| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the B-coordinate vector of x with respect to the basis B = {v_1, v_2}, we need to express x as a linear combination of the basis vectors v_1 and v_2. In other words, we need to find scalars a and b such that: x = a * v_1 + b * v_2 We can set up a system of linear equations by substituting the given values: [10, 10] = a * [3, 1] + b * [-1, 3] This can be rewritten as: 10 = 3a - b 10 = a + 3b Solving this system of linear equations for a and b, we get: a = 4 b = 2 Therefore, the B-coordinate vector of x is: [x]_B = [4, 2] This means that x can be expressed as a linear combination of the basis vectors v_1 and v_2 with coefficients 4 and 2, respectively. |
[4, 2] |
| deepseekr1 | To find the B-coordinate vector of \( \mathbf{x} = [10, 10] \) with respect to the basis \( B = \{\mathbf{v}_1 = [3, 1], \mathbf{v}_2 = [-1, 3]\} \), we solve for scalars \( c_1 \) and \( c_2 \) such that: \[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{x} \] This translates to the system of equations: \[ \begin{cases} 3c_1 - c_2 = 10 \\ c_1 + 3c_2 = 10 \end{cases} \] **Solving the system:** 1. Multiply the first equation by 3: \[ 9c_1 - 3c_2 = 30 \] 2. Add to the second equation: \[ (9c_1 - 3c_2) + (c_1 + 3c_2) = 30 + 10 \implies 10c_1 = 40 \implies c_1 = 4 \] 3. Substitute \( c_1 = 4 \) into \( c_1 + 3c_2 = 10 \): \[ 4 + 3c_2 = 10 \implies 3c_2 = 6 \implies c_2 = 2 \] **Verification:** \[ 4\mathbf{v}_1 + 2\mathbf{v}_2 = 4[3, 1] + 2[-1, 3] = [12, 4] + [-2, 6] = [10, 10] = \mathbf{x} \] Thus, the B-coordinate vector of \( \mathbf{x} \) is \(\boxed{[4, 2]}\). |
[4, 2] |
| gpto1 | To find the \( B \)-coordinate vector of \( x \) with respect to the basis \( B = \{ \mathbf{v}_1, \mathbf{v}_2 \} \), where \( \mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \), we need to express \( x \) as a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). Let \( x = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \), where \( c_1 \) and \( c_2 \) are scalars to be determined. Given \( x = \begin{bmatrix} 10 \\ 10 \end{bmatrix} \), we set up the equation: \[ \begin{bmatrix} 10 \\ 10 \end{bmatrix} = c_1 \begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 3 \end{bmatrix} \] Breaking this into component equations: 1. \( 10 = 3c_1 - c_2 \) 2. \( 10 = c_1 + 3c_2 \) Now, solve the system of equations: **From equation 1:** \[ - c_2 = 10 - 3c_1 \\ \Rightarrow c_2 = 3c_1 - 10 \quad (1) \] **Substitute \( c_2 \) into equation 2:** \[ 10 = c_1 + 3(3c_1 - 10) \\ 10 = c_1 + 9c_1 - 30 \\ 10 = 10c_1 - 30 \\ 10c_1 = 40 \\ c_1 = 4 \] **Find \( c_2 \) using equation (1):** \[ c_2 = 3(4) - 10 = 12 - 10 = 2 \] Therefore, the \( B \)-coordinate vector of \( x \) is: \[ [x]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} \] **Answer:** The B-coordinate vector of x is [4 2]ᵗ—that is, [x]_B = [4, 2]ᵗ. |
[4, 2] |
