[THOUGHT] Let me break down this problem:

1. $A$ is a $6 \times 6$ matrix over $\mathbb{F}_2$
2. $Z(A)$ is the set of column indices whose columns are all-zero
3. We need:
   • $A$ has no zero rows
   • $|Z(A)| = 2$ (exactly 2 columns are all-zero)
   • In $G$, no vertex in $Z(A)$ is isolated
   • $G$ has no edges between $Z(A)$ and its complement

Let me count the pairs $(A, G)$:

Counting matrices $A$:

• Choose which 2 columns are all-zero: $\binom{6}{2} = 15$ ways
• The remaining 4 columns must be non-zero columns (each column has at least one 1)
• Each row must be non-zero

For the 4 non-zero columns: each column can be any of $2^6 - 1 = 63$ non-zero vectors. But we also need each row to be non-zero.

Actually, let me think more carefully. If columns $i, j$ are the zero columns, then:

• The 4 remaining columns (call them the "active" columns) must each be non-zero
• Each row, when restricted to these 4 active columns, must be non-zero (since the other 2 columns are zero)

So we need to count $6 \times 4$ matrices over $\mathbb{F}_2$ where:

• No row is all-zero
• No column is all-zero

This is an inclusion-exclusion problem. Let me count matrices with no zero rows and no zero columns.

Counting graphs $G$: Given $Z(A) = \{i, j\}$:

• No edges between $Z(A)$ and its complement
• No vertex in $Z(A)$ is isolated (so there must be an edge between $i$ and $j$)

So the graph decomposes into:

• A graph on $Z(A) = \{i, j\}$ where the edge $(i,j)$ must exist (1 way)
• A graph on the complement $\{1,...,6\} \setminus Z(A)$ (4 vertices), any graph is allowed: $2^{\binom{4}{2}} = 2^6 = 64$ ways

Let me compute this with Python. [/THOUGHT] [PYTHON] from itertools import combinations import numpy as np

Count 6x4 matrices over F_2 with no zero rows and no zero columns

Using inclusion-exclusion

def count_matrices_no_zero_rows_cols(n_rows, n_cols): """Count n_rows x n_cols matrices over F_2 with no zero rows and no zero columns""" # Total matrices: 2^(n_rows * n_cols) # Use inclusion-exclusion on rows and columns

total = 0
for num_zero_rows in range(n_rows + 1):
    for num_zero_cols in range(n_cols + 1):
        # Choose which rows are forced to be zero
        # Choose which columns are forced to be zero
        # The remaining (n_rows - num_zero_rows) x (n_cols - num_zero_cols) submatrix is free
        sign = (-1) ** (num_zero_rows + num_zero_cols)
        ways_choose_rows = int(np.math.comb(n_rows, num_zero_rows))
        ways_choose_cols = int(np.math.comb(n_cols, num_zero_cols))
        remaining_entries = (n_rows - num_zero_rows) * (n_cols - num_zero_cols)
        total += sign * ways_choose_rows * ways_choose_cols * (2 ** remaining_entries)

return total

For our problem: 6 rows, 4 columns (the non-zero columns)

num_matrices_per_choice = count_matrices_no_zero_rows_cols(6, 4) print(f"Number of 6x4 matrices with no zero rows and no zero columns: {num_matrices_per_choice}")

Number of ways to choose which 2 columns are zero

num_column_choices = int(np.math.comb(6, 2)) print(f"Number of ways to choose Z(A): {num_column_choices}")

Total number of valid matrices A

total_matrices = num_column_choices * num_matrices_per_choice print(f"Total valid matrices A: {total_matrices}")

For graphs G:

- Edge between the 2 vertices in Z(A) must exist: 1 way

- Any graph on the 4 vertices in complement: 2^C(4,2) = 2^6 = 64 ways

num_graphs_per_ZA = 1 * (2 ** 6) print(f"Number of valid graphs G for each Z(A): {num_graphs_per_ZA}")

But wait - the graph G is on the full vertex set, and Z(A) depends on A

For each matrix A with a specific Z(A), we count graphs G satisfying the conditions

Total pairs (A, G)

total_pairs = total_matrices * num_graphs_per_ZA print(f"Total pairs (A, G): {total_pairs}") [/PYTHON]

[THOUGHT] Let me fix the code - I need to use math.comb instead of np.math.comb. [/THOUGHT] [PYTHON] from itertools import combinations from math import comb

Count 6x4 matrices over F_2 with no zero rows and no zero columns

Using inclusion-exclusion

def count_matrices_no_zero_rows_cols(n_rows, n_cols): """Count n_rows x n_cols matrices over F_2 with no zero rows and no zero columns""" total = 0 for num_zero_rows in range(n_rows + 1): for num_zero_cols in range(n_cols + 1): sign = (-1) ** (num_zero_rows + num_zero_cols) ways_choose_rows = comb(n_rows, num_zero_rows) ways_choose_cols = comb(n_cols, num_zero_cols) remaining_entries = (n_rows - num_zero_rows) * (n_cols - num_zero_cols) total += sign * ways_choose_rows * ways_choose_cols * (2 ** remaining_entries)

return total

For our problem: 6 rows, 4 columns (the non-zero columns)

num_matrices_per_choice = count_matrices_no_zero_rows_cols(6, 4) print(f"Number of 6x4 matrices with no zero rows and no zero columns: {num_matrices_per_choice}")

Number of ways to choose which 2 columns are zero

num_column_choices = comb(6, 2) print(f"Number of ways to choose Z(A): {num_column_choices}")

Total number of valid matrices A

total_matrices = num_column_choices * num_matrices_per_choice print(f"Total valid matrices A: {total_matrices}")

For graphs G:

- Edge between the 2 vertices in Z(A) must exist: 1 way

- Any graph on the 4 vertices in complement: 2^C(4,2) = 2^6 = 64 ways

num_graphs_per_ZA = 1 * (2 ** 6) print(f"Number of valid graphs G for each Z(A): {num_graphs_per_ZA}")

Total pairs (A, G)

total_pairs = total_matrices * num_graphs_per_ZA print(f"Total pairs (A, G): {total_pairs}") [/PYTHON]

[THOUGHT] The calculation shows:

• Number of 6×4 matrices over F_2 with no zero rows and no zero columns: 10,924,399
• Number of ways to choose which 2 columns are zero (Z(A)): 15
• Total valid matrices A: 163,865,985
• Number of valid graphs G for each choice of Z(A): 64 (the edge between the two vertices in Z(A) must exist, and any graph on the 4 complement vertices is allowed)
• Total pairs (A, G): 10,487,423,040 [/THOUGHT]

\boxed{10487423040}