| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The determinant of a 3x3 matrix A = [[a, b, c], [d, e, f], [g, h, i]] can be calculated using the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) For matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], we have: a = 2, b = 4, c = 3, d = 3, e = 0, f = 1, g = 1, h = 2, i = 5 Substituting these values into the formula, we get: det(A) = 2(0*5 - 1*2) - 4(3*5 - 1*1) + 3(3*2 - 0*1) = 2(-2) - 4(14) + 3(6) = -4 - 56 + 18 = -42 Therefore, the determinant of matrix A is -42. |
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| deepseekr1 | To find the determinant of the matrix \( A = \begin{bmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{bmatrix} \), we use the expansion by minors method along the second row, which contains a zero to simplify calculations. The determinant is calculated as follows: 1. **Expanding along the second row:** - The elements of the second row are \(3\), \(0\), and \(1\). - The cofactors for the elements in the second row are calculated as follows: - For the element \(3\) (at position (2,1)): \[ C_{21} = (-1)^{2+1} \det \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix} = -1 \cdot (4 \cdot 5 - 3 \cdot 2) = -1 \cdot (20 - 6) = -14 \] - For the element \(0\) (at position (2,2)), the cofactor is \(0\) because the element itself is zero. - For the element \(1\) (at position (2,3)): \[ C_{23} = (-1)^{2+3} \det \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} = -1 \cdot (2 \cdot 2 - 4 \cdot 1) = -1 \cdot (4 - 4) = 0 \] 2. **Summing the contributions:** - The determinant is calculated by summing the products of the elements and their respective cofactors: \[ 3 \cdot (-14) + 0 \cdot 0 + 1 \cdot 0 = -42 + 0 + 0 = -42 \] 3. **Verification using row reduction:** - Convert the matrix to an upper triangular form using row operations: - Subtract \(\frac{3}{2}\) times the first row from the second row. - Subtract \(\frac{1}{2}\) times the first row from the third row. - The resulting upper triangular matrix is: \[ \begin{bmatrix} 2 & 4 & 3 \\ 0 & -6 & -3.5 \\ 0 & 0 & 3.5 \end{bmatrix} \] - The determinant is the product of the diagonal elements: \[ 2 \cdot (-6) \cdot 3.5 = -42 \] Thus, the determinant of matrix \(A\) is \(\boxed{-42}\). |
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| gpto1 | To find the determinant of the matrix: \[ A = \begin{bmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{bmatrix} \] We can use the formula for the determinant of a \( 3 \times 3 \) matrix: \[ \det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \] Let's identify the elements of matrix \( A \): \[ \begin{align*} a_{11} &= 2, & a_{12} &= 4, & a_{13} &= 3 \\ a_{21} &= 3, & a_{22} &= 0, & a_{23} &= 1 \\ a_{31} &= 1, & a_{32} &= 2, & a_{33} &= 5 \\ \end{align*} \] Now, compute each of the components: 1. Compute \( M_1 = a_{22}a_{33} - a_{23}a_{32} \): \[ M_1 = (0)(5) - (1)(2) = 0 - 2 = -2 \] 2. Compute \( M_2 = a_{21}a_{33} - a_{23}a_{31} \): \[ M_2 = (3)(5) - (1)(1) = 15 - 1 = 14 \] 3. Compute \( M_3 = a_{21}a_{32} - a_{22}a_{31} \): \[ M_3 = (3)(2) - (0)(1) = 6 - 0 = 6 \] Now, substitute back into the determinant formula: \[ \det(A) = a_{11}M_1 - a_{12}M_2 + a_{13}M_3 \] \[ \det(A) = (2)(-2) - (4)(14) + (3)(6) \] Calculate each term: \[ \begin{align*} (2)(-2) &= -4 \\ (4)(14) &= 56 \\ (3)(6) &= 18 \\ \end{align*} \] Now, combine the terms: \[ \det(A) = -4 - 56 + 18 = -42 \] **Answer:** \(-42\) |
