| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D. I, II, and III. If v is an eigenvector of A, then Av = λv for some scalar λ. I. 2Av = 2λv, so v is also an eigenvector of 2A. II. A^2v = A(Av) = A(λv) = λAv = λ^2v, so v is also an eigenvector of A^2. III. A^(-1)v = (1/λ)v, so v is also an eigenvector of A^(-1). Therefore, all three statements are true. |
D |
| deepseekr1 | The question asks which statements about the eigenvector \( v \) of an invertible matrix \( A \) are necessarily true. 1. **Statement I**: If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( 2A \cdot v = 2\lambda v \), confirming \( v \) is also an eigenvector of \( 2A \). **True**. 2. **Statement II**: \( A^2 \cdot v = A(Av) = A(\lambda v) = \lambda^2 v \), so \( v \) is an eigenvector of \( A^2 \). **True**. 3. **Statement III**: Since \( A \) is invertible, \( \lambda \neq 0 \). Multiplying \( Av = \lambda v \) by \( A^{-1} \) gives \( v = \lambda A^{-1}v \), so \( A^{-1}v = \frac{1}{\lambda}v \), proving \( v \) is an eigenvector of \( A^{-1} \). **True**. All statements I, II, and III hold. **Answer: D**. \boxed{D} |
D |
| gpto1 | To determine which statements are necessarily true, let's analyze each one using the properties of eigenvectors and eigenvalues. **Given**: \( A \) is an invertible matrix, and \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), so: \[ A v = \lambda v \] Since \( A \) is invertible, \( \lambda \neq 0 \). --- **Statement I**: *\( v \) is also an eigenvector of \( 2A \).* **Verification**: \[ (2A) v = 2(A v) = 2(\lambda v) = (2\lambda) v \] This shows that \( v \) is an eigenvector of \( 2A \) with eigenvalue \( 2\lambda \). **Conclusion**: True. --- **Statement II**: *\( v \) is also an eigenvector of \( A^2 \).* **Verification**: \[ A^2 v = A(A v) = A(\lambda v) = \lambda (A v) = \lambda (\lambda v) = \lambda^2 v \] This shows that \( v \) is an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). **Conclusion**: True. --- **Statement III**: *\( v \) is also an eigenvector of \( A^{-1} \).* **Verification**: Starting from \( A v = \lambda v \), multiply both sides by \( A^{-1} \): \[ A^{-1} (A v) = A^{-1} (\lambda v) \] Simplify: \[ (I) v = \lambda (A^{-1} v) \] \[ v = \lambda (A^{-1} v) \] Solve for \( A^{-1} v \): \[ A^{-1} v = \frac{1}{\lambda} v \] This shows that \( v \) is an eigenvector of \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda} \). **Conclusion**: True. --- **Final Answer**: All statements I, II, and III are necessarily true. **Answer: D** |
D |
| GPT-4o | D. I, II, and III | D |
| llama3.1-70B-Instruct | D |
D |
| llama3.1-405B-Instruct | D. I, II, and III. If v is an eigenvector of A, then Av = λv for some scalar λ. I. 2Av = 2λv, so v is also an eigenvector of 2A. II. A^2v = A(Av) = A(λv) = λAv = λ^2v, so v is also an eigenvector of A^2. III. A^(-1)v = (1/λ)v, so v is also an eigenvector of A^(-1). Therefore, all three statements are true. |
D |
| DeepSeek-Coder-V2 | D. I, II, and III | D |
| Qwen2-72B-Instruct | D. I, II, and III | D |
| baichuan4 | B. II only | D |
| doubao | D If v is an eigenvector of A with eigenvalue λ, then Av = λv. For 2A, 2Av = 2λv, so v is an eigenvector of 2A. For A^2, A^2v = A(Av) = A(λv) = λ(Av) = λ^2v, so v is an eigenvector of A^2. For A^(-1), A^(-1)v = (1/λ)v, so v is also an eigenvector of A^(-1). So I, II, and III are all true. |
D |
| minimax | D. I, II, and III Explanation: Let's consider each statement: I. If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( Av = \lambda v \). Multiplying both sides by 2, we get \( 2Av = 2\lambda v \), which means \( 2A \) has eigenvalue \( 2\lambda \) for the same eigenvector \( v \). So, statement I is true. II. If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( A^2v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda^2 v \). This shows that \( A^2 \) has eigenvalue \( \lambda^2 \) for the same eigenvector \( v \). So, statement II is true. III. If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( A^{-1}v = \frac{1}{\lambda} v \) provided \( \lambda \neq 0 \) (since \( A \) is invertible, \( \lambda \) cannot be zero). This means \( A^{-1} \) has eigenvalue \( \frac{1}{\lambda} \) for the same eigenvector \( v \). So, statement III is true. Therefore, all three statements are true. |
D |
| kimi | D. I, II, and III Explanation: 1. If \( v \) is an eigenvector of \( A \), then \( A \cdot v = \lambda v \) for some scalar \( \lambda \). If we multiply \( A \) by a scalar, say 2, then \( 2A \cdot v = 2(A \cdot v) = 2(\lambda v) = (2\lambda)v \). So \( v \) is also an eigenvector of \( 2A \) with eigenvalue \( 2\lambda \). 2. For \( A^2 \), if \( A \cdot v = \lambda v \), then \( A^2 \cdot v = A \cdot (A \cdot v) = A \cdot (\lambda v) = \lambda(\lambda v) = \lambda^2 v \). So \( v \) is also an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). 3. For \( A^{-1} \), if \( A \cdot v = \lambda v \), then \( A^{-1} \cdot (A \cdot v) = A^{-1} \cdot (\lambda v) = \lambda(A^{-1} \cdot v) = \frac{1}{\lambda}v \). So \( v \) is also an eigenvector of \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda} \), provided that \( \lambda \neq 0 \) (which it cannot be, because \( v \) is an eigenvector and eigenvectors are non-zero). All three statements are true if \( v \) is an eigenvector of an invertible matrix \( A \). |
