| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the determinant of the given matrix, we use the formula for the determinant of a 2x2 matrix: \[ \text{det} \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) = ad - bc \] Given the matrix: \[ \left( \begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \end{array} \right) \] we identify the elements as follows: - \( a = -\frac{13}{5} \) - \( b = -\frac{37}{10} \) - \( c = -\frac{39}{10} \) - \( d = \frac{1}{10} \) Now, we calculate the determinant: \[ \text{det} = ad - bc = \left( -\frac{13}{5} \right) \left( \frac{1}{10} \right) - \left( -\frac{37}{10} \right) \left( -\frac{39}{10} \right) \] First, compute \( ad \): \[ ad = \left( -\frac{13}{5} \right) \left( \frac{1}{10} \right) = -\frac{13}{50} \] Next, compute \( bc \): \[ bc = \left( -\frac{37}{10} \right) \left( -\frac{39}{10} \right) = \frac{37 \times 39}{100} = \frac{1443}{100} \] Now, subtract \( bc \) from \( ad \): \[ \text{det} = -\frac{13}{50} - \frac{1443}{100} \] To subtract these fractions, we need a common denominator. The common denominator of 50 and 100 is 100. Convert \(-\frac{13}{50}\) to a fraction with denominator 100: \[ -\frac{13}{50} = -\frac{13 \times 2}{50 \times 2} = -\frac{26}{100} \] Now, perform the subtraction: \[ \text{det} = -\frac{26}{100} - \frac{1443}{100} = -\frac{26 + 1443}{100} = -\frac{1469}{100} \] Thus, the determinant of the matrix is: \[ \boxed{-\frac{1469}{100}} \] |
-\frac{1469}{100} |
| deepseek-r1 | To find the determinant of the matrix \[ \left( \begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array} \right), \] we use the formula for the determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), which is \(ad - bc\). 1. Calculate \(a \times d\): \[ \left(-\frac{13}{5}\right) \times \left(\frac{1}{10}\right) = -\frac{13}{50} \] 2. Calculate \(b \times c\): \[ \left(-\frac{37}{10}\right) \times \left(-\frac{39}{10}\right) = \frac{1443}{100} \] 3. Subtract the second result from the first: \[ -\frac{13}{50} - \frac{1443}{100} \] 4. Convert \(-\frac{13}{50}\) to a fraction with a denominator of 100: \[ -\frac{13}{50} = -\frac{26}{100} \] 5. Perform the subtraction: \[ -\frac{26}{100} - \frac{1443}{100} = -\frac{26 + 1443}{100} = -\frac{1469}{100} \] Thus, the determinant of the matrix is \(\boxed{-\dfrac{1469}{100}}\). |
-\frac{1469}{100} |
| gpto1 | To find the determinant of the 2x2 matrix: \[ \begin{pmatrix} -\dfrac{13}{5} & -\dfrac{37}{10} \\ -\dfrac{39}{10} & \dfrac{1}{10} \end{pmatrix} \] We use the formula for the determinant of a 2x2 matrix: \[ \det = ad - bc \] Where: \( a = -\dfrac{13}{5} \), \( b = -\dfrac{37}{10} \), \( c = -\dfrac{39}{10} \), \( d = \dfrac{1}{10} \). Compute \( ad \): \[ \left(-\dfrac{13}{5}\right) \left( \dfrac{1}{10} \right) = -\dfrac{13}{50} \] Compute \( bc \): \[ \left( -\dfrac{37}{10} \right) \left( -\dfrac{39}{10} \right) = \dfrac{37 \times 39}{10 \times 10} = \dfrac{1443}{100} \] Now, calculate the determinant: \[ \det = -\dfrac{13}{50} - \dfrac{1443}{100} \] Convert \(-\dfrac{13}{50}\) to have a common denominator of 100: \[ -\dfrac{13}{50} = -\dfrac{26}{100} \] Therefore: \[ \det = -\dfrac{26}{100} - \dfrac{1443}{100} = -\dfrac{1469}{100} \] So, the determinant is: \[ \boxed{ -\dfrac{1469}{100} } \] \boxed{\, -\dfrac{1469}{100} \,} |
-\frac{1469}{100} |
