| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the determinant of the given matrix: \[ \left( \begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right) \] The determinant, calculated and rounded to two decimal places, is: \[ \boxed{-38.03} \] |
-\frac{1369}{36} |
| deepseek-r1 | To find the determinant of the matrix \[ \begin{pmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \\ \end{pmatrix} \] we first factor out common denominators from the rows. Factoring out \(\frac{1}{6}\) from the first and second rows, and \(\frac{1}{3}\) from the third row, the determinant becomes \(\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{3} = \frac{1}{108}\) times the determinant of the matrix with integer entries: \[ \begin{pmatrix} -8 & 24 & -11 \\ -26 & 23 & -25 \\ 5 & -2 & -5 \\ \end{pmatrix} \] We compute the determinant of this integer matrix using the rule of Sarrus or cofactor expansion. Expanding along the third row for simplicity: \[ \text{det} = 5 \cdot \left( 24 \cdot (-25) - (-11) \cdot 23 \right) - (-2) \cdot \left( -8 \cdot (-25) - (-11) \cdot (-26) \right) + (-5) \cdot \left( -8 \cdot 23 - 24 \cdot (-26) \right) \] Calculating each term: 1. \(5 \cdot (-600 + 253) = 5 \cdot (-347) = -1735\) 2. \(-(-2) \cdot (200 - 286) = -2 \cdot (-86) = -172\) 3. \((-5) \cdot (-184 + 624) = -5 \cdot 440 = -2200\) Summing these terms: \[ -1735 - 172 - 2200 = -4107 \] Thus, the determinant of the integer matrix is \(-4107\). Multiplying by the factor \(\frac{1}{108}\) gives: \[ \frac{-4107}{108} = -\frac{1369}{36} \] Therefore, the determinant of the original matrix is \[ \boxed{-\dfrac{1369}{36}} \] |
-\frac{1369}{36} |
| gpto1 | To find the determinant of the given matrix: \[ A = \begin{pmatrix} -\dfrac{4}{3} & 4 & -\dfrac{11}{6} \\ -\dfrac{13}{3} & \dfrac{23}{6} & -\dfrac{25}{6} \\ \dfrac{5}{3} & -\dfrac{2}{3} & -\dfrac{5}{3} \end{pmatrix} \] We will compute the determinant using cofactor expansion along the first row. The determinant is given by: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Let's compute the minors and cofactors: 1. **Compute \( M_{11} \):** \[ M_{11} = \begin{vmatrix} \dfrac{23}{6} & -\dfrac{25}{6} \\ -\dfrac{2}{3} & -\dfrac{5}{3} \end{vmatrix} = \left( \dfrac{23}{6} \cdot -\dfrac{5}{3} \right) - \left( -\dfrac{25}{6} \cdot -\dfrac{2}{3} \right) = -\dfrac{115}{18} - \dfrac{50}{18} = -\dfrac{65}{18} \] 2. **Compute \( M_{12} \):** \[ M_{12} = \begin{vmatrix} -\dfrac{13}{3} & -\dfrac{25}{6} \\ \dfrac{5}{3} & -\dfrac{5}{3} \end{vmatrix} = \left( -\dfrac{13}{3} \cdot -\dfrac{5}{3} \right) - \left( -\dfrac{25}{6} \cdot \dfrac{5}{3} \right) = \dfrac{65}{9} + \dfrac{125}{18} = \dfrac{255}{18} = \dfrac{85}{6} \] 3. **Compute \( M_{13} \):** \[ M_{13} = \begin{vmatrix} -\dfrac{13}{3} & \dfrac{23}{6} \\ \dfrac{5}{3} & -\dfrac{2}{3} \end{vmatrix} = \left( -\dfrac{13}{3} \cdot -\dfrac{2}{3} \right) - \left( \dfrac{23}{6} \cdot \dfrac{5}{3} \right) = \dfrac{26}{9} - \dfrac{115}{18} = -\dfrac{7}{2} \] Now, plug these back into the determinant: \[ \det(A) = \left( -\dfrac{4}{3} \right) \left( -\dfrac{65}{18} \right) - 4 \left( \dfrac{85}{6} \right) + \left( -\dfrac{11}{6} \right) \left( -\dfrac{7}{2} \right) \] Compute each term: 1. **First term:** \[ \left( -\dfrac{4}{3} \right) \left( -\dfrac{65}{18} \right) = \dfrac{260}{54} = \dfrac{130}{27} \] 2. **Second term:** \[ - 4 \left( \dfrac{85}{6} \right) = - \dfrac{340}{6} = -\dfrac{170}{3} \] 3. **Third term:** \[ \left( -\dfrac{11}{6} \right) \left( -\dfrac{7}{2} \right) = \dfrac{77}{12} \] Add them together: \[ \det(A) = \dfrac{130}{27} - \dfrac{170}{3} + \dfrac{77}{12} \] Compute with a common denominator (108): \[ \det(A) = \dfrac{520}{108} - \dfrac{6120}{108} + \dfrac{693}{108} = \dfrac{(520 - 6120 + 693)}{108} = \dfrac{-4907}{108} \] Thus, the determinant of the given matrix is: \[ \boxed{\dfrac{-4907}{108}} \] \boxed{\dfrac{-4907}{108}} |