| GPT-4o | To find the B-coordinate vector of \( x \) with respect to the basis \( B \) consisting of vectors \( v_1 = [3,1] \) and \( v_2 = [-1, 3] \), we need to express \( x \) as a linear combination of \( v_1 \) and \( v_2 \). This means we need to find scalars \( c_1 \) and \( c_2 \) such that: \[ x = c_1 v_1 + c_2 v_2 \] Given: \[ x = [10, 10] \] \[ v_1 = [3, 1] \] \[ v_2 = [-1, 3] \] We need to solve: \[ [10, 10] = c_1 [3, 1] + c_2 [-1, 3] \] This translates to the following system of linear equations: \[ 3c_1 - c_2 = 10 \] \[ c_1 + 3c_2 = 10 \] We can solve this system using various methods, such as substitution, elimination, or matrix methods. Here, we will use the elimination method. First, let's isolate \( c_1 \) in the first equation: \[ c_2 = 3c_1 - 10 \] Next, substitute \( c_2 \) into the second equation: \[ c_1 + 3(3c_1 - 10) = 10 \] \[ c_1 + 9c_1 - 30 = 10 \] \[ 10c_1 - 30 = 10 \] \[ 10c_1 = 40 \] \[ c_1 = 4 \] Now, substitute \( c_1 = 4 \) back into the equation \( c_2 = 3c_1 - 10 \): \[ c_2 = 3(4) - 10 \] \[ c_2 = 12 - 10 \] \[ c_2 = 2 \] Therefore, the B-coordinate vector of \( x \) is: \[ [c_1, c_2] = [4, 2] \] So, the B-coordinate vector of \( x \) with respect to the basis \( B \) is: \[ \boxed{[4, 2]} \] |
[4, 2] |
| llama3.1-70B-Instruct | To find the B-coordinate vector of x, we need to express x as a linear combination of the basis vectors v_1 and v_2. Let's write x as a linear combination of v_1 and v_2: x = [10, 10] = a*v_1 + b*v_2 where a and b are scalars. Substituting the values of v_1 and v_2, we get: [10, 10] = a*[3, 1] + b*[-1, 3] Equating the corresponding components, we get: 10 = 3a - b 10 = a + 3b Solving this system of equations, we get: a = 3 b = 1 Therefore, the B-coordinate vector of x is: [x]_B = [3, 1] So, the B-coordinate vector of x is [3, 1]. |
[4, 2] |
| llama3.1-405B-Instruct | To find the B-coordinate vector of x, we need to express x as a linear combination of the basis vectors v_1 and v_2. Let's write x as a linear combination of v_1 and v_2: x = a_1 v_1 + a_2 v_2 where a_1 and a_2 are scalars. We can write this equation in matrix form as: [10, 10] = a_1 [3, 1] + a_2 [-1, 3] Now, we can solve for a_1 and a_2 by equating the corresponding components: 10 = 3a_1 - a_2 10 = a_1 + 3a_2 Solving this system of equations, we get: a_1 = 4 a_2 = 2 Therefore, the B-coordinate vector of x is: [x]_B = [a_1, a_2] = [4, 2] So, the B-coordinate vector of x is [4, 2]. |
[4, 2] |
| DeepSeek-Coder-V2 | To find the B-coordinate vector of \( x = [10, 10] \) with respect to the basis \( B \) consisting of vectors \( v_1 = [3, 1] \) and \( v_2 = [-1, 3] \), we need to express \( x \) as a linear combination of \( v_1 \) and \( v_2 \). This means we need to find scalars \( c_1 \) and \( c_2 \) such that: \[ x = c_1 v_1 + c_2 v_2 \] Substituting the given vectors, we get: \[ [10, 10] = c_1 [3, 1] + c_2 [-1, 3] \] This can be written as a system of linear equations: \[ 3c_1 - c_2 = 10 \] \[ c_1 + 3c_2 = 10 \] We will solve this system using the method of elimination. First, we multiply the second equation by 3 to align the coefficients of \( c_1 \): \[ 3c_1 + 9c_2 = 30 \] Now, we subtract the first equation from this new equation: \[ (3c_1 + 9c_2) - (3c_1 - c_2) = 30 - 10 \] \[ 10c_2 = 20 \] \[ c_2 = 2 \] Next, we substitute \( c_2 = 2 \) back into the first equation: \[ 3c_1 - 2 = 10 \] \[ 3c_1 = 12 \] \[ c_1 = 4 \] Thus, the B-coordinate vector of \( x \) is: \[ [c_1, c_2] = [4, 2] \] So, the B-coordinate vector of \( x = [10, 10] \) is \( [4, 2] \). |