| gpt-4o-2024-08-06 | To find the determinant of a \(2 \times 2\) matrix \[ \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \] we use the formula: \[ \text{det} = ad - bc. \] For the given matrix \[ \begin{pmatrix} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \end{pmatrix}, \] we identify the elements as follows: - \(a = -\frac{13}{5}\), - \(b = -\frac{37}{10}\), - \(c = -\frac{39}{10}\), - \(d = \frac{1}{10}\). Substitute these values into the determinant formula: \[ \text{det} = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right). \] Calculate each term separately: 1. Calculate \(ad\): \[ ad = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) = -\frac{13}{50}. \] 2. Calculate \(bc\): \[ bc = \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right) = \frac{37 \times 39}{100}. \] Now, compute \(37 \times 39\): \[ 37 \times 39 = 37 \times (40 - 1) = 37 \times 40 - 37 \times 1 = 1480 - 37 = 1443. \] Thus, \[ bc = \frac{1443}{100}. \] Substitute back into the determinant formula: \[ \text{det} = -\frac{13}{50} - \frac{1443}{100}. \] To combine these fractions, find a common denominator, which is 100: \[ -\frac{13}{50} = -\frac{26}{100}. \] Now, compute: \[ \text{det} = -\frac{26}{100} - \frac{1443}{100} = -\frac{26 + 1443}{100} = -\frac{1469}{100}. \] Thus, the determinant of the matrix is \[ \boxed{-\frac{1469}{100}}. \] |
-\frac{1469}{100} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the determinant of the given 2x2 matrix, we use the formula for the determinant of a 2x2 matrix \(\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\), which is \(ad - bc\). Given matrix: \[ \left( \begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array} \right) \] Here, \(a = -\frac{13}{5}\), \(b = -\frac{37}{10}\), \(c = -\frac{39}{10}\), and \(d = \frac{1}{10}\). Step-by-step calculation: 1. Calculate \(ad\): \[ ad = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) = -\frac{13}{50} \] 2. Calculate \(bc\): \[ bc = \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right) = \frac{37 \times 39}{100} = \frac{1443}{100} \] 3. Subtract \(bc\) from \(ad\): \[ \text{Determinant} = -\frac{13}{50} - \frac{1443}{100} \] 4. To subtract these fractions, we need a common denominator. The common denominator for 50 and 100 is 100: \[ -\frac{13}{50} = -\frac{26}{100} \] 5. Now subtract: \[ -\frac{26}{100} - \frac{1443}{100} = -\frac{26 + 1443}{100} = -\frac{1469}{100} \] Thus, the determinant of the matrix is: \[ \boxed{-\frac{1469}{100}} \] |
-\frac{1469}{100} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Recall the formula for the determinant of a 2x2 matrix The determinant of a 2x2 matrix $\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right)$ is given by the formula $ad - bc$. ## Step 2: Apply the formula to the given matrix Using the formula, we can calculate the determinant of the given matrix as $\left(-\frac{13}{5}\right)\left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right)\left(-\frac{39}{10}\right)$. ## Step 3: Simplify the expression Simplifying the expression, we get $-\frac{13}{50} - \frac{1443}{100}$. ## Step 4: Convert the fractions to have a common denominator Converting the fractions to have a common denominator of 100, we get $-\frac{26}{100} - \frac{1443}{100}$. ## Step 5: Combine the fractions Combining the fractions, we get $-\frac{1469}{100}$. The final answer is: $\boxed{-\frac{1469}{100}}$ |