-\frac{1369}{36} |
| gpt-4o-2024-08-06 | To find the determinant of the given \(3 \times 3\) matrix, we will use the formula for the determinant of a \(3 \times 3\) matrix: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where the matrix \(A\) is: \[ A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \] For the given matrix: \[ A = \begin{pmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{pmatrix} \] we have: - \(a = -\frac{4}{3}\) - \(b = 4\) - \(c = -\frac{11}{6}\) - \(d = -\frac{13}{3}\) - \(e = \frac{23}{6}\) - \(f = -\frac{25}{6}\) - \(g = \frac{5}{3}\) - \(h = -\frac{2}{3}\) - \(i = -\frac{5}{3}\) Now, we calculate each of the terms in the determinant formula: 1. Calculate \(ei - fh\): \[ ei = \frac{23}{6} \cdot \left(-\frac{5}{3}\right) = -\frac{115}{18} \] \[ fh = -\frac{25}{6} \cdot \left(-\frac{2}{3}\right) = \frac{50}{18} \] \[ ei - fh = -\frac{115}{18} - \frac{50}{18} = -\frac{165}{18} = -\frac{55}{6} \] 2. Calculate \(di - fg\): \[ di = -\frac{13}{3} \cdot \left(-\frac{5}{3}\right) = \frac{65}{9} \] \[ fg = -\frac{25}{6} \cdot \frac{5}{3} = -\frac{125}{18} \] \[ di - fg = \frac{65}{9} + \frac{125}{18} = \frac{130}{18} + \frac{125}{18} = \frac{255}{18} = \frac{85}{6} \] 3. Calculate \(dh - eg\): \[ dh = -\frac{13}{3} \cdot \left(-\frac{2}{3}\right) = \frac{26}{9} \] \[ eg = \frac{23}{6} \cdot \frac{5}{3} = \frac{115}{18} \] \[ dh - eg = \frac{26}{9} - \frac{115}{18} = \frac{52}{18} - \frac{115}{18} = -\frac{63}{18} = -\frac{7}{2} \] Now substitute these into the determinant formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] \[ = -\frac{4}{3} \cdot \left(-\frac{55}{6}\right) - 4 \cdot \frac{85}{6} + \left(-\frac{11}{6}\right) \cdot \left(-\frac{7}{2}\right) \] Calculate each term: \[ -\frac{4}{3} \cdot \left(-\frac{55}{6}\right) = \frac{220}{18} = \frac{110}{9} \] \[ -4 \cdot \frac{85}{6} = -\frac{340}{6} = -\frac{170}{3} \] \[ -\frac{11}{6} \cdot \left(-\frac{7}{2}\right) = \frac{77}{12} \] Combine these results: \[ \text{det}(A) = \frac{110}{9} - \frac{170}{3} + \frac{77}{12} \] Convert to a common denominator (36): \[ \frac{110}{9} = \frac{440}{36}, \quad -\frac{170}{3} = -\frac{2040}{36}, \quad \frac{77}{12} = \frac{231}{36} \] \[ \text{det}(A) = \frac{440}{36} - \frac{2040}{36} + \frac{231}{36} = \frac{440 + 231 - 2040}{36} = \frac{-1369}{36} \] Thus, the determinant of the matrix is: \[ \boxed{-\frac{1369}{36}} \] |
-\frac{1369}{36} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the determinant of the given matrix, we will use the method of cofactor expansion along the first row. The matrix is: \[ \left( \begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right) \] The determinant of a \(3 \times 3\) matrix \(\left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \\ \end{array} \right)\) is given by: \[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \] For our matrix, we have: - \(a = -\frac{4}{3}\) - \(b = 4\) - \(c = -\frac{11}{6}\) - \(d = -\frac{13}{3}\) - \(e = \frac{23}{6}\) - \(f = -\frac{25}{6}\) - \(g = \frac{5}{3}\) - \(h = -\frac{2}{3}\) - \(i = -\frac{5}{3}\) Now, we calculate each term: 1. \(a(ei - fh)\): \[ -\frac{4}{3} \left( \frac{23}{6} \cdot -\frac{5}{3} - \left( -\frac{25}{6} \right) \cdot -\frac{2}{3} \right) \] \[ = -\frac{4}{3} \left( -\frac{115}{18} - \frac{50}{18} \right) \] \[ = -\frac{4}{3} \left( -\frac{165}{18} \right) \] \[ = -\frac{4}{3} \cdot -\frac{55}{6} \] \[ = \frac{220}{18} \] \[ = \frac{110}{9} \] 2. \(-b(di - fg)\): \[ -4 \left( -\frac{13}{3} \cdot -\frac{5}{3} - \left( -\frac{25}{6} \right) \cdot \frac{5}{3} \right) \] \[ = -4 \left( -\frac{65}{9} - \left( -\frac{125}{18} \right) \right) \] \[ = -4 \left( -\frac{65}{9} + \frac{125}{18} \right) \] \[ = -4 \left( -\frac{130}{18} + \frac{125}{18} \right) \] \[ = -4 \left( -\frac{5}{18} \right) \] \[ = \frac{20}{18} \] \[ = \frac{10}{9} \] 3. \(c(dh - eg)\): \[ -\frac{11}{6} \left( -\frac{13}{3} \cdot -\frac{2}{3} - \frac{23}{6} \cdot \frac{5}{3} \right) \] \[ = -\frac{11}{6} \left( -\frac{26}{9} - \frac{115}{18} \right) \] \[ = -\frac{11}{6} \left( -\frac{52}{18} - \frac{115}{18} \right) \] \[ = -\frac{11}{6} \left( -\frac{167}{18} \right) \] \[ = \frac{1837}{108} \] \[ = \frac{11}{6} \cdot \frac{167}{18} \] \[ = \frac{1837}{108} \] Adding these results together: \[ \frac{110}{9} - \frac{10}{9} + \frac{1837}{108} \] To add these fractions, we need a common denominator. The common denominator is 108: \[ \frac{110}{9} = \frac{110 \times 12}{108} = \frac{1320}{108} \] \[ \frac{10}{9} = \frac{10 \times 12}{108} = \frac{120}{108} \] So, \[ \frac{1320}{108} - \frac{120}{108} + \frac{1837}{108} = \frac{1320 - 120 + 1837}{108} = \frac{2037}{108} = \frac{679}{36} \] Thus, the determinant of the matrix is: \[ \boxed{\frac{679}{36}} \] |
-\frac{1369}{36} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: To find the determinant of the given 3x3 matrix, we will use the formula for the determinant of a 3x3 matrix. The formula for the determinant of a 3x3 matrix is: \[\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)\] ## Step 2: Identify the elements of the matrix. The given matrix is: \[\begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{array}\] So, a = -\frac{4}{3}, b = 4, c = -\frac{11}{6}, d = -\frac{13}{3}, e = \frac{23}{6}, f = -\frac{25}{6}, g = \frac{5}{3}, h = -\frac{2}{3}, and i = -\frac{5}{3}. ## Step 3: Plug the elements of the matrix into the determinant formula. Substituting the elements of the matrix into the determinant formula, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = -\frac{4}{3}(\frac{23}{6}(-\frac{5}{3}) - (-\frac{25}{6})(-\frac{2}{3})) - 4((-\frac{13}{3})(-\frac{5}{3}) - (-\frac{25}{6})(\frac{5}{3})) + (-\frac{11}{6})((-\frac{13}{3})(-\frac{2}{3}) - (\frac{23}{6})(\frac{5}{3}))\] ## Step 4: Simplify the expression. Expanding and simplifying the expression, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = -\frac{4}{3}(-\frac{115}{18} - \frac{25}{9}) - 4(\frac{65}{9} + \frac{125}{18}) - \frac{11}{6}(\frac{26}{9} - \frac{115}{18})\] ## Step 5: Continue simplifying the expression. Continuing to simplify the expression, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = -\frac{4}{3}(-\frac{115}{18} - \frac{50}{18}) - 4(\frac{130}{18} + \frac{125}{18}) - \frac{11}{6}(\frac{52}{18} - \frac{115}{18})\] ## Step 6: Further simplify the expression. Further simplifying the expression, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = -\frac{4}{3}(-\frac{165}{18}) - 4(\frac{255}{18}) - \frac{11}{6}(-\frac{63}{18})\] ## Step 7: Continue simplifying the expression. Continuing to simplify the expression, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = \frac{4}{3}(\frac{165}{18}) - \frac{4}{1}(\frac{255}{18}) + \frac{11}{6}(\frac{63}{18})\] ## Step 8: Further simplify the expression. Further simplifying the