[4, 2] |
| Qwen2-72B-Instruct | To find the B-coordinate vector of x with respect to the basis B = {v_1, v_2}, we need to express x as a linear combination of the basis vectors v_1 and v_2. In other words, we need to find scalars a and b such that: x = a * v_1 + b * v_2 We can set up a system of linear equations by substituting the given values: [10, 10] = a * [3, 1] + b * [-1, 3] This can be rewritten as: 10 = 3a - b 10 = a + 3b Solving this system of linear equations for a and b, we get: a = 4 b = 2 Therefore, the B-coordinate vector of x is: [x]_B = [4, 2] This means that x can be expressed as a linear combination of the basis vectors v_1 and v_2 with coefficients 4 and 2, respectively. |
[4, 2] |
| baichuan4 | To find the B-coordinate vector of x with respect to the basis B consisting of vectors v₁ = [3, 1] and v₂ = [-1, 3], we need to express x as a linear combination of v₁ and v₂. Given x = [10, 10], we want to find scalars a and b such that: x = av₁ + bv₂ Substituting the values for x, v₁, and v₂, we get: [10, 10] = a[3, 1] + b[-1, 3] This gives us a system of linear equations: 3a - b = 10 a + 3b = 10 We can solve this system to find a and b. First, let's multiply the second equation by 3 to get: 3a + 9b = 30 Now we can subtract the first equation from this one to eliminate a: 3a + 9b - (3a - b) = 30 - 10 10b = 20 Solving for b, we get: b = 20 / 10 b = 2 Now we can substitute this value of b into the second equation to find a: a + 3(2) = 10 a + 6 = 10 a = 10 - 6 a = 4 So, we have found the scalars a = 4 and b = 2. Therefore, the B-coordinate vector of x is: [4, 2] |
[4, 2] |
| doubao | To find the B-coordinate vector of x, we need to find the coefficients c1 and c2 such that x = c1v1 + c2v2. Let's set up the equation: [10, 10] = c1[3, 1] + c2[-1, 3] This gives us the system of equations: 3c1 - c2 = 10 c1 + 3c2 = 10 Multiplying the second equation by 3 and subtracting it from the first equation gives: 3c1 - c2 - 3(c1 + 3c2) = 10 - 3(10) 3c1 - c2 - 3c1 - 9c2 = 10 - 30 -10c2 = -20 c2 = 2. Substituting c2 = 2 into the second equation gives: c1 + 3(2) = 10, c1 = 10 - 6 = 4. So the B-coordinate vector of x is [4, 2]. |
[4, 2] |
| minimax | To find the B-coordinate vector of \( x \) with respect to the basis \( B = \{v_1, v_2\} \), we need to express \( x \) as a linear combination of \( v_1 \) and \( v_2 \). In other words, we want to find scalars \( c_1 \) and \( c_2 \) such that: \[ x = c_1 v_1 + c_2 v_2 \] Given \( x = [10, 10] \), \( v_1 = [3, 1] \), and \( v_2 = [-1, 3] \), we can set up the following system of equations: \[ c_1 \cdot 3 + c_2 \cdot (-1) = 10 \] \[ c_1 \cdot 1 + c_2 \cdot 3 = 10 \] This can be written in matrix form as: \[ \begin{bmatrix} 3 & -1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 10 \\ 10 \end{bmatrix} \] Now we can solve for \( c_1 \) and \( c_2 \) using matrix methods such as Gaussian elimination or by finding the inverse of the coefficient matrix (if it exists) and multiplying it by the right-hand side vector. Let's solve this system using matrix multiplication: \[ \begin{bmatrix} 3 & -1 \\ 1 & 3 \end{bmatrix}^{-1} \begin{bmatrix} 10 \\ 10 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \] First, we need to find the inverse of the 2x2 matrix: \[ A = \begin{bmatrix} 3 & -1 \\ 1 & 3 \end{bmatrix} \] The inverse of a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For our matrix, \( a = 3 \), \( b = -1 \), \( c = 1 \), and \( d = 3 \). The determinant \( ad - bc \) is: \[ \text{det}(A) = (3)(3) - (-1)(1) = 9 + 1 = 10 \] So the inverse of \( A \) is: \[ A^{-1} = \frac{1}{10} \begin{bmatrix} 3 & 1 \\ -1 & 3 \end{bmatrix} \] Now we can find \( \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} \): \[ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 3 & 1 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 10 \\ 10 \end{bmatrix} \] \[ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 3 \cdot 10 + 1 \cdot 10 \\ -1 \cdot 10 + 3 \cdot 10 \end{bmatrix} \] \[ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 40 \\ 20 \end{bmatrix} \] \[ \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} \] Therefore, the B-coordinate vector of \( x \) is \( [c_1, c_2] = [4, 2] \). |
[4, 2] |
| kimi | To find the B-coordinate vector of x = [10, 10] with respect to the basis B consisting of vectors v_1 = [3, 1] and v_2 = [-1, 3], we need to express x as a linear combination of v_1 and v_2. Let's denote the B-coordinate vector of x as [a, b], where a and b are scalars. We want to find a and b such that: x = a * v_1 + b * v_2 Substituting the given vectors, we get: [10, 10] = a * [3, 1] + b * [-1, 3] This gives us a system of linear equations: 3a - b = 10 a + 3b = 10 Now we can solve this system for a and b. From the first equation, we can express b in terms of a: b = 3a - 10 Substitute this expression for b into the second equation: a + 3(3a - 10) = 10 a + 9a - 30 = 10 10a = 40 a = 4 Now that we have a, we can find b: b = 3(4) - 10 b = 12 - 10 b = 2 So the B-coordinate vector of x is [4, 2], which means: x = 4 * [3, 1] + 2 * [-1, 3] |
[4, 2] |
| wenxinyiyan | 首先，我们需要找到向量x在基B下的坐标。基B由向量v_1 = [3,1]和v_2 = [-1,3]组成。 我们可以将向量x表示为基B的线性组合，即： x = c_1 * v_1 + c_2 * v_2 其中，c_1和c_2是我们要找的B-坐标向量。 将v_1和v_2代入上式，得到： [10, 10] = c_1 * [3, 1] + c_2 * [-1, 3] 这可以转化为一个线性方程组： 3*c_1 - c_2 = 10 c_1 + 3*c_2 = 10 解这个方程组，我们得到： c_1 = 4 c_2 = 2 因此，向量x在基B下的B-坐标向量为[4, 2]。 |
[4, 2] |
| zhipu | To find the B-coordinate vector of \( x \) with respect to the basis \( B \), we need to express \( x \) as a linear combination of the basis vectors \( v_1 \) and \( v_2 \). That is, we need to find scalars \( a \) and \( b \) such that: \[ x = a v_1 + b v_2 \] Given: \[ v_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix}, \quad v_2 = \begin{bmatrix} -1 \\ 3 \end{bmatrix}, \quad x = \begin{bmatrix} 10 \\ 10 \end{bmatrix} \] We need to solve for \( a \) and \( b \) in the equation: \[ \begin{bmatrix} 10 \\ 10 \end{bmatrix} = a \begin{bmatrix} 3 \\ 1 \end{bmatrix} + b \begin{bmatrix} -1 \\ 3 \end{bmatrix} \] This can be written as a system of linear equations: \[ \begin{cases} 3a - b = 10 \\ a + 3b = 10 \end{cases} \] We can solve this system using methods such as substitution, elimination, or matrix techniques. Here, we'll use the elimination