| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the characteristic polynomial of the given matrix, we need to compute the determinant of the matrix \( A - \lambda I \), where \( A \) is the given matrix and \( I \) is the identity matrix. The matrix \( A \) is: \[ A = \left( \begin{array}{ccc} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{array} \right) \] First, we form the matrix \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{ccc} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{array} \right) \] Next, we need to compute the determinant of this matrix. The determinant of a \( 3 \times 3 \) matrix \( \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by: \[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \] Applying this formula to our matrix \( A - \lambda I \): \[ \text{det}(A - \lambda I) = (-4 - \lambda) \left[ (10 - \lambda)(-14 - \lambda) - (-5)(4) \right] - (-13) \left[ (-9)(-14 - \lambda) - (12)(4) \right] + (10) \left[ (-9)(-5) - (12)(10 - \lambda) \right] \] We now compute each term inside the brackets: 1. \((10 - \lambda)(-14 - \lambda) - (-5)(4)\): \[ (10 - \lambda)(-14 - \lambda) = 10 \cdot (-14) + 10 \cdot (-\lambda) - \lambda \cdot (-14) - \lambda \cdot \lambda = -140 - 10\lambda + 14\lambda + \lambda^2 = \lambda^2 + 4\lambda - 140 \] \[ - (-5)(4) = 20 \] \[ (10 - \lambda)(-14 - \lambda) - (-5)(4) = \lambda^2 + 4\lambda - 140 + 20 = \lambda^2 + 4\lambda - 120 \] 2. \((-9)(-14 - \lambda) - (12)(4)\): \[ (-9)(-14 - \lambda) = 126 + 9\lambda \] \[ (12)(4) = 48 \] \[ (-9)(-14 - \lambda) - (12)(4) = 126 + 9\lambda - 48 = 78 + 9\lambda \] 3. \((-9)(-5) - (12)(10 - \lambda)\): \[ (-9)(-5) = 45 \] \[ (12)(10 - \lambda) = 120 - 12\lambda \] \[ (-9)(-5) - (12)(10 - \lambda) = 45 - (120 - 12\lambda) = 45 - 120 + 12\lambda = -75 + 12\lambda \] Now, substituting these back into the determinant formula: \[ \text{det}(A - \lambda I) = (-4 - \lambda)(\lambda^2 + 4\lambda - 120) - (-13)(78 + 9\lambda) + 10(-75 + 12\lambda) \] Expanding each term: \[ (-4 - \lambda)(\lambda^2 + 4\lambda - 120) = -4\lambda^2 - 16\lambda + 480 - \lambda^3 - 4\lambda^2 + 120\lambda = -\lambda^3 - 8\lambda^2 + 104\lambda + 480 \] \[ - (-13)(78 + 9\lambda) = 1014 + 117\lambda \] \[ 10(-75 + 12\lambda) = -750 + 120\lambda \] Combining all terms: \[ \text{det}(A - \lambda I) = -\lambda^3 - 8\lambda^2 + 104\lambda + 480 + 1014 + 117\lambda - 750 + 120\lambda \] \[ = -\lambda^3 - 8\lambda^2 + (104 + 117 + 120)\lambda + (480 + 1014 - 750) \] \[ = -\lambda^3 - 8\lambda^2 + 341\lambda + 744 \] Thus, the characteristic polynomial is: \[ \boxed{-\lambda^3 - 8\lambda^2 + 341\lambda + 744} \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| deepseek-r1 | To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \end{pmatrix}, \] we need to compute the determinant of \(A - \lambda I\). The matrix \(A - \lambda I\) is: \[ A - \lambda I = \begin{pmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \end{pmatrix}. \] Using the formula for the determinant of a 3x3 matrix, we expand along the first row: \[ \det(A - \lambda I) = (-4 - \lambda) \cdot \det\begin{pmatrix} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{pmatrix} - (-13) \cdot \det\begin{pmatrix} -9 & 4 \\ 12 & -14 - \lambda \end{pmatrix} + 10 \cdot \det\begin{pmatrix} -9 & 10 - \lambda \\ 12 & -5 \end{pmatrix}. \] Calculating each minor: 1. \(\det\begin{pmatrix} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{pmatrix} = (10 - \lambda)(-14 - \lambda) + 20 = \lambda^2 + 4\lambda - 120\) 2. \(\det\begin{pmatrix} -9 & 4 \\ 12 & -14 - \lambda \end{pmatrix} = 78 + 9\lambda\) 3. \(\det\begin{pmatrix} -9 & 10 - \lambda \\ 12 & -5 \end{pmatrix} = -75 + 12\lambda\) Substituting these minors back into the determinant expression: \[ \det(A - \lambda I) = (-4 - \lambda)(\lambda^2 + 4\lambda - 120) + 13(78 + 9\lambda) + 10(-75 + 12\lambda). \] Expanding and simplifying each term: 1. \((-4 - \lambda)(\lambda^2 + 4\lambda - 120) = -\lambda^3 - 8\lambda^2 + 104\lambda + 480\) 2. \(13(78 + 9\lambda) = 1014 + 117\lambda\) 3. \(10(-75 + 12\lambda) = -750 + 120\lambda\) Combining all terms: \[ \det(A - \lambda I) = -\lambda^3 - 8\lambda^2 + 341\lambda + 744. \] Since the characteristic polynomial is typically defined as \(\det(\lambda I - A)\), we multiply by \(-1\): \[ \det(\lambda I - A) = \lambda^3 + 8\lambda^2 - 341\lambda - 744. \] Thus, the characteristic polynomial of the matrix \(A\) is: \[ \boxed{x^3 + 8x^2 - 341x - 744} \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| gpto1 | To find the characteristic polynomial \( p(\lambda) \) of the matrix \( A \): \[ A = \begin{pmatrix} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{pmatrix} \] We compute \( \det(A - \lambda I) \), where \( I \) is the identity matrix. First, compute \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{pmatrix} \] Next, compute the determinant: \[ \begin{align*} \det(A - \lambda I) &= (-4 - \lambda)\left( (10 - \lambda)(-14 - \lambda) - (4)(-5) \right) \\ &\quad{} - (-13)\left( (-9)(-14 - \lambda) - (4)(12) \right) \\ &\quad{} + 10\left( (-9)(-5) - (10 - \lambda)(12) \right) \end{align*} \] Simplify each term: 1. Compute the minors: \[ \begin{align*} M_1 &= (10 - \lambda)(-14 - \lambda) - (4)(-5) = (-140 + 4\lambda + \lambda^2) + 20 = -120 + 4\lambda + \lambda^2 \\ M_2 &= (-9)(-14 - \lambda) - (4)(12) = (126 + 9\lambda) - 48 = 78 + 9\lambda \\ M_3 &= (-9)(-5) - (12)(10 - \lambda) = 45 - (120 - 12\lambda) = -75 + 12\lambda \end{align*} \] 2. Multiply each minor by the corresponding element of \( A - \lambda I \): \[ \begin{align*} T_1 &= (-4 - \lambda)(-120 + 4\lambda + \lambda^2) \\ T_2 &= -(-13)(78 + 9\lambda) = 13(78 + 9\lambda) \\ T_3 &= 10(-75 + 12\lambda) \end{align*} \] 3. Expand and simplify: \[ \begin{align*} T_1 &= (-4 - \lambda)(-120 + 4\lambda + \lambda^2) = 480 + 104\lambda - 8\lambda^2 - \lambda^3 \\ T_2 &= 1014 + 117\lambda \\ T_3 &= -750 + 120\lambda \end{align*} \] 4. Sum the terms: \[ \det(A - \lambda I) = T_1 + T_2 + T_3 = (480 + 1014 - 750) + (104\lambda + 117\lambda + 120\lambda) - 8\lambda^2 - \lambda^3 \] 5. Simplify: \[ \begin{align*} \det(A - \lambda I) &= 744 + 341\lambda - 8\lambda^2 - \lambda^3 = -\lambda^3 - 8\lambda^2 + 341\lambda + 744 \end{align*} \] Therefore, the characteristic polynomial is: \[ \boxed{-\lambda^3\ -\ 8\lambda^2\ +\ 341\lambda\ +\ 744} \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| gpt-4o-2024-08-06 | To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \end{pmatrix}, \] we need to compute the determinant of the matrix \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar. The matrix \( A - \lambda I \) is: \[ A - \lambda I = \begin{pmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \end{pmatrix}. \] The characteristic polynomial is given by the determinant of this matrix: \[ \det(A - \lambda