expression, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = \frac{660}{54} - \frac{1020}{18} + \frac{693}{108}\] ## Step 9: Simplify the fractions. Simplifying the fractions, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = \frac{110}{9} - \frac{170}{3} + \frac{77}{12}\] ## Step 10: Find a common denominator for the fractions. Finding a common denominator for the fractions, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = \frac{440}{36} - \frac{2040}{36} + \frac{231}{36}\] ## Step 11: Combine the fractions. Combining the fractions, we get: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = -\frac{1369}{36}\] The final answer is: $\boxed{-\frac{1369}{36}}$ |
-\frac{1369}{36} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: To find the determinant of the given 3x3 matrix, we will use the formula for the determinant of a 3x3 matrix. The formula for the determinant of a 3x3 matrix is given by: \[\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a(ei - fh) - b(di - fg) + c(dh - eg)\] ## Step 2: Identify the elements of the matrix. From the given matrix, we can identify the elements as follows: a = -\frac{4}{3}, b = 4, c = -\frac{11}{6}, d = -\frac{13}{3}, e = \frac{23}{6}, f = -\frac{25}{6}, g = \frac{5}{3}, h = -\frac{2}{3}, i = -\frac{5}{3} ## Step 3: Apply the formula for the determinant of a 3x3 matrix. Substitute the identified elements into the formula: \[\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix} = -\frac{4}{3}(\frac{23}{6}(-\frac{5}{3}) - (-\frac{25}{6})(-\frac{2}{3})) - 4(-\frac{13}{3}(-\frac{5}{3}) - (-\frac{25}{6})(\frac{5}{3})) + (-\frac{11}{6})(-\frac{13}{3}(-\frac{2}{3}) - \frac{23}{6}(\frac{5}{3}))\] ## Step 4: Simplify the expression. Simplify the expression by performing the arithmetic operations: \[= -\frac{4}{3}(-\frac{115}{18} - \frac{50}{18}) - 4(\frac{65}{9} + \frac{125}{18}) - \frac{11}{6}(\frac{26}{9} - \frac{115}{18})\] \[= -\frac{4}{3}(-\frac{165}{18}) - 4(\frac{65}{9} + \frac{125}{18}) - \frac{11}{6}(\frac{26}{9} - \frac{115}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - 4(\frac{65}{9} + \frac{125}{18}) - \frac{11}{6}(\frac{26}{9} - \frac{115}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - 4(\frac{130}{18} + \frac{125}{18}) - \frac{11}{6}(\frac{52}{18} - \frac{115}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - 4(\frac{255}{18}) - \frac{11}{6}(-\frac{63}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - \frac{255}{9} \cdot \frac{4}{2} + \frac{11}{6}(\frac{63}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - \frac{255}{9} \cdot \frac{2}{1} + \frac{11}{6}(\frac{63}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - \frac{510}{9} + \frac{11}{6}(\frac{63}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - \frac{170}{3} + \frac{11}{6}(\frac{63}{18})\] \[= \frac{4}{3}(\frac{165}{18}) - \frac{170}{3} + \frac{11}{6}(\frac{7}{2})\] \[= \frac{4}{3}(\frac{165}{18}) - \frac{170}{3} + \frac{77}{12}\] \[= \frac{4}{3}(\frac{55}{6}) - \frac{170}{3} + \frac{77}{12}\] \[= \frac{220}{18} - \frac{170}{3} + \frac{77}{12}\] \[= \frac{220}{18} - \frac{1020}{18} + \frac{77}{12}\] \[= \frac{220 - 1020}{18} + \frac{77}{12}\] \[= \frac{-800}{18} + \frac{77}{12}\] \[= \frac{-800}{18} + \frac{77 \cdot 3}{12 \cdot 3}\] \[= \frac{-800}{18} + \frac{231}{36}\] \[= \frac{-800 \cdot 2}{18 \cdot 2} + \frac{231}{36}\] \[= \frac{-1600}{36} + \frac{231}{36}\] \[= \frac{-1600 + 231}{36}\] \[= \frac{-1369}{36}\] \[= -\frac{1369}{36}\] The final answer is: $\boxed{-\frac{1369}{36}}$ |
-\frac{1369}{36} |
