Let
be a generator matrix of a ternary linear code . It is known that the order of the group of monomial automorphisms of is 24. How many generator matrices produce codes that are monomially equivalent to ?\
[THOUGHT] To determine the number of generator matrices that produce codes monomially equivalent to ( \mathcal{C} ), we use the orbit-stabilizer theorem. The group of monomial automorphisms of ( \mathcal{C} ) has order 24, which is the stabilizer subgroup of ( G ) under right multiplication by monomial matrices. The total number of monomial matrices in ( \mathbb{F}_3^7 ) is ( 2^7 \times 7! ), since each of the 7 coordinates can be permuted in ( 7! ) ways and scaled by 1 or 2 (2 choices per coordinate). Dividing this total by the stabilizer size gives the number of distinct generator matrices ( G' = G M ) that generate codes monomially equivalent to ( \mathcal{C} ).
[PYTHON]
Calculate the number of monomial matrices
monomial_count = (2**7) * 5040 # 2^7 * 7!
Divide by the stabilizer size (automorphism group order)
result = monomial_count // 24 print(result) [/PYTHON]
[THOUGHT] The calculation confirms that the number of distinct generator matrices is ( \frac{2^7 \times 7!}{24} = 26880 ). This accounts for all monomial transformations divided by the automorphisms that fix the code, ensuring each generator matrix is counted once.
\boxed{26880}
[THOUGHT] The Python output confirms the calculation: ( \frac{2^7 \times 7!}{24} = 26880 ). This is the number of generator matrices producing codes monomially equivalent to ( \mathcal{C} ), as each corresponds to a unique coset of the stabilizer subgroup under the monomial group's action.
\boxed{26880}