| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The determinant of a 3x3 matrix can be calculated using the formula: 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. In this case, the matrix is: A = [[0, 1, 2], [7, 8, 3], [6, 5, 4]] So, a = 0, b = 1, c = 2, d = 7, e = 8, f = 3, g = 6, h = 5, and i = 4. Substituting these values into the formula, we get: det(A) = 0(8*4 - 3*5) - 1(7*4 - 3*6) + 2(7*5 - 8*6) = 0(32 - 15) - 1(28 - 18) + 2(35 - 48) = 0 - 10 - 26 = -36 Therefore, the determinant of the matrix is -36. |
-36 |
| deepseekr1 | The determinant of the matrix [[0, 1, 2], [7, 8, 3], [6, 5, 4]] can be calculated using the cofactor expansion method. Expanding along the first row: \[ \begin{align*} \text{det} &= 0 \cdot \begin{vmatrix}8 & 3 \\ 5 & 4\end{vmatrix} - 1 \cdot \begin{vmatrix}7 & 3 \\ 6 & 4\end{vmatrix} + 2 \cdot \begin{vmatrix}7 & 8 \\ 6 & 5\end{vmatrix} \\ &= 0 \cdot (8 \cdot 4 - 3 \cdot 5) - 1 \cdot (7 \cdot 4 - 3 \cdot 6) + 2 \cdot (7 \cdot 5 - 8 \cdot 6) \\ &= 0 - 1 \cdot (28 - 18) + 2 \cdot (35 - 48) \\ &= 0 - 10 + 2 \cdot (-13) \\ &= -10 - 26 \\ &= -36 \end{align*} \] **Answer:** The determinant is \boxed{-36}. |
-36 |
| gpto1 | To find the determinant of the given \( 3 \times 3 \) matrix: \[ \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \\ \end{bmatrix} \] We can use the method of expanding the determinant by minors (also known as the cofactor expansion). The determinant of a \( 3 \times 3 \) matrix: \[ \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \] is calculated as: \[ \begin{align*} \det(A) = &\ a_{11}(a_{22}a_{33} - a_{23}a_{32}) \\ & - a_{12}(a_{21}a_{33} - a_{23}a_{31}) \\ & + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \end{align*} \] Let's identify the elements from the given matrix: - \( a_{11} = 0 \) - \( a_{12} = 1 \) - \( a_{13} = 2 \) - \( a_{21} = 7 \) - \( a_{22} = 8 \) - \( a_{23} = 3 \) - \( a_{31} = 6 \) - \( a_{32} = 5 \) - \( a_{33} = 4 \) Now, we compute each term step by step. **First Term:** \[ T_1 = a_{11}(a_{22}a_{33} - a_{23}a_{32}) = 0 \times (8 \times 4 - 3 \times 5) \] Calculating inside the parentheses: \[ 8 \times 4 - 3 \times 5 = 32 - 15 = 17 \] But since \( a_{11} = 0 \): \[ T_1 = 0 \times 17 = 0 \] **Second Term:** \[ T_2 = -a_{12}(a_{21}a_{33} - a_{23}a_{31}) = -1 \times (7 \times 4 - 3 \times 6) \] Calculating inside the parentheses: \[ 7 \times 4 - 3 \times 6 = 28 - 18 = 10 \] Thus: \[ T_2 = -1 \times 10 = -10 \] **Third Term:** \[ T_3 = a_{13}(a_{21}a_{32} - a_{22}a_{31}) = 2 \times (7 \times 5 - 8 \times 6) \] Calculating inside the parentheses: \[ 7 \times 5 - 8 \times 6 = 35 - 48 = -13 \] Thus: \[ T_3 = 2 \times (-13) = -26 \] **Adding All Terms:** \[ \det(A) = T_1 + T_2 + T_3 = 0 + (-10) + (-26) = -36 \] **Conclusion:** The determinant of the matrix is **\(-36\)**. |
-36 |
| GPT-4o | To find the determinant of a 3x3 matrix like \(\begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix}\), we can use the formula for the determinant of a 3x3 matrix given by: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where the matrix is: \[ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \] For the matrix \( \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix} \): - \(a = 0\) - \(b = 1\) - \(c = 2\) - \(d = 7\) - \(e = 8\) - \(f = 3\) - \(g = 6\) - \(h = 5\) - \(i = 4\) Now, plug these values into the formula: \[ \text{det}(A) = 0 \cdot (8 \cdot 4 - 3 \cdot 5) - 1 \cdot (7 \cdot 4 - 3 \cdot 6) + 2 \cdot (7 \cdot 5 - 8 \cdot 6) \] Calculate each term separately: 1. \(0 \cdot (32 - 15) = 0 \cdot 17 = 0\) 2. \(-1 \cdot (28 - 18) = -1 \cdot 10 = -10\) 3. \(2 \cdot (35 - 48) = 2 \cdot (-13) = -26\) Add them together: \[ \text{det}(A) = 0 - 10 - 26 = -36 \] Therefore, the determinant of the matrix \(\begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix}\) is \(-36\). |