| Qwen2-72B-Instruct | To find the determinant of the given matrix, we can use the formula for the determinant of a 3x3 matrix: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] where \(a, b, c\) are the elements of the first row, and \(d, e, f\) are the elements of the second row, and \(g, h, i\) are the elements of the third row. Given matrix: \[ \left( \begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right) \] Let's calculate the determinant using the formula: \[ \text{det}(A) = \left(-\frac{4}{3}\right)\left(\left(\frac{23}{6}\right)\left(-\frac{5}{3}\right) - \left(-\frac{2}{3}\right)\left(-\frac{25}{6}\right)\right) - 4\left(\left(-\frac{13}{3}\right)\left(-\frac{5}{3}\right) - \left(\frac{5}{3}\right)\left(-\frac{25}{6}\right)\right) + \left(-\frac{11}{6}\right)\left(\left(-\frac{13}{3}\right)\left(-\frac{2}{3}\right) - \left(\frac{5}{3}\right)\left(\frac{23}{6}\right)\right) \] Simplifying each term: \[ \text{det}(A) = \left(-\frac{4}{3}\right)\left(-\frac{115}{18} - \frac{50}{18}\right) - 4\left(\frac{65}{9} + \frac{125}{18}\right) + \left(-\frac{11}{6}\right)\left(\frac{26}{9} - \frac{115}{18}\right) \] \[ \text{det}(A) = \left(-\frac{4}{3}\right)\left(-\frac{165}{18}\right) - 4\left(\frac{130 + 125}{18}\right) - \frac{11}{6}\left(\frac{52 - 115}{18}\right) \] \[ \text{det}(A) = \frac{4}{3}\cdot\frac{165}{18} - 4\cdot\frac{255}{18} - \frac{11}{6}\cdot\frac{-63}{18} \] \[ \text{det}(A) = \frac{660}{54} - \frac{1020}{18} + \frac{693}{108} \] \[ \text{det}(A) = \frac{660}{54} - \frac{2040}{54} + \frac{693}{108} \] \[ \text{det}(A) = \frac{660 - 2040}{54} + \frac{693}{108} \] \[ \text{det}(A) = -\frac{1380}{54} + \frac{693}{108} \] \[ \text{det}(A) = -\frac{2760}{108} + \frac{693}{108} \] \[ \text{det}(A) = -\frac{2067}{108} \] \[ \text{det}(A) = -\frac{689}{36} \] Therefore, the determinant of the given matrix is \(-\frac{689}{36}\). \[ \boxed{-\frac{689}{36}} \] |
-\frac{1369}{36} |
| baichuan4 | To find the determinant of the given matrix, we can use the standard method of expanding the determinant along any row or column. In this case, we will expand along the first row for convenience. The matrix is: $$ A = \left( \begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right) $$ The determinant of $A$ can be calculated as: $$ \det(A) = -\frac{4}{3} \cdot \left| \begin{array}{cc} \frac{23}{6} & -\frac{25}{6} \\ -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right| - 4 \cdot \left| \begin{array}{cc} -\frac{13}{3} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{5}{3} \\ \end{array} \right| - \frac{11}{6} \cdot \left| \begin{array}{cc} -\frac{13}{3} & \frac{23}{6} \\ \frac{5}{3} & -\frac{2}{3} \\ \end{array} \right| $$ Now, we calculate each of these 2x2 determinants: $$ \left| \begin{array}{cc} \frac{23}{6} & -\frac{25}{6} \\ -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right| = \frac{23}{6} \cdot (-\frac{5}{3}) - (-\frac{25}{6}) \cdot (-\frac{2}{3}) = -\frac{115}{18} - \frac{50}{18} = -\frac{165}{18} = -\frac{55}{6} $$ $$ \left| \begin{array}{cc} -\frac{13}{3} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{5}{3} \\ \end{array} \right| = -\frac{13}{3} \cdot (-\frac{5}{3}) - (-\frac{25}{6}) \cdot \frac{5}{3} = \frac{65}{9} - \frac{125}{18} = \frac{130}{18} - \frac{125}{18} = \frac{5}{18} $$ $$ \left| \begin{array}{cc} -\frac{13}{3} & \frac{23}{6} \\ \frac{5}{3} & -\frac{2}{3} \\ \end{array} \right| = -\frac{13}{3} \cdot (-\frac{2}{3}) - \frac{23}{6} \cdot \frac{5}{3} = \frac{26}{9} - \frac{115}{18} = \frac{52}{18} - \frac{115}{18} = -\frac{63}{18} = -\frac{7}{2} $$ Now, we substitute these values back into the determinant of $A$: $$ \det(A) = -\frac{4}{3} \cdot (-\frac{55}{6}) - 4 \cdot \frac{5}{18} - \frac{11}{6} \cdot (-\frac{7}{2}) $$ $$ \det(A) = \frac{220}{18} - \frac{20}{18} + \frac{77}{12} $$ $$ \det(A) = \frac{200}{18} + \frac{77}{12} $$ To add these fractions, we need a common denominator. The least common multiple of 18 and 12 is 36, so we convert both fractions to have a denominator of 36: $$ \det(A) = \frac{200 \cdot 2}{36} + \frac{77 \cdot 3}{36} $$ $$ \det(A) = \frac{400}{36} + \frac{231}{36} $$ $$ \det(A) = \frac{631}{36} $$ Therefore, the determinant of the given matrix is: $$ \boxed{\frac{631}{36}} $$ |
-\frac{1369}{36} |
| doubao | The determinant of the matrix is: \[ \begin{align*} &\begin{vmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{vmatrix}\\ =&-\frac{4}{3}\times\begin{vmatrix} \frac{23}{6} & -\frac{25}{6} \\ -\frac{2}{3} & -\frac{5}{3} \end{vmatrix}-4\times\begin{vmatrix} -\frac{13}{3} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{5}{3} \end{vmatrix}-\frac{11}{6}\times\begin{vmatrix} -\frac{13}{3} & \frac{23}{6} \\ \frac{5}{3} & -\frac{2}{3} \end{vmatrix}\\ =&-\frac{4}{3}\times((\frac{23}{6}\times(-\frac{5}{3}))-(-\frac{25}{6}\times(-\frac{2}{3})))-4\times((-\frac{13}{3}\times(-\frac{5}{3}))-(-\frac{25}{6}\times\frac{5}{3}))-\frac{11}{6}\times((-\frac{13}{3}\times(-\frac{2}{3}))-(\frac{23}{6}\times\frac{5}{3}))\\ =&-\frac{4}{3}\times((-\frac{115}{18})-(\frac{25}{9}))-4\times((\frac{65}{9})-(-\frac{125}{18}))-\frac{11}{6}\times((\frac{26}{9})-(\frac{115}{18}))\\ =&-\frac{4}{3}\times(-\frac{115}{18}-\frac{50}{18})-4\times(\frac{65}{9}+\frac{125}{18})-\frac{11}{6}\times(\frac{52}{18}-\frac{115}{18})\\ =&-\frac{4}{3}\times(-\frac{165}{18})-4\times(\frac{130}{18}+\frac{125}{18})-\frac{11}{6}\times(-\frac{63}{18})\\ =&\frac{220}{9}-4\times\frac{255}{18}+\frac{77}{12}\\ =&\frac{220}{9}-\frac{510}{9}+\frac{77}{12}\\ =&-\frac{290}{9}+\frac{77}{12}\\ =&-\frac{1160}{36}+\frac{231}{36}\\ =&-\frac{929}{36} \end{align*} \] So the answer is $\boxed{-\frac{929}{36}}$. |
-\frac{1369}{36} |
| gpto1 | To find the determinant of the given matrix: \[ A = \begin{pmatrix} -\dfrac{4}{3} & 4 & -\dfrac{11}{6} \\ -\dfrac{13}{3} & \dfrac{23}{6} & -\dfrac{25}{6} \\ \dfrac{5}{3} & -\dfrac{2}{3} & -\dfrac{5}{3} \end{pmatrix} \] We will compute the determinant using cofactor expansion along the first row. The determinant is given by: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Let's compute the minors and cofactors: 1. **Compute \( M_{11} \):** \[ M_{11} = \begin{vmatrix} \dfrac{23}{6} & -\dfrac{25}{6} \\ -\dfrac{2}{3} & -\dfrac{5}{3} \end{vmatrix} = \left( \dfrac{23}{6} \cdot -\dfrac{5}{3} \right) - \left( -\dfrac{25}{6} \cdot -\dfrac{2}{3} \right) = -\dfrac{115}{18} - \dfrac{50}{18} = -\dfrac{65}{18} \] 2. **Compute \( M_{12} \):** \[ M_{12} = \begin{vmatrix} -\dfrac{13}{3} & -\dfrac{25}{6} \\ \dfrac{5}{3} & -\dfrac{5}{3} \end{vmatrix} = \left( -\dfrac{13}{3} \cdot -\dfrac{5}{3} \right) - \left( -\dfrac{25}{6} \cdot \dfrac{5}{3} \right) = \dfrac{65}{9} + \dfrac{125}{18} = \dfrac{255}{18} = \dfrac{85}{6} \] 3. **Compute \( M_{13} \):** \[ M_{13} = \begin{vmatrix} -\dfrac{13}{3} & \dfrac{23}{6} \\ \dfrac{5}{3} & -\dfrac{2}{3} \end{vmatrix} = \left( -\dfrac{13}{3} \cdot -\dfrac{2}{3} \right) - \left( \dfrac{23}{6} \cdot \dfrac{5}{3} \right) = \dfrac{26}{9} - \dfrac{115}{18} = -\dfrac{7}{2} \] Now, plug these back into the determinant: \[ \det(A) = \left( -\dfrac{4}{3} \right) \left( -\dfrac{65}{18} \right) - 4 \left( \dfrac{85}{6} \right) + \left( -\dfrac{11}{6} \right) \left( -\dfrac{7}{2} \right) \] Compute each term: 1. **First term:** \[ \left( -\dfrac{4}{3} \right) \left( -\dfrac{65}{18} \right) = \dfrac{260}{54} = \dfrac{130}{27} \] 2. **Second term:** \[ - 4 \left( \dfrac{85}{6} \right) = - \dfrac{340}{6} = -\dfrac{170}{3} \] 3. **Third term:** \[ \left( -\dfrac{11}{6} \right) \left( -\dfrac{7}{2} \right) = \dfrac{77}{12} \] Add them together: \[ \det(A) = \dfrac{130}{27} - \dfrac{170}{3} + \dfrac{77}{12} \] Compute with a common denominator (108): \[ \det(A) = \dfrac{520}{108} - \dfrac{6120}{108} + \dfrac{693}{108} = \dfrac{(520 - 6120 + 693)}{108} = \dfrac{-4907}{108} \] Thus, the determinant of the given matrix is: \[ \boxed{\dfrac{-4907}{108}} \] \boxed{\dfrac{-4907}{108}} |
-\frac{1369}{36} |
| kimi | To find the determinant of the given matrix, we will use the cofactor expansion along the first row. The matrix is: $\left(\begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{array}\right)$ Let's denote the matrix as A: $A = \left(\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right)$ where $a_{11} = -\frac{4}{3}$, $a_{12} = 4$, $a_{13} = -\frac{11}{6}$, $a_{21} = -\frac{13}{3}$, $a_{22} = \frac{23}{6}$, $a_{23} = -\frac{25}{6}$, $a_{31} = \frac{5}{3}$, $a_{32} = -\frac{2}{3}$, and $a_{33} = -\frac{5}{3}$. The determinant of A, denoted as $|A|$, can be calculated using the cofactor expansion along the first row: $|A| = a_{11}C_{11} - a_{12}C_{12} + a_{13}C_{13}$ where $C_{ij}$ is the cofactor of the element $a_{ij}$. First, we need to find the 2x2 determinants for each cofactor: $C_{11} = \left|\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right| = \left(\frac{23}{6} \cdot -\frac{5}{3}\right) - \left(-\frac{25}{6} \cdot -\frac{2}{3}\right) = -\frac{115}{18} - \frac{50}{18} = -\frac{165}{18}$ $C_{12} = -\left|\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right| = -\left(-\frac{13}{3} \cdot -\frac{5}{3} - \left(-\frac{25}{6} \cdot \frac{5}{3}\right)\right) = -\left(-\frac{65}{9} - \frac{125}{18}\right) = \frac{65}{9} + \frac{125}{18} = \frac{130}{18}$ $C_{13} = \left|\begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right| = \left(-\frac{13}{3} \cdot -\frac{2}{3}\right) - \left(\frac{23}{6} \cdot \frac{5}{3}\right) = \frac{26}{9} - \frac{115}{18} = -\frac{37}{18}$ Now, we can calculate the determinant: $|A| = a_{11}C_{11} - a_{12}C_{12} + a_{13}C_{13} = -\frac{4}{3} \cdot -\frac{165}{18} - 4 \cdot \frac{130}{18} - \frac{11}{6} \cdot -\frac{37}{18}$ $|A| = \frac{660}{54} - \frac{520}{54} + \frac{407}{108} = \frac{140}{54} + \frac{407}{108} = \frac{280}{108} + \frac{407}{108} = \frac{687}{108}$ So, the determinant of the given matrix is: $\boxed{\frac{687}{108}}$ |
-\frac{1369}{36} |
| minimax | To find the determinant of the given \(3 \times 3\) matrix, we can use the rule of Sarrus or the cofactor expansion method. For this example, let's use the cofactor expansion along the first row. The matrix is: \[ \left( \begin{array}{ccc} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right) \] The determinant, denoted as \( \text{det}(A) \), can be calculated as follows: \[ \text{det}(A) = -\frac{4}{3} \cdot \left| \begin{array}{cc} \frac{23}{6} & -\frac{25}{6} \\ -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right| + 4 \cdot \left| \ |
-\frac{1369}{36} |