-\frac{1469}{100} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Recall the formula for the determinant of a 2x2 matrix The determinant of a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by the formula $ad - bc$. ## Step 2: Identify the elements of the given matrix The given matrix is $\begin{pmatrix} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \end{pmatrix}$. So, $a = -\frac{13}{5}$, $b = -\frac{37}{10}$, $c = -\frac{39}{10}$, and $d = \frac{1}{10}$. ## Step 3: Apply the determinant formula Substitute the values of $a$, $b$, $c$, and $d$ into the determinant formula: $ad - bc = \left(-\frac{13}{5}\right)\left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right)\left(-\frac{39}{10}\right)$. ## Step 4: Simplify the expression First, calculate the products: $\left(-\frac{13}{5}\right)\left(\frac{1}{10}\right) = -\frac{13}{50}$ and $\left(-\frac{37}{10}\right)\left(-\frac{39}{10}\right) = \frac{37 \cdot 39}{100} = \frac{1443}{100}$. Then, subtract these two results: $-\frac{13}{50} - \frac{1443}{100}$. ## Step 5: Find a common denominator and subtract The common denominator is 100. Convert $-\frac{13}{50}$ to have a denominator of 100: $-\frac{13}{50} \cdot \frac{2}{2} = -\frac{26}{100}$. Now subtract: $-\frac{26}{100} - \frac{1443}{100} = -\frac{1469}{100}$. The final answer is: $\boxed{-\frac{1469}{100}}$ |
-\frac{1469}{100} |
| Qwen2-72B-Instruct | To find the determinant of a 2x2 matrix, you can use the formula: \[ \text{det}(A) = ad - bc \] where \(A = \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)\). Given the matrix \[ A = \left( \begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array} \right) \] we have \(a = -\frac{13}{5}\), \(b = -\frac{37}{10}\), \(c = -\frac{39}{10}\), and \(d = \frac{1}{10}\). Thus, \[ \text{det}(A) = \left(-\frac{13}{5}\right)\left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right)\left(-\frac{39}{10}\right) \] \[ = -\frac{13}{50} - \frac{1443}{100} \] \[ = -\frac{13}{50} - \frac{1443}{100} \] \[ = -\frac{26}{100} - \frac{1443}{100} \] \[ = -\frac{1469}{100} \] So, the determinant of the given matrix is \(-\frac{1469}{100}\). \[ \boxed{-\frac{1469}{100}} \] |
-\frac{1469}{100} |
| baichuan4 | To find the determinant of a $2 \times 2$ matrix, we use the formula: $$\text{det}\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) = ad - bc$$ Given the matrix: $$\left(\begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array}\right)$$ We identify the elements as: $$a = -\frac{13}{5}, \quad b = -\frac{37}{10}, \quad c = -\frac{39}{10}, \quad d = \frac{1}{10}$$ Now, we calculate the determinant: $$\text{det} = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right)$$ Simplifying the first term: $$\left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) = -\frac{13}{50}$$ Simplifying the second term: $$\left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right) = \frac{37 \cdot 39}{100}$$ Calculating the product: $$\frac{37 \cdot 39}{100} = \frac{1443}{100}$$ Now, we subtract the two terms: $$\text{det} = -\frac{13}{50} - \frac{1443}{100}$$ To subtract, we need to have the same denominator. Multiplying the first term by $\frac{2}{2}$: $$-\frac{13}{50} \cdot \frac{2}{2} = -\frac{26}{100}$$ Now, we can subtract: $$\text{det} = -\frac{26}{100} - \frac{1443}{100} = -\frac{1469}{100}$$ The determinant is: $$\boxed{-\frac{1469}{100}}$$ |