I) = \begin{vmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \end{vmatrix}. \] We can compute this determinant using the cofactor expansion along the first row: \[ \det(A - \lambda I) = (-4 - \lambda) \begin{vmatrix} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{vmatrix} - (-13) \begin{vmatrix} -9 & 4 \\ 12 & -14 - \lambda \end{vmatrix} + 10 \begin{vmatrix} -9 & 10 - \lambda \\ 12 & -5 \end{vmatrix}. \] Now, we compute each of these 2x2 determinants: 1. \(\begin{vmatrix} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{vmatrix} = (10 - \lambda)(-14 - \lambda) - (4)(-5)\). \[ = (10 - \lambda)(-14 - \lambda) + 20 = (10\lambda + 14\lambda + \lambda^2) - 140 + 20 = \lambda^2 + 4\lambda - 120. \] 2. \(\begin{vmatrix} -9 & 4 \\ 12 & -14 - \lambda \end{vmatrix} = (-9)(-14 - \lambda) - (4)(12)\). \[ = 126 + 9\lambda - 48 = 9\lambda + 78. \] 3. \(\begin{vmatrix} -9 & 10 - \lambda \\ 12 & -5 \end{vmatrix} = (-9)(-5) - (10 - \lambda)(12)\). \[ = 45 - (120 - 12\lambda) = 45 - 120 + 12\lambda = 12\lambda - 75. \] Substituting these back into the determinant expression, we have: \[ \det(A - \lambda I) = (-4 - \lambda)(\lambda^2 + 4\lambda - 120) + 13(9\lambda + 78) + 10(12\lambda - 75). \] Expanding each term: - \((-4 - \lambda)(\lambda^2 + 4\lambda - 120) = (-4)(\lambda^2 + 4\lambda - 120) - \lambda(\lambda^2 + 4\lambda - 120)\). \[ = -4\lambda^2 - 16\lambda + 480 - (\lambda^3 + 4\lambda^2 - 120\lambda) \] \[ = -\lambda^3 - 8\lambda^2 + 104\lambda + 480. \] - \(13(9\lambda + 78) = 117\lambda + 1014\). - \(10(12\lambda - 75) = 120\lambda - 750\). Combining all terms: \[ -\lambda^3 - 8\lambda^2 + 104\lambda + 480 + 117\lambda + 1014 + 120\lambda - 750. \] Combine like terms: - The \(\lambda^3\) term: \(-\lambda^3\). - The \(\lambda^2\) term: \(-8\lambda^2\). - The \(\lambda\) term: \(104\lambda + 117\lambda + 120\lambda = 341\lambda\). - The constant term: \(480 + 1014 - 750 = 744\). Thus, the characteristic polynomial is: \[ \boxed{-\lambda^3 - 8\lambda^2 + 341\lambda + 744}. \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| DeepSeek-Coder-V2-Lite-Instruct | To find the characteristic polynomial of the given matrix, we need to compute the determinant of the matrix \( A - \lambda I \), where \( A \) is the given matrix and \( \lambda \) is a scalar. The characteristic polynomial is given by \( \det(A - \lambda I) \). Given matrix \( A \): \[ A = \left( \begin{array}{ccc} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{array} \right) \] First, we form the matrix \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{ccc} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{array} \right) \] Next, we compute the determinant of \( A - \lambda I \). We will use the cofactor expansion along the first row: \[ \det(A - \lambda I) = (-4 - \lambda) \left| \begin{array}{cc} 10 - \lambda & 4 \\ -5 & -14 - \lambda \\ \end{array} \right| - (-13) \left| \begin{array}{cc} -9 & 4 \\ 12 & -14 - \lambda \\ \end{array} \right| + 10 \left| \begin{array}{cc} -9 & 10 - \lambda \\ 12 & -5 \\ \end{array} \right| \] We now compute each of the 2x2 determinants: 1. \[ \left| \begin{array}{cc} 10 - \lambda & 4 \\ -5 & -14 - \lambda \\ \end{array} \right| = (10 - \lambda)(-14 - \lambda) - (4)(-5) = -140 - 10\lambda + \lambda^2 + 20 = \lambda^2 - 10\lambda - 120 \] 2. \[ \left| \begin{array}{cc} -9 & 4 \\ 12 & -14 - \lambda \\ \end{array} \right| = (-9)(-14 - \lambda) - (4)(12) = 126 + 9\lambda - 48 = 78 + 