| wenxinyiyan | To find the determinant of the given matrix, we will use the cofactor expansion method along the first row. The determinant is calculated as follows: $\text{det}(A) = (-\frac{4}{3}) \cdot \text{det}(A_{11}) + 4 \cdot \text{det}(A_{12}) + (-\frac{11}{6}) \cdot \text{det}(A_{13})$ where $A_{11}, A_{12}, \text{and} A_{13}$ are the submatrices obtained by removing the first row and the first, second, and third columns, respectively. Now, let's calculate the determinants of the submatrices: $\text{det}(A_{11}) = \left| \begin{array}{cc} \frac{23}{6} & -\frac{25}{6} \\ -\frac{2}{3} & -\frac{5}{3} \\ \end{array} \right| = \frac{23}{6} \times (-\frac{5}{3}) - (-\frac{25}{6}) \times (-\frac{2}{3}) = -\frac{115}{18} - \frac{25}{9} = -\frac{55}{6}$ $\text{det}(A_{12}) = \left| \begin{array}{cc} -\frac{13}{3} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{5}{3} \\ \end{array} \right| = -\frac{13}{3} \times (-\frac{5}{3}) - (-\frac{25}{6}) \times \frac{5}{3} = \frac{65}{9} + \frac{125}{18} = \frac{255}{18} = \frac{85}{6}$ $\text{det}(A_{13}) = \left| \begin{array}{cc} -\frac{13}{3} & \frac{23}{6} \\ \frac{5}{3} & -\frac{2}{3} \\ \end{array} \right| = -\frac{13}{3} \times (-\frac{2}{3}) - \frac{23}{6} \times \frac{5}{3} = \frac{26}{9} - \frac{115}{18} = -\frac{63}{18} = -\frac{7}{2}$ Now, substituting these values into the original determinant expression: $\text{det}(A) = (-\frac{4}{3}) \times (-\frac{55}{6}) + 4 \times \frac{85}{6} + (-\frac{11}{6}) \times (-\frac{7}{2})$ $= \frac{220}{18} + \frac{1020}{18} + \frac{77}{12}$ $= \frac{1240}{18} + \frac{77}{12}$ $= \frac{2480}{36} + \frac{231}{36}$ $= \frac{2711}{36}$ $= \boxed{\frac{2711}{36}}$ |
-\frac{1369}{36} |
| zhipu | To find the determinant of the matrix \[ A = \begin{pmatrix} -\frac{4}{3} & 4 & -\frac{11}{6} \\ -\frac{13}{3} & \frac{23}{6} & -\frac{25}{6} \\ \frac{5}{3} & -\frac{2}{3} & -\frac{5}{3} \end{pmatrix}, \] we use the formula for the determinant of a \(3 \times 3\) matrix: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg), \] where \(a, b, c\) are the elements of the first row, and \(d, e, f, g, h, i\) are the elements of the second and third rows respectively. Plugging in the values from matrix \(A\): - \(a = -\frac{4}{3}\), \(b = 4\), \(c = -\frac{11}{6}\) - \(d = -\frac{13}{3}\), \(e = \frac{23}{6}\), \(f = -\frac{25}{6}\) - \(g = \frac{5}{3}\), \(h = -\frac{2}{3}\), \(i = -\frac{5}{3}\) Now calculate each term in the determinant formula: 1. \(ei - fh = \left(\frac{23}{6}\right)\left(-\frac{5}{3}\right) - \left(-\frac{25}{6}\right)\left(-\frac{2}{3}\right)\) \[ ei - fh = -\frac{115}{18} - \frac{50}{18} = -\frac{165}{18} = -\frac{55}{6} \] 2. \(di - fg = \left(-\frac{13}{3}\right)\left(-\frac{5}{3}\right) - \left(\frac{23}{6}\right)\left(\frac{5}{3}\right)\) \[ di - fg = \frac{65}{9} - \frac{115}{18} = \frac{130}{18} - \frac{115}{18} = \frac{15}{18} = \frac{5}{6} \] 3. \(dh - eg = \left(-\frac{13}{3}\right)\left(-\frac{2}{3}\right) - \left(\frac{23}{6}\right)\left(\frac{5}{3}\right)\) \[ dh - eg = \frac{26}{9} - \frac{115}{18} = \frac{52}{18} - \frac{115}{18} = -\frac{63}{18} = -\frac{7}{2} \] Putting these into the determinant formula: \[ \det(A) = -\frac{4}{3}\left(-\frac{55}{6}\right) - 4\left(\frac{5}{6}\right) - \frac{11}{6}\left(-\frac{7}{2}\right) \] Calculate each product: \[ -\frac{4}{3} \times -\frac{55}{6} = \frac{220}{18} = \frac{110}{9}, \] \[ 4 \times \frac{5}{6} = \frac{20}{6} = \frac{10}{3} = \frac{30}{9}, \] \[ -\frac{11}{6} \times -\frac{7}{2} = \frac{77}{12} = \frac{154}{24} = \frac{77}{12} = \frac{308}{36} = \frac{154}{18} = \frac{77}{9}. \] Now sum these up: \[ \det(A) = \frac{110}{9} - \frac{30}{9} + \frac{77}{9} = \frac{110 - 30 + 77}{9} = \frac{157}{9}. \] Thus, the determinant of the matrix is \[ \boxed{\frac{157}{9}}. \] |
-\frac{1369}{36} |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-178a032f-7a3e-9879-9221-175155efabeb', 'request_id': '178a032f-7a3e-9879-9221-175155efabeb'} | -\frac{1369}{36} |