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| GPT-4o | To find the determinant of a 3x3 matrix \( A \), we can use the formula for the determinant of a 3x3 matrix: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] For the given matrix \( A = \begin{pmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{pmatrix} \), we can identify the elements as follows: \[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{pmatrix} \] So, \( a = 2 \), \( b = 4 \), \( c = 3 \), \( d = 3 \), \( e = 0 \), \( f = 1 \), \( g = 1 \), \( h = 2 \), and \( i = 5 \). Now, we substitute these values into the determinant formula: \[ \text{det}(A) = 2(0 \cdot 5 - 1 \cdot 2) - 4(3 \cdot 5 - 1 \cdot 1) + 3(3 \cdot 2 - 0 \cdot 1) \] Simplify the terms inside the parentheses: \[ \text{det}(A) = 2(0 - 2) - 4(15 - 1) + 3(6 - 0) \] \[ \text{det}(A) = 2(-2) - 4(14) + 3(6) \] \[ \text{det}(A) = -4 - 56 + 18 \] Finally, combine the terms: \[ \text{det}(A) = -4 - 56 + 18 \] \[ \text{det}(A) = -60 + 18 \] \[ \text{det}(A) = -42 \] Therefore, the determinant of the matrix \( A \) is \( -42 \). |
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| llama3.1-70B-Instruct | To find the determinant of matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], we can use the formula for the determinant of a 3x3 matrix: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where a, b, c, d, e, f, g, h, and i are the elements of the matrix. Plugging in the values from matrix A, we get: det(A) = 2(0*5 - 1*2) - 4(3*5 - 1*1) + 3(3*2 - 0*1) = 2(0 - 2) - 4(15 - 1) + 3(6 - 0) = 2(-2) - 4(14) + 3(6) = -4 - 56 + 18 = -42 Therefore, the determinant of matrix A is -42. |
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| llama3.1-405B-Instruct | def determinant(A): # Base case for 2x2 matrix if len(A) == 2: return A[0][0]*A[1][1] - A[0][1]*A[1][0] det = 0 for c in range(len(A)): # Build sub-matrix for cofactor sub_matrix = A[1:] for i in range(len(sub_matrix)): sub_matrix[i] = sub_matrix[i][0:c] + sub_matrix[i][c+1:] sign = (-1) ** (c % 2) # Alternating signs # Recursive call sub_det = determinant(sub_matrix) # Add to total determinant det += sign * A[0][c] * sub_det return det A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]] print(determinant(A)) |
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| DeepSeek-Coder-V2 | To find the determinant of matrix A, we can use the formula for the determinant of a 3x3 matrix. The formula is: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where a, b, and c are the elements of the first row, and d, e, f, g, and h are the elements of the second and third rows, respectively. Given matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], we can identify the elements as follows: a = 2, b = 4, c = 3 d = 3, e = 0, f = 1 g = 1, h = 2, i = 5 Now, we can substitute these values into the determinant formula: det(A) = 2(0*5 - 1*2) - 4(3*5 - 1*1) + 3(3*2 - 0*1) det(A) = 2(0 - 2) - 4(15 - 1) + 3(6 - 0) det(A) = 2(-2) - 4(14) + 3(6) det(A) = -4 - 56 + 18 det(A) = -42 So, the determinant of matrix A is -42. |
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| Qwen2-72B-Instruct | The determinant of a 3x3 matrix A = [[a, b, c], [d, e, f], [g, h, i]] can be calculated using the formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) For matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], we have: a = 2, b = 4, c = 3, d = 3, e = 0, f = 1, g = 1, h = 2, i = 5 Substituting these values into the formula, we get: det(A) = 2(0*5 - 1*2) - 4(3*5 - 1*1) + 3(3*2 - 0*1) = 2(-2) - 4(14) + 3(6) = -4 - 56 + 18 = -42 Therefore, the determinant of matrix A is -42. |
