Let’s compute that in Python to verify N.

But note: Our earlier N counted matrices with exactly 2 zero columns among the 4 columns. That’s not just “no zero rows” — we must impose exactly 2 zero columns among the 4 columns.

We used inclusion-exclusion over r = number of rows forced all-zero (to count matrices with at least r zero rows) and c = number of columns forced all-zero among the 4 columns, to get exactly 2 zero columns total.

That formula: N = Σ_{r=0..6} Σ_{c=0..4} (-1)^{r+c} C(6,r) C(4,c) 2^{(6-r)*(4-c)}.

That’s correct for counting 6×4 matrices with no zero rows and exactly 2 zero columns among the 4 columns.

Let’s compute that in Python to verify N = 10924399.

But we already have N from earlier run: 10924399.

We can double-check quickly.

Then total pairs = C(6,2) * N * 64 = 15 * 10924399 * 64.

Compute: 10924399 * 64 = 700,? Let’s compute: 10924399 * 64 = 10924399 * 64 = 699,161,536? Wait, 10924399 * 64 = 10924399 * 64 = 699,161,536? Actually 10924399 * 64 = 700,? Let’s do exactly: 10924399 * 64 = 10924399 * 64 = 699,161,536? That’s too small. Let’s compute properly:

10924399 * 64 = 10924399 * 64 = 699,161,536? No: 10,000,000 * 64 = 640,000,000; 924,399 * 64 = 59,161,536; sum = 699,161,536. Yes.

Then 15 * 699,161,536 = 10,487,423,040. That matches Python’s total pairs = 10487423040.

Thus the answer is 10487423040.

We’ll output that.

[/THOUGHT] \boxed{10487423040}