-36 |
| llama3.1-70B-Instruct | To find the determinant of the matrix [[0, 1, 2], [7, 8, 3], [6, 5, 4]], 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. In this case, we have: a = 0, b = 1, c = 2 d = 7, e = 8, f = 3 g = 6, h = 5, i = 4 Plugging these values into the formula, we get: det(A) = 0(8*4 - 3*5) - 1(7*4 - 3*6) + 2(7*5 - 8*6) = 0(32 - 15) - 1(28 - 18) + 2(35 - 48) = 0(17) - 1(10) + 2(-13) = 0 - 10 - 26 = -36 Therefore, the determinant of the matrix [[0, 1, 2], [7, 8, 3], [6, 5, 4]] is -36. |
-36 |
| llama3.1-405B-Instruct | def determinant(matrix): # Base case for 2x2 matrix if len(matrix) == 2: return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0] det = 0 for c in range(len(matrix)): # Build sub-matrix for cofactor sub_matrix = matrix[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 * matrix[0][c] * sub_det return det matrix = [[0, 1, 2], [7, 8, 3], [6, 5, 4]] print(determinant(matrix)) |
-36 |
| DeepSeek-Coder-V2 | To find the determinant of the matrix [[0, 1, 2], [7, 8, 3], [6, 5, 4]], we will use the method of cofactor expansion along the first row. The matrix is: \[ A = \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix} \] The determinant of a 3x3 matrix \( A \) is given by: \[ \text{det}(A) = a_{11} \cdot \text{det}(A_{11}) - a_{12} \cdot \text{det}(A_{12}) + a_{13} \cdot \text{det}(A_{13}) \] Where \( a_{ij} \) are the elements of the first row, and \( A_{ij} \) are the 2x2 submatrices obtained by removing the i-th row and j-th column. 1. Calculate \( \text{det}(A_{11}) \): \[ A_{11} = \begin{bmatrix} 8 & 3 \\ 5 & 4 \end{bmatrix} \] \[ \text{det}(A_{11}) = 8 \cdot 4 - 3 \cdot 5 = 32 - 15 = 17 \] 2. Calculate \( \text{det}(A_{12}) \): \[ A_{12} = \begin{bmatrix} 7 & 3 \\ 6 & 4 \end{bmatrix} \] \[ \text{det}(A_{12}) = 7 \cdot 4 - 3 \cdot 6 = 28 - 18 = 10 \] 3. Calculate \( \text{det}(A_{13}) \): \[ A_{13} = \begin{bmatrix} 7 & 8 \\ 6 & 5 \end{bmatrix} \] \[ \text{det}(A_{13}) = 7 \cdot 5 - 8 \cdot 6 = 35 - 48 = -13 \] Now, substitute these determinants back into the formula: \[ \text{det}(A) = 0 \cdot 17 - 1 \cdot 10 + 2 \cdot (-13) \] \[ \text{det}(A) = 0 - 10 - 26 \] \[ \text{det}(A) = -36 \] Therefore, the determinant of the matrix [[0, 1, 2], [7, 8, 3], [6, 5, 4]] is: \[ \boxed{-36} \] |
-36 |
| Qwen2-72B-Instruct | The determinant of a 3x3 matrix can be calculated using the formula: 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. In this case, the matrix is: A = [[0, 1, 2], [7, 8, 3], [6, 5, 4]] So, a = 0, b = 1, c = 2, d = 7, e = 8, f = 3, g = 6, h = 5, and i = 4. Substituting these values into the formula, we get: det(A) = 0(8*4 - 3*5) - 1(7*4 - 3*6) + 2(7*5 - 8*6) = 0(32 - 15) - 1(28 - 18) + 2(35 - 48) = 0 - 10 - 26 = -36 Therefore, the determinant of the matrix is -36. |
-36 |