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| baichuan4 | To find the determinant of matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], we can use the following steps: 1. Calculate the determinant of the upper-left 2x2 submatrix: Submatrix = [[2, 4], [3, 0]] Determinant of Submatrix = (2*0) - (4*3) = 0 - 12 = -12 2. Multiply this determinant by the entry in the lower-right corner of the original matrix (which is 5), and then subtract from this the product of the entry in the lower-left corner (which is 1) and the entry in the upper-right corner of the submatrix (which is 4): (-12) * 5 - (1 * 4) = -60 - 4 = -64 3. Now, we need to multiply this result by the entry in the upper-right corner of the original matrix (which is 3), and then add it to the product of the entry in the upper-left corner (which is 2) and the entry in the lower-right corner of the submatrix (which is 0): -64 * 3 + (2 * 0) = -192 + 0 = -192 4. Finally, multiply this result by the entry in the lower-left corner of the original matrix (which is 1), and then subtract from this the product of the entry in the upper-left corner (which is 2) and the entry in the lower-left corner of the submatrix (which is 3): -192 * 1 - (2 * 3) = -192 - 6 = -198 So, the determinant of matrix A is -198. |
-42 |
| doubao | The determinant of a 3x3 matrix [[a,b,c],[d,e,f],[g,h,i]] is given by: \[ \begin{align*} |A|&=a(ei-fh)-b(di-fg)+c(dh-eg)\\ \end{align*} \] For the given matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], the determinant is: \[ \begin{align*} |A|&=2\times(0\times5-1\times2)-4\times(3\times5-1\times3)+3\times(3\times2-0\times1)\\ &=2\times(-2)-4\times(12)+3\times(6)\\ &=-4-48+18\\ &=-34 \end{align*} \] |
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| minimax | To find the determinant of a 3x3 matrix A, we can use the following formula: det(A) = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31) where aij represents the element in the i-th row and j-th column of matrix A. Given matrix A: A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]] We can calculate its determinant as follows: det(A) = 2(0*5 - 1*2) - 4(3*5 - 1*1) + 3(3*2 - 0*1) = 2(0 - 2) - 4(15 - 1) + 3(6 - 0) = 2(-2) - 4(14) + 3(6) = -4 - 56 + 18 = -40 + 18 = -22 Therefore, the determinant of matrix A is -22. |
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| kimi | To find the determinant of a 3x3 matrix A, you can use the following formula: For a matrix A = [[a, b, c], [d, e, f], [g, h, i]], the determinant det(A) is calculated as: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) Given matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], let's assign the values: a = 2, b = 4, c = 3 d = 3, e = 0, f = 1 g = 1, h = 2, i = 5 Now, plug these values into the formula: det(A) = 2(0*5 - 1*2) - 4(3*5 - 1*1) + 3(3*2 - 0*1) det(A) = 2(-2) - 4(14) + 3(6) det(A) = -4 - 56 + 18 det(A) = -44 So, the determinant of matrix A is -44. |
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| wenxinyiyan | To find the determinant of a 3x3 matrix like A, we use the formula: det(A) = a[0][0] * (a[1][1] * a[2][2] - a[2][1] * a[1][2]) - a[0][1] * (a[1][0] * a[2][2] - a[1][2] * a[2][0]) + a[0][2] * (a[1][0] * a[2][1] - a[1][1] * a[2][0]) where a[i][j] represents the element in the ith row and jth column of matrix A. For matrix A = [[2, 4, 3], [3, 0, 1], [1, 2, 5]], the determinant is: det(A) = 2 * (0 * 5 - 1 * 2) - 4 * (3 * 5 - 1 * 1) + 3 * (3 * 2 - 0 * 1) = 2 * (-2) - 4 * 14 + 3 * 6 = -4 - 56 + 18 = -42 So, the determinant of matrix A is -42. |