D |
| wenxinyiyan | D. I, II, and III Explanation: Let's say v is an eigenvector of A with eigenvalue λ (λ ≠ 0 since A is invertible). I. For 2A, we have: (2A)v = 2(Av) = 2(λv) = (2λ)v So, v is an eigenvector of 2A with eigenvalue 2λ. II. For A^2, we have: A^2v = A(Av) = A(λv) = λ(Av) = λ(λv) = λ^2v So, v is an eigenvector of A^2 with eigenvalue λ^2. III. For A^(-1), we have: A^(-1)v = A^(-1)(λA^(-1)Av) = λA^(-1)v Dividing both sides by λ (since λ ≠ 0), we get: A^(-1)v = (1/λ)v So, v is an eigenvector of A^(-1) with eigenvalue 1/λ. |
D |
| zhipu | To determine which statements are necessarily true regarding the eigenvector \( v \) of an invertible matrix \( A \), let's analyze each statement individually. ### Statement I: \( v \) is also an eigenvector of \( 2A \). - If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( A v = \lambda v \). - For \( 2A \), we have \( (2A)v = 2(Av) = 2(\lambda v) = (2\lambda)v \). - This means \( v \) is also an eigenvector of \( 2A \) with eigenvalue \( 2\lambda \). Thus, Statement I is true. ### Statement II: \( v \) is also an eigenvector of \( A^2 \). - If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( A v = \lambda v \). - For \( A^2 \), we have \( (A^2)v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda (\lambda v) = \lambda^2 v \). - This means \( v \) is also an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). Thus, Statement II is true. ### Statement III: \( v \) is also an eigenvector of \( A^{-1} \). - If \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), then \( A v = \lambda v \). - Since \( A \) is invertible, \( \lambda \neq 0 \). - For \( A^{-1} \), we have \( A^{-1}(Av) = A^{-1}(\lambda v) \). - Since \( A^{-1}A = I \), it follows that \( v = \lambda A^{-1}v \). - Dividing both sides by \( \lambda \), we get \( A^{-1}v = \frac{1}{\lambda}v \). - This means \( v \) is also an eigenvector of \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda} \). Thus, Statement III is true. Since all three statements (I, II, and III) are necessarily true, the correct answer is: **D. I, II, and III** |
D |
| gpto1 | To determine which statements are necessarily true, let's analyze each one using the properties of eigenvectors and eigenvalues. **Given**: \( A \) is an invertible matrix, and \( v \) is an eigenvector of \( A \) with eigenvalue \( \lambda \), so: \[ A v = \lambda v \] Since \( A \) is invertible, \( \lambda \neq 0 \). --- **Statement I**: *\( v \) is also an eigenvector of \( 2A \).* **Verification**: \[ (2A) v = 2(A v) = 2(\lambda v) = (2\lambda) v \] This shows that \( v \) is an eigenvector of \( 2A \) with eigenvalue \( 2\lambda \). **Conclusion**: True. --- **Statement II**: *\( v \) is also an eigenvector of \( A^2 \).* **Verification**: \[ A^2 v = A(A v) = A(\lambda v) = \lambda (A v) = \lambda (\lambda v) = \lambda^2 v \] This shows that \( v \) is an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). **Conclusion**: True. --- **Statement III**: *\( v \) is also an eigenvector of \( A^{-1} \).* **Verification**: Starting from \( A v = \lambda v \), multiply both sides by \( A^{-1} \): \[ A^{-1} (A v) = A^{-1} (\lambda v) \] Simplify: \[ (I) v = \lambda (A^{-1} v) \] \[ v = \lambda (A^{-1} v) \] Solve for \( A^{-1} v \): \[ A^{-1} v = \frac{1}{\lambda} v \] This shows that \( v \) is an eigenvector of \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda} \). **Conclusion**: True. --- **Final Answer**: All statements I, II, and III are necessarily true. **Answer: D** |
D |
| deepseekv3 | Let's analyze each statement one by one: **Given**: \( v \) is an eigenvector of an invertible matrix \( A \). This means there exists a scalar \( \lambda \) (the eigenvalue) such that: \[ A v = \lambda v \] --- **I. \( v \) is also an eigenvector of \( 2A \).** Multiply both sides of \( A v = \lambda v \) by 2: \[ 2A v = 2\lambda v \] This shows that \( v \) is an eigenvector of \( 2A \) with eigenvalue \( 2\lambda \). **Statement I is true.** --- **II. \( v \) is also an eigenvector of \( A^2 \).** Multiply both sides of \( A v = \lambda v \) by \( A \): \[ A^2 v = A (\lambda v) = \lambda (A v) = \lambda (\lambda v) = \lambda^2 v \] This shows that \( v \) is an eigenvector of \( A^2 \) with eigenvalue \( \lambda^2 \). **Statement II is true.** --- **III. \( v \) is also an eigenvector of \( A^{-1} \).** Since \( A \) is invertible and \( A v = \lambda v \), multiply both sides by \( A^{-1} \): \[ v = \lambda A^{-1} v \] Rearrange to solve for \( A^{-1} v \): \[ A^{-1} v = \frac{1}{\lambda} v \] This shows that \( v \) is an eigenvector of \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda} \). **Statement III is true.** --- **Conclusion**: All three statements (I, II, and III) are necessarily true. **Answer**: **D** |
D |