-\frac{1469}{100} |
| doubao | The determinant of a 2x2 matrix $\begin{pmatrix}a & b\\c & d\end{pmatrix}$ is $ad-bc$. For the given matrix $\begin{pmatrix}-\frac{13}{5} & -\frac{37}{10}\\-\frac{39}{10} & \frac{1}{10}\end{pmatrix}$, the determinant is: \[ \begin{align*} &\left(-\frac{13}{5}\right)\times\frac{1}{10}-\left(-\frac{37}{10}\right)\times\left(-\frac{39}{10}\right)\\ =&-\frac{13}{50}-\frac{1443}{100}\\ =&-\frac{26}{100}-\frac{1443}{100}\\ =&-\frac{1469}{100} \end{align*} \] So the answer is $\boxed{-\frac{1469}{100}}$. |
-\frac{1469}{100} |
| gpto1 | To find the determinant of the 2x2 matrix: \[ \begin{pmatrix} -\dfrac{13}{5} & -\dfrac{37}{10} \\ -\dfrac{39}{10} & \dfrac{1}{10} \end{pmatrix} \] We use the formula for the determinant of a 2x2 matrix: \[ \det = ad - bc \] Where: \( a = -\dfrac{13}{5} \), \( b = -\dfrac{37}{10} \), \( c = -\dfrac{39}{10} \), \( d = \dfrac{1}{10} \). Compute \( ad \): \[ \left(-\dfrac{13}{5}\right) \left( \dfrac{1}{10} \right) = -\dfrac{13}{50} \] Compute \( bc \): \[ \left( -\dfrac{37}{10} \right) \left( -\dfrac{39}{10} \right) = \dfrac{37 \times 39}{10 \times 10} = \dfrac{1443}{100} \] Now, calculate the determinant: \[ \det = -\dfrac{13}{50} - \dfrac{1443}{100} \] Convert \(-\dfrac{13}{50}\) to have a common denominator of 100: \[ -\dfrac{13}{50} = -\dfrac{26}{100} \] Therefore: \[ \det = -\dfrac{26}{100} - \dfrac{1443}{100} = -\dfrac{1469}{100} \] So, the determinant is: \[ \boxed{ -\dfrac{1469}{100} } \] \boxed{\, -\dfrac{1469}{100} \,} |
-\frac{1469}{100} |
| kimi | To find the determinant of the given 2x2 matrix, we will use the formula for the determinant of a 2x2 matrix: \[ \text{det}\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right) = ad - bc \] Given the matrix: \[ \left(\begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array}\right) \] We identify the elements as follows: - \( a = -\frac{13}{5} \) - \( b = -\frac{37}{10} \) - \( c = -\frac{39}{10} \) - \( d = \frac{1}{10} \) Now, we substitute these values into the determinant formula: \[ \text{det} = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right) \] First, calculate \( ad \): \[ ad = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) = -\frac{13}{50} \] Next, calculate \( bc \): \[ bc = \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right) = \frac{37 \times 39}{100} = \frac{1443}{100} \] Now, substitute these results back into the determinant formula: \[ \text{det} = -\frac{13}{50} - \frac{1443}{100} \] To subtract these fractions, we need a common denominator. The common denominator for 50 and 100 is 100. Convert \(-\frac{13}{50}\) to a fraction with a denominator of 100: \[ -\frac{13}{50} = -\frac{13 \times 2}{50 \times 2} = -\frac{26}{100} \] Now, subtract the fractions: \[ \text{det} = -\frac{26}{100} - \frac{1443}{100} = -\frac{26 + 1443}{100} = -\frac{1469}{100} \] Thus, the determinant of the matrix is: \[ \boxed{-\frac{1469}{100}} \] |
-\frac{1469}{100} |
| minimax | To find the determinant of a 2x2 matrix, we use the formula: $$\text{det}(A) = ad - bc$$ where the matrix $A$ is: $$\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right)$$ For the given matrix: $$\left(\begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array}\right)$$ we have: $$a = -\frac{13}{5}, b = -\frac{37}{10}, c = -\frac{39}{10}, d = \frac{1}{10}$$ Now, we calculate the determinant: $$\text{det}(A) = \left(-\frac{13}{5}\right)\left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right)\left(-\frac{39}{10}\right)$$ $$\text{det}( |