9\lambda \] 3. \[ \left| \begin{array}{cc} -9 & 10 - \lambda \\ 12 & -5 \\ \end{array} \right| = (-9)(-5) - (12)(10 - \lambda) = 45 - 120 + 12\lambda = 12\lambda - 75 \] Substituting these back into the determinant expression: \[ \det(A - \lambda I) = (-4 - \lambda)(\lambda^2 - 10\lambda - 120) - (-13)(78 + 9\lambda) + 10(12\lambda - 75) \] Expanding and simplifying: \[ \det(A - \lambda I) = (-4 - \lambda)(\lambda^2 - 10\lambda - 120) + 13(78 + 9\lambda) + 10(12\lambda - 75) \] \[ = -4\lambda^2 + 40\lambda + 480 - \lambda^3 + 10\lambda^2 + 120\lambda + 1044 + 108\lambda - 750 \] Combining like terms: \[ = -\lambda^3 + 6\lambda^2 + 268\lambda + 774 \] Thus, the characteristic polynomial is: \[ \boxed{-\lambda^3 + 6\lambda^2 + 268\lambda + 774} \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: To find the characteristic polynomial of the given matrix, we first need to understand what the characteristic polynomial is. It is defined as the determinant of the matrix obtained by subtracting a variable, usually denoted as lambda (λ), from the diagonal elements of the original matrix. ## Step 2: The given matrix is a 3x3 matrix. To find its characteristic polynomial, we will subtract λ from the diagonal elements and then calculate the determinant of the resulting matrix. ## Step 3: The original matrix is: $\left(\n\begin{array}{ccc}\n -4 & -13 & 10 \\\n -9 & 10 & 4 \\\n 12 & -5 & -14 \\\n\end{array}\n\right)$. Subtracting λ from the diagonal elements, we get: $\left(\n\begin{array}{ccc}\n -4-λ & -13 & 10 \\\n -9 & 10-λ & 4 \\\n 12 & -5 & -14-λ \\\n\end{array}\n\right)$. ## Step 4: Now, we calculate the determinant of this matrix by expanding along the first row. The determinant is given by: $(-4-λ) \times \left|\begin{array}{cc} 10-λ & 4 \\ -5 & -14-λ \end{array}\right| - (-13) \times \left|\begin{array}{cc} -9 & 4 \\ 12 & -14-λ \end{array}\right| + 10 \times \left|\begin{array}{cc} -9 & 10-λ \\ 12 & -5 \end{array}\right|$. ## Step 5: Calculating each of the 2x2 determinants, we get: $\left|\begin{array}{cc} 10-λ & 4 \\ -5 & -14-λ \end{array}\right| = (10-λ)(-14-λ) - (-5)(4)$, $\left|\begin{array}{cc} -9 & 4 \\ 12 & -14-λ \end{array}\right| = (-9)(-14-λ) - (12)(4)$, $\left|\begin{array}{cc} -9 & 10-λ \\ 12 & -5 \end{array}\right| = (-9)(-5) - (12)(10-λ)$. ## Step 6: After calculating the determinants, we simplify the expressions and combine like terms to obtain the characteristic polynomial. ## Step 7: The final characteristic polynomial will be in the form of $aλ^3 + bλ^2 + cλ + d$, where a, b, c, and d are constants obtained from the simplified expression. ## Step 8: After simplifying and combining like terms, we get the characteristic polynomial as $-λ^3 - 8λ^2 + 71λ + 420$. The final answer is: $\boxed{-λ^3 - 8λ^2 + 71λ + 420}$ |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Understand the problem To find the characteristic polynomial of the given matrix, we need to calculate the determinant of the matrix $A - \lambda I$, where $A$ is the given matrix, $\lambda$ is the eigenvalue, and $I$ is the identity matrix. ## Step 2: Write down the matrix $A - \lambda I$ The matrix $A - \lambda I$ is given by: $\left(\begin{array}{ccc} -4-\lambda & -13 & 10 \\ -9 & 10-\lambda & 4 \\ 12 & -5 & -14-\lambda \end{array}\right)$. ## Step 3: Calculate the determinant of $A - \lambda I$ Using the formula for the determinant of a 3x3 matrix, we get: $\det(A - \lambda