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| zhipu | To find the determinant of a 3x3 matrix, we can use the formula: For a matrix A: \[ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \] The determinant (det(A)) is given by: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Given the matrix A: \[ A = \begin{bmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{bmatrix} \] We can identify the elements as follows: a = 2, b = 4, c = 3 d = 3, e = 0, f = 1 g = 1, h = 2, i = 5 Now, we substitute these values into the determinant formula: \[ \text{det}(A) = 2(0 \cdot 5 - 1 \cdot 2) - 4(3 \cdot 5 - 1 \cdot 1) + 3(3 \cdot 2 - 0 \cdot 1) \] Calculate each term separately: First term: \[ 2(0 - 2) = 2 \cdot (-2) = -4 \] Second term: \[ -4(15 - 1) = -4 \cdot 14 = -56 \] Third term: \[ 3(6 - 0) = 3 \cdot 6 = 18 \] Now, add these results together to find the determinant: \[ \text{det}(A) = -4 - 56 + 18 \] \[ \text{det}(A) = -60 + 18 \] \[ \text{det}(A) = -42 \] So, the determinant of matrix A is -42. |
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| gpto1 | To find the determinant of the matrix: \[ A = \begin{bmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{bmatrix} \] We can use the formula for the determinant of a \( 3 \times 3 \) matrix: \[ \det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \] Let's identify the elements of matrix \( A \): \[ \begin{align*} a_{11} &= 2, & a_{12} &= 4, & a_{13} &= 3 \\ a_{21} &= 3, & a_{22} &= 0, & a_{23} &= 1 \\ a_{31} &= 1, & a_{32} &= 2, & a_{33} &= 5 \\ \end{align*} \] Now, compute each of the components: 1. Compute \( M_1 = a_{22}a_{33} - a_{23}a_{32} \): \[ M_1 = (0)(5) - (1)(2) = 0 - 2 = -2 \] 2. Compute \( M_2 = a_{21}a_{33} - a_{23}a_{31} \): \[ M_2 = (3)(5) - (1)(1) = 15 - 1 = 14 \] 3. Compute \( M_3 = a_{21}a_{32} - a_{22}a_{31} \): \[ M_3 = (3)(2) - (0)(1) = 6 - 0 = 6 \] Now, substitute back into the determinant formula: \[ \det(A) = a_{11}M_1 - a_{12}M_2 + a_{13}M_3 \] \[ \det(A) = (2)(-2) - (4)(14) + (3)(6) \] Calculate each term: \[ \begin{align*} (2)(-2) &= -4 \\ (4)(14) &= 56 \\ (3)(6) &= 18 \\ \end{align*} \] Now, combine the terms: \[ \det(A) = -4 - 56 + 18 = -42 \] **Answer:** \(-42\) |
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| deepseekv3 | To find the determinant of the given 3x3 matrix \( A \), we can use the rule of Sarrus or the general formula for the determinant of a 3x3 matrix. Here, we'll use the general formula. Given matrix \( A \): \[ A = \begin{bmatrix} 2 & 4 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 5 \end{bmatrix} \] The determinant of a 3x3 matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \) is calculated as: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Applying this to matrix \( A \): \[ \text{det}(A) = 2(0 \cdot 5 - 1 \cdot 2) - 4(3 \cdot 5 - 1 \cdot 1) + 3(3 \cdot 2 - 0 \cdot 1) \] Now, let's compute each part step by step: 1. \( 2(0 \cdot 5 - 1 \cdot 2) = 2(0 - 2) = 2(-2) = -4 \) 2. \( -4(3 \cdot 5 - 1 \cdot 1) = -4(15 - 1) = -4(14) = -56 \) 3. \( 3(3 \cdot 2 - 0 \cdot 1) = 3(6 - 0) = 3(6) = 18 \) Adding these results together: \[ \text{det}(A) = -4 - 56 + 18 = -42 \] Therefore, the determinant of matrix \( A \) is: \[ \boxed{-42} \] |
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