-\frac{1469}{100} |
| wenxinyiyan | To find the determinant of a 2x2 matrix, we use the formula: $$\text{Determinant} = (a \times d) - (b \times c)$$where $a, b, c,$ and $d$ are the elements of the matrix: $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$For the given matrix: $$\begin{pmatrix} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{pmatrix}$$we have $a = -\frac{13}{5}$, $b = -\frac{37}{10}$, $c = -\frac{39}{10}$, and $d = \frac{1}{10}$. Plugging these values into the determinant formula, we get: $$\text{Determinant} = \left(-\frac{13}{5} \times \frac{1}{10}\right) - \left(-\frac{37}{10} \times -\frac{39}{10}\right)$$$$= -\frac{13}{50} - \frac{1443}{100}$$$$= -\frac{2886}{100} - \frac{130}{100}$$$$= -\frac{3016}{100}$$$$= -\frac{1508}{50}$$$$= -\frac{754}{25}$$$$= \boxed{-\frac{754}{25}}$$ |
-\frac{1469}{100} |
| zhipu | To find the determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), we use the formula: \[ \text{det} = ad - bc \] Given the matrix: \[ \begin{pmatrix} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \end{pmatrix} \] we identify \(a = -\frac{13}{5}\), \(b = -\frac{37}{10}\), \(c = -\frac{39}{10}\), and \(d = \frac{1}{10}\). Now, we calculate \(ad\) and \(bc\): 1. \(ad = \left(-\frac{13}{5}\right) \left(\frac{1}{10}\right) = -\frac{13}{50}\) 2. \(bc = \left(-\frac{37}{10}\right) \left(-\frac{39}{10}\right) = \frac{1443}{100}\) Next, we find the determinant: \[ \text{det} = ad - bc = -\frac{13}{50} - \frac{1443}{100} \] To subtract these fractions, we need a common denominator. The common denominator of 50 and 100 is 100. We convert \(-\frac{13}{50}\) to \(-\frac{26}{100}\): \[ -\frac{13}{50} = -\frac{26}{100} \] Now, we perform the subtraction: \[ \text{det} = -\frac{26}{100} - \frac{1443}{100} = \frac{-26 - 1443}{100} = \frac{-1469}{100} \] Thus, the determinant of the matrix is: \[ \boxed{-\frac{1469}{100}} \] |
-\frac{1469}{100} |
| deepseek-v3 | To find the determinant of the given \( 2 \times 2 \) matrix: \[ \left( \begin{array}{cc} -\frac{13}{5} & -\frac{37}{10} \\ -\frac{39}{10} & \frac{1}{10} \\ \end{array} \right) \] we use the formula for the determinant of a \( 2 \times 2 \) matrix: \[ \text{det} = ad - bc \] where the matrix is: \[ \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \] **Step 1: Identify the elements of the matrix** \[ a = -\frac{13}{5}, \quad b = -\frac{37}{10}, \quad c = -\frac{39}{10}, \quad d = \frac{1}{10} \] **Step 2: Apply the determinant formula** \[ \text{det} = \left(-\frac{13}{5}\right) \times \left(\frac{1}{10}\right) - \left(-\frac{37}{10}\right) \times \left(-\frac{39}{10}\right) \] **Step 3: Calculate each part separately** \[ \text{First term: } -\frac{13}{5} \times \frac{1}{10} = -\frac{13}{50} \] \[ \text{Second term: } -\frac{37}{10} \times -\frac{39}{10} = \frac{1443}{100} \] **Step 4: Combine the terms** \[ \text{det} = -\frac{13}{50} - \frac{1443}{100} \] To subtract these fractions, convert them to have a common denominator: \[ -\frac{13}{50} = -\frac{26}{100} \] \[ \text{det} = -\frac{26}{100} - \frac{1443}{100} = -\frac{1469}{100} \] **Final Answer:** \[ \boxed{-\dfrac{1469}{100}} \] |
-\frac{1469}{100} |