金声玉亮2.0模型评测报告

问题

There is a single choice question about abstract algebra. Answer the question by replying A, B, C or D.
Question: Let G denoted the set of all n x n non-singular matrices with rational numbers as entries. Then under multiplication G is a/an
A. subgroup
B. finite abelian group
C. infinite, non abelian group
D. ininite, abelian
Answer:

答案

模型	模型答案	正确答案
金声玉亮2.0	C. infinite, non abelian group Reasoning: The set of all n x n non-singular matrices with rational numbers as entries forms a group under multiplication, as it satisfies the group properties (closure, associativity, identity, and inverse). However, this group is not abelian, as matrix multiplication is not commutative in general. Additionally, the group is infinite, as there are infinitely many non-singular matrices with rational entries. Therefore, the correct answer is C. infinite, non abelian group.	C
deepseekr1	The set \( G \) consists of all \( n \times n \) non-singular matrices with rational entries, forming the general linear group \( \text{GL}(n, \mathbb{Q}) \) under matrix multiplication. 1. Group Structure: \( G \) is a group because it satisfies closure (product of invertible matrices is invertible), associativity, has an identity matrix, and every element has an inverse (non-singularity ensures invertibility). 2. Finiteness: \( G \) is infinite because there are infinitely many rational numbers, allowing infinitely many distinct matrices. This eliminates options B (finite abelian) and A (subgroup, as \( G \) is a standalone group). 3. Abelian vs. Non-Abelian: - For \( n = 1 \), \( G \) reduces to non-zero rationals under multiplication, which is abelian. However, for \( n \geq 2 \), matrix multiplication is non-commutative (e.g., counterexamples exist with simple \( 2 \times 2 \) matrices). - Since the question does not specify \( n = 1 \), the default assumption in abstract algebra is \( n \geq 2 \), making \( G \) non-abelian. Thus, \( G \) is an infinite, non-abelian group. Answer: C	C
gpto1	C	C
GPT-4o	C. infinite, non abelian group	C
llama3.1-70B-Instruct	C	C
llama3.1-405B-Instruct	C. infinite, non abelian group Reasoning: The set of all n x n non-singular matrices with rational numbers as entries forms a group under multiplication, as it satisfies the group properties (closure, associativity, identity, and inverse). However, this group is not abelian, as matrix multiplication is not commutative in general. Additionally, the group is infinite, as there are infinitely many non-singular matrices with rational entries. Therefore, the correct answer is C. infinite, non abelian group.	C
DeepSeek-Coder-V2	C	C
Qwen2-72B-Instruct	C. infinite, non abelian group	C
baichuan4	C. infinite, non abelian group	C
doubao	C The set of all non-singular matrices with rational entries is an infinite group under multiplication. And in general, it is non-abelian (matrix multiplication is not commutative in general).	C
minimax	C. infinite, non abelian group	C
kimi	C. infinite, non abelian group Explanation: The set G of all n x n non-singular matrices with rational numbers as entries, under multiplication, forms a group. This is because: 1. Closure: The product of two non-singular matrices is again a non-singular matrix. 2. Associativity: Matrix multiplication is associative. 3. Identity: The n x n identity matrix is in G, and multiplying any matrix by the identity matrix leaves it unchanged. 4. Inverses: Every non-singular matrix has an inverse in G, which is also a non-singular matrix with rational entries. Since G is not necessarily commutative (matrix multiplication is not commutative in general), it is non-abelian. Also, G is infinite because there are infinitely many n x n non-singular matrices with rational entries. Therefore, the correct answer is C. Note that the term "abelian" is misspelled in the options as "ininite, abelian" and should be "infinite, abelian".	C
wenxinyiyan	C Explanation: 1. Closure under multiplication: Since the product of two non-singular matrices is again non-singular, and the product of matrices with rational entries has rational entries, G is closed under multiplication. 2. Associativity: Matrix multiplication is associative, so this property holds in G. 3. Identity element: The identity matrix I (with 1s on the diagonal and 0s elsewhere) is an element of G and serves as the identity element under multiplication. 4. Inverse element: Since G consists of non-singular matrices, every element A in G has an inverse A^-1 that is also in G (and has rational entries). These four properties show that G is a group under multiplication. 5. Infinite group: There are infinitely many non-singular n x n matrices with rational entries. For example, consider diagonal matrices with rational entries on the diagonal; there are infinitely many such matrices. 6. Non-abelian: For n > 1, matrix multiplication is generally not commutative. That means there exist matrices A and B in G such that AB ≠ BA. This shows that G is not abelian. Therefore, G is an infinite, non-abelian group under multiplication.	C
zhipu	C. infinite, non abelian group Explanation: The set \( G \) of all \( n \times n \) non-singular matrices with rational numbers as entries forms a group under matrix multiplication. This group is infinite because there are infinitely many such matrices. Additionally, matrix multiplication is generally not commutative, meaning \( AB \neq BA \) for some matrices \( A \) and \( B \) in \( G \). Hence, \( G \) is a non-abelian group. Therefore, the correct answer is C.	C
gpto1	C	C
deepseekv3	The set \( G \) of all \( n \times n \) non-singular matrices with rational numbers as entries forms a group under matrix multiplication. Here's why: 1. Closure: The product of two non-singular matrices is also non-singular. 2. Associativity: Matrix multiplication is associative. 3. Identity: The identity matrix \( I_n \) is in \( G \) and serves as the identity element. 4. Inverse: Every non-singular matrix has an inverse, which is also a non-singular matrix with rational entries. Now, addressing the specific options: - A. subgroup: This is incorrect because \( G \) is a group, not just a subgroup. - B. finite abelian group: This is incorrect because \( G \) is infinite (there are infinitely many non-singular matrices with rational entries). - C. infinite, non-abelian group: This is correct. \( G \) is infinite, and matrix multiplication is generally non-commutative for \( n \geq 2 \). - D. infinite, abelian: This is incorrect because \( G \) is non-abelian. Thus, the correct answer is: C	C