| baichuan4 | To find the determinant of the matrix: ``` | 0 1 2 | | 7 8 3 | | 6 5 4 | ``` We can use the rule of Sarrus to calculate the determinant of a 3x3 matrix. The rule of Sarrus involves writing the first two columns of the matrix to the right of the matrix and then summing up the products of the diagonals going from bottom-left to top-right (positive diagonals), and subtracting the products of the diagonals going from top-left to bottom-right (negative diagonals). So, we have: ``` 0 1 2 0 1 7 8 3 7 8 6 5 4 6 5 ``` Now, let's calculate the diagonal products: Positive diagonals: - 0 * 8 * 4 = 0 - 1 * 3 * 6 = 18 - 2 * 7 * 5 = 70 Negative diagonals: - 2 * 8 * 6 = 96 - 1 * 3 * 0 = 0 - 0 * 7 * 5 = 0 Sum of positive diagonals: 0 + 18 + 70 = 88 Sum of negative diagonals: 96 + 0 + 0 = 96 Determinant = 88 - 96 = -8 Therefore, the determinant of the matrix is -8. |
-36 |
| doubao | The determinant of a 3x3 matrix [[a,b,c],[d,e,f],[g,h,i]] is given by: \[ \begin{align*} &a(ei-fh)-b(di-fg)+c(dh-eg)\\ \end{align*} \] For the given matrix [[0,1,2],[7,8,3],[6,5,4]], the determinant is: \[ \begin{align*} &0\times(8\times4-3\times5)-1\times(7\times4-3\times6)+2\times(7\times5-8\times6)\\ &=0-1\times(28-18)+2\times(35-48)\\ &=-1\times10+2\times(-13)\\ &=-10-26\\ &=-36 \end{align*} \] So the determinant is -36. |
-36 |
| minimax | To find the determinant of a 3x3 matrix, you can use the following formula: For a matrix A = [[a, b, c], [d, e, f], [g, h, i]], the determinant (denoted as det(A) or |A|) is calculated as: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) Let's apply this formula to the given matrix: A = [[0, 1, 2], [7, 8, 3], [6, 5, 4]] det(A) = 0(8*4 - 3*5) - 1(7*4 - 3*6) + 2(7*5 - 8*6) = 0(32 - 15) - 1(28 - 18) + 2(35 - 48) = 0(17) - 1(10) + 2(-13) = 0 - 10 - 26 = -10 - 26 = -36 So, the determinant of the matrix A is -36. |
-36 |
| kimi | To find the determinant of a 3x3 matrix, you can use the following formula: For a matrix A: \[ A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \] The determinant of A, denoted as |A|, is calculated as: \[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \] Given the matrix: \[ A = \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix} \] We can plug in the values into the formula: \[ |A| = 0(8*4 - 3*5) - 1(7*4 - 3*6) + 2(7*5 - 8*6) \] \[ |A| = 0 - 1(28 - 18) + 2(35 - 48) \] \[ |A| = 0 - 1(10) + 2(-13) \] \[ |A| = 0 - 10 - 26 \] \[ |A| = -36 \] So, the determinant of the given matrix is -36. |
-36 |
| wenxinyiyan | To find the determinant of the given matrix, we will use the cofactor expansion method. We will expand along the first row for this example. Given matrix: | 0 1 2 | | 7 8 3 | | 6 5 4 | We will expand along the first row, which means we will multiply each element in the first row by its cofactor and then add them together. Cofactor of a matrix element is the determinant of the matrix formed by deleting the row and column containing that element. It also includes a sign, which is positive for elements in the first row if their column number is odd and negative if their column number is even. So, the determinant calculation will be: 0 * cofactor(0,1,2) + 1 * cofactor(1,8,3) + 2 * cofactor(2,7,3) Let's calculate each cofactor: 1. Cofactor of element (0,1,2) is not needed since 0 multiplied by any number is 0. 2. Cofactor of element (1,8,3) = determinant of |7 3| = 7*4 - 6*3 = 28 - 18 = 10 3. Cofactor of element (2,7,3) = determinant of |7 8| = 7*4 - 6*5 = 28 - 30 = -2 Now, let's multiply each element by its cofactor and add them: 0 * 0 + 1 * 10 + 2 * (-2) = 0 + 10 - 4 = 6 So, the determinant of the given matrix is 6. |
-36 |