method. First, multiply the second equation by 3 to align the coefficients of \( a \): \[ \begin{cases} 3a - b = 10 \\ 3a + 9b = 30 \end{cases} \] Now, subtract the first equation from the second: \[ (3a + 9b) - (3a - b) = 30 - 10 \] \[ 3a + 9b - 3a + b = 20 \] \[ 10b = 20 \] \[ b = 2 \] Substitute \( b = 2 \) back into the first equation: \[ 3a - 2 = 10 \] \[ 3a = 12 \] \[ a = 4 \] So, the scalars \( a \) and \( b \) are 4 and 2, respectively. Therefore, the B-coordinate vector of \( x \) is: \[ \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} \] Thus, the B-coordinate vector of \( x = \begin{bmatrix} 10 \\ 10 \end{bmatrix} \) with respect to the basis \( B \) is \( \begin{bmatrix} 4 \\ 2 \end{bmatrix} \). |
[4, 2] |
| gpto1 | To find the \( B \)-coordinate vector of \( x \) with respect to the basis \( B = \{ \mathbf{v}_1, \mathbf{v}_2 \} \), where \( \mathbf{v}_1 = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} -1 \\ 3 \end{bmatrix} \), we need to express \( x \) as a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). Let \( x = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \), where \( c_1 \) and \( c_2 \) are scalars to be determined. Given \( x = \begin{bmatrix} 10 \\ 10 \end{bmatrix} \), we set up the equation: \[ \begin{bmatrix} 10 \\ 10 \end{bmatrix} = c_1 \begin{bmatrix} 3 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 3 \end{bmatrix} \] Breaking this into component equations: 1. \( 10 = 3c_1 - c_2 \) 2. \( 10 = c_1 + 3c_2 \) Now, solve the system of equations: **From equation 1:** \[ - c_2 = 10 - 3c_1 \\ \Rightarrow c_2 = 3c_1 - 10 \quad (1) \] **Substitute \( c_2 \) into equation 2:** \[ 10 = c_1 + 3(3c_1 - 10) \\ 10 = c_1 + 9c_1 - 30 \\ 10 = 10c_1 - 30 \\ 10c_1 = 40 \\ c_1 = 4 \] **Find \( c_2 \) using equation (1):** \[ c_2 = 3(4) - 10 = 12 - 10 = 2 \] Therefore, the \( B \)-coordinate vector of \( x \) is: \[ [x]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix} \] **Answer:** The B-coordinate vector of x is [4 2]ᵗ—that is, [x]_B = [4, 2]ᵗ. |
[4, 2] |
| deepseekv3 | To find the B-coordinate vector of \( \mathbf{x} = [10, 10] \) with respect to the basis \( B = \{\mathbf{v}_1, \mathbf{v}_2\} \), where \( \mathbf{v}_1 = [3, 1] \) and \( \mathbf{v}_2 = [-1, 3] \), we need to express \( \mathbf{x} \) as a linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). That is, we need to find scalars \( c_1 \) and \( c_2 \) such that: \[ \mathbf{x} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \] Substituting the given vectors, we have: \[ [10, 10] = c_1 [3, 1] + c_2 [-1, 3] \] This gives us the following system of equations: \[ \begin{cases} 3c_1 - c_2 = 10 \\ c_1 + 3c_2 = 10 \end{cases} \] Let's solve this system step by step. 1. **From the first equation:** \[ 3c_1 - c_2 = 10 \quad \Rightarrow \quad c_2 = 3c_1 - 10 \] 2. **Substitute \( c_2 \) into the second equation:** \[ c_1 + 3(3c_1 - 10) = 10 \] \[ c_1 + 9c_1 - 30 = 10 \] \[ 10c_1 = 40 \quad \Rightarrow \quad c_1 = 4 \] 3. **Now, find \( c_2 \) using \( c_1 = 4 \):** \[ c_2 = 3(4) - 10 = 12 - 10 = 2 \] Thus, the B-coordinate vector of \( \mathbf{x} = [10, 10] \) is \( [c_1, c_2] = [4, 2] \). \[ \boxed{[4, 2]} \] |
[4, 2] |