模型

模型答案

正确答案

金声玉亮2.0

C. infinite, non abelian group

Reasoning: The set of all n x n non-singular matrices with rational numbers as entries forms a group under multiplication, as it satisfies the group properties (closure, associativity, identity, and inverse). However, this group is not abelian, as matrix multiplication is not commutative in general. Additionally, the group is infinite, as there are infinitely many non-singular matrices with rational entries. Therefore, the correct answer is C. infinite, non abelian group.

deepseekr1

The set \( G \) consists of all \( n \times n \) non-singular matrices with rational entries, forming the general linear group \( \text{GL}(n, \mathbb{Q}) \) under matrix multiplication.

1. **Group Structure**: \( G \) is a group because it satisfies closure (product of invertible matrices is invertible), associativity, has an identity matrix, and every element has an inverse (non-singularity ensures invertibility).

2. **Finiteness**: \( G \) is infinite because there are infinitely many rational numbers, allowing infinitely many distinct matrices. This eliminates options B (finite abelian) and A (subgroup, as \( G \) is a standalone group).

3. **Abelian vs. Non-Abelian**:
- For \( n = 1 \), \( G \) reduces to non-zero rationals under multiplication, which is abelian. However, for \( n \geq 2 \), matrix multiplication is non-commutative (e.g., counterexamples exist with simple \( 2 \times 2 \) matrices).
- Since the question does not specify \( n = 1 \), the default assumption in abstract algebra is \( n \geq 2 \), making \( G \) non-abelian.

Thus, \( G \) is an **infinite, non-abelian group**.

Answer: **C**

gpto1

GPT-4o

C. infinite, non abelian group

llama3.1-70B-Instruct

llama3.1-405B-Instruct

DeepSeek-Coder-V2

Qwen2-72B-Instruct

C. infinite, non abelian group

baichuan4

C. infinite, non abelian group

doubao

C

The set of all non-singular matrices with rational entries is an infinite group under multiplication. And in general, it is non-abelian (matrix multiplication is not commutative in general).

minimax

C. infinite, non abelian group

kimi

C. infinite, non abelian group

Explanation: The set G of all n x n non-singular matrices with rational numbers as entries, under multiplication, forms a group. This is because:

1. Closure: The product of two non-singular matrices is again a non-singular matrix.
2. Associativity: Matrix multiplication is associative.
3. Identity: The n x n identity matrix is in G, and multiplying any matrix by the identity matrix leaves it unchanged.
4. Inverses: Every non-singular matrix has an inverse in G, which is also a non-singular matrix with rational entries.

Since G is not necessarily commutative (matrix multiplication is not commutative in general), it is non-abelian. Also, G is infinite because there are infinitely many n x n non-singular matrices with rational entries. Therefore, the correct answer is C.

Note that the term "abelian" is misspelled in the options as "ininite, abelian" and should be "infinite, abelian".

wenxinyiyan

C

Explanation:

1. **Closure under multiplication**: Since the product of two non-singular matrices is again non-singular, and the product of matrices with rational entries has rational entries, G is closed under multiplication.

2. **Associativity**: Matrix multiplication is associative, so this property holds in G.

3. **Identity element**: The identity matrix I (with 1s on the diagonal and 0s elsewhere) is an element of G and serves as the identity element under multiplication.

4. **Inverse element**: Since G consists of non-singular matrices, every element A in G has an inverse A^-1 that is also in G (and has rational entries).

These four properties show that G is a group under multiplication.

5. **Infinite group**: There are infinitely many non-singular n x n matrices with rational entries. For example, consider diagonal matrices with rational entries on the diagonal; there are infinitely many such matrices.

6. **Non-abelian**: For n > 1, matrix multiplication is generally not commutative. That means there exist matrices A and B in G such that AB ≠ BA. This shows that G is not abelian.

Therefore, G is an infinite, non-abelian group under multiplication.

zhipu

C. infinite, non abelian group

Explanation: The set \( G \) of all \( n \times n \) non-singular matrices with rational numbers as entries forms a group under matrix multiplication. This group is infinite because there are infinitely many such matrices. Additionally, matrix multiplication is generally not commutative, meaning \( AB \neq BA \) for some matrices \( A \) and \( B \) in \( G \). Hence, \( G \) is a non-abelian group. Therefore, the correct answer is C.

gpto1

deepseekv3

The set \( G \) of all \( n \times n \) non-singular matrices with rational numbers as entries forms a group under matrix multiplication. Here's why:

1. **Closure**: The product of two non-singular matrices is also non-singular.
2. **Associativity**: Matrix multiplication is associative.
3. **Identity**: The identity matrix \( I_n \) is in \( G \) and serves as the identity element.
4. **Inverse**: Every non-singular matrix has an inverse, which is also a non-singular matrix with rational entries.

Now, addressing the specific options:

- **A. subgroup**: This is incorrect because \( G \) is a group, not just a subgroup.
- **B. finite abelian group**: This is incorrect because \( G \) is infinite (there are infinitely many non-singular matrices with rational entries).
- **C. infinite, non-abelian group**: This is correct. \( G \) is infinite, and matrix multiplication is generally non-commutative for \( n \geq 2 \).
- **D. infinite, abelian**: This is incorrect because \( G \) is non-abelian.

Thus, the correct answer is:

**C**