I) = (-4-\lambda)((10-\lambda)(-14-\lambda) - 4(-5)) - (-13)((-9)(-14-\lambda) - 4(12)) + 10((-9)(-5) - (10-\lambda)(12))$. ## Step 4: Expand and simplify the determinant Expanding and simplifying the determinant, we get: $\det(A - \lambda I) = (-4-\lambda)(-140 + 10\lambda + 14\lambda - \lambda^2 + 20) + 13(126 + 9\lambda - 48) + 10(45 - 120 + 12\lambda)$. ## Step 5: Combine like terms and simplify further Combining like terms and simplifying further, we get: $\det(A - \lambda I) = (-4-\lambda)(-120 + 24\lambda - \lambda^2) + 13(78 + 9\lambda) + 10(-75 + 12\lambda)$. ## Step 6: Expand and simplify again Expanding and simplifying again, we get: $\det(A - \lambda I) = 480 - 96\lambda + 4\lambda^2 - 24\lambda^2 + \lambda^3 + 1014 + 117\lambda - 750 + 120\lambda$. ## Step 7: Combine like terms and simplify further Combining like terms and simplifying further, we get: $\det(A - \lambda I) = \lambda^3 - 20\lambda^2 + 141\lambda + 744$. The final answer is: $\boxed{\lambda^3 - 20\lambda^2 + 141\lambda + 744}$ |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| Qwen2-72B-Instruct | To find the characteristic polynomial of a matrix, we need to compute its determinant after subtracting $\lambda I$ from it, where $I$ is the identity matrix and $\lambda$ is an eigenvalue. The given matrix is: $$A = \left(\begin{array}{ccc} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{array}\right)$$ The characteristic equation is given by $\det(A-\lambda I)=0$, where $I$ is the identity matrix of the same size as $A$. So, we have: $$A-\lambda I = \left(\begin{array}{ccc} -4-\lambda & -13 & 10 \\ -9 & 10-\lambda & 4 \\ 12 & -5 & -14-\lambda \\ \end{array}\right)$$ Now, we calculate the determinant of this matrix: \[ \begin{vmatrix} -4-\lambda & -13 & 10 \\ -9 & 10-\lambda & 4 \\ 12 & -5 & -14-\lambda \\ \end{vmatrix} = 0 \] Expanding along the first row (or any other row/column that makes the calculation easier): \[ = (-4-\lambda)\begin{vmatrix} 10-\lambda & 4 \\ -5 & -14-\lambda \\ \end{vmatrix} - (-13)\begin{vmatrix} -9 & 4 \\ 12 & -14-\lambda \\ \end{vmatrix} + 10\begin{vmatrix} -9 & 10-\lambda \\ 12 & -5 \\ \end{vmatrix} \] \[ = (-4-\lambda)((10-\lambda)(-14-\lambda)-(-5)(4)) + 13((-9)(-14-\lambda)-(4)(12)) + 10((-9)(-5)-(10-\lambda)(12)) \] \[ = (-4-\lambda)((10-\lambda)(-14-\lambda)+20) + 13(126+9\lambda-48) + 10(45-(120-12\lambda)) \] \[ = (-4-\lambda)(-140-10\lambda+14\lambda+\lambda^2+20) + 13(78+9\lambda) + 10(-75+12\lambda) \] \[ = (-4-\lambda)(\lambda^2+4\lambda-120) + 1014+117\lambda - 750+120\lambda \] \[ = (-4-\lambda)(\lambda^2+4\lambda-120) + 264+237\lambda \] \[ = -\lambda^3-4\lambda^2+120\lambda-4\lambda^2-16\lambda+480+264+237\lambda \] \[ = -\lambda^3-8\lambda^2+341\lambda+744 \] So, the characteristic polynomial of the given matrix is: \[ p(\lambda) = -\lambda^3-8\lambda^2+341\lambda+744 \] Therefore, the final answer is $\boxed{-\lambda^3-8\lambda^2+341\lambda+744}$. |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| baichuan4 | To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix minus $x$ times the identity matrix. The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. Given the matrix $A$: $$ A = \left( \begin{array}{ccc} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{array} \right) $$ We first construct the matrix $A - xI$, where $I$ is the $3 \times 3$ identity matrix: $$ A - xI = \left( \begin{array}{ccc} -4-x & -13 & 10 \\ -9 & 10-x & 4 \\ 12 & -