| zhipu | To find the determinant of a 3x3 matrix, you can use the following formula: For a matrix \( A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \), the determinant (det(A)) is given by: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Given the matrix: \[ A = \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix} \] we can label the elements as follows: - \( a = 0 \) - \( b = 1 \) - \( c = 2 \) - \( d = 7 \) - \( e = 8 \) - \( f = 3 \) - \( g = 6 \) - \( h = 5 \) - \( i = 4 \) Now plug these values into the determinant formula: \[ \text{det}(A) = 0 \cdot (8 \cdot 4 - 3 \cdot 5) - 1 \cdot (7 \cdot 4 - 3 \cdot 6) + 2 \cdot (7 \cdot 5 - 8 \cdot 6) \] Calculate each term separately: - \( 0 \cdot (8 \cdot 4 - 3 \cdot 5) = 0 \cdot (32 - 15) = 0 \cdot 17 = 0 \) - \( -1 \cdot (7 \cdot 4 - 3 \cdot 6) = -1 \cdot (28 - 18) = -1 \cdot 10 = -10 \) - \( 2 \cdot (7 \cdot 5 - 8 \cdot 6) = 2 \cdot (35 - 48) = 2 \cdot (-13) = -26 \) Now sum these results: \[ \text{det}(A) = 0 - 10 - 26 = -36 \] So the determinant of the matrix \( \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix} \) is \( -36 \). |
-36 |
| gpto1 | To find the determinant of the given \( 3 \times 3 \) matrix: \[ \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \\ \end{bmatrix} \] We can use the method of expanding the determinant by minors (also known as the cofactor expansion). The determinant of a \( 3 \times 3 \) matrix: \[ \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \] is calculated as: \[ \begin{align*} \det(A) = &\ a_{11}(a_{22}a_{33} - a_{23}a_{32}) \\ & - a_{12}(a_{21}a_{33} - a_{23}a_{31}) \\ & + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \end{align*} \] Let's identify the elements from the given matrix: - \( a_{11} = 0 \) - \( a_{12} = 1 \) - \( a_{13} = 2 \) - \( a_{21} = 7 \) - \( a_{22} = 8 \) - \( a_{23} = 3 \) - \( a_{31} = 6 \) - \( a_{32} = 5 \) - \( a_{33} = 4 \) Now, we compute each term step by step. **First Term:** \[ T_1 = a_{11}(a_{22}a_{33} - a_{23}a_{32}) = 0 \times (8 \times 4 - 3 \times 5) \] Calculating inside the parentheses: \[ 8 \times 4 - 3 \times 5 = 32 - 15 = 17 \] But since \( a_{11} = 0 \): \[ T_1 = 0 \times 17 = 0 \] **Second Term:** \[ T_2 = -a_{12}(a_{21}a_{33} - a_{23}a_{31}) = -1 \times (7 \times 4 - 3 \times 6) \] Calculating inside the parentheses: \[ 7 \times 4 - 3 \times 6 = 28 - 18 = 10 \] Thus: \[ T_2 = -1 \times 10 = -10 \] **Third Term:** \[ T_3 = a_{13}(a_{21}a_{32} - a_{22}a_{31}) = 2 \times (7 \times 5 - 8 \times 6) \] Calculating inside the parentheses: \[ 7 \times 5 - 8 \times 6 = 35 - 48 = -13 \] Thus: \[ T_3 = 2 \times (-13) = -26 \] **Adding All Terms:** \[ \det(A) = T_1 + T_2 + T_3 = 0 + (-10) + (-26) = -36 \] **Conclusion:** The determinant of the matrix is **\(-36\)**. |
-36 |
| deepseekv3 | To find the determinant of the matrix: \[ A = \begin{bmatrix} 0 & 1 & 2 \\ 7 & 8 & 3 \\ 6 & 5 & 4 \end{bmatrix} \] We can use the **cofactor expansion** method along the first row. The determinant of a \( 3 \times 3 \) matrix is calculated as: \[ \det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \] Substitute the values from matrix \( A \): \[ \det(A) = 0 \cdot (8 \cdot 4 - 3 \cdot 5) - 1 \cdot (7 \cdot 4 - 3 \cdot 6) + 2 \cdot (7 \cdot 5 - 8 \cdot 6) \] Simplify each term: 1. \( 0 \cdot (32 - 15) = 0 \) 2. \( -1 \cdot (28 - 18) = -10 \) 3. \( 2 \cdot (35 - 48) = 2 \cdot (-13) = -26 \) Add the results: \[ \det(A) = 0 - 10 - 26 = -36 \] Thus, the determinant of the matrix is: \[ \boxed{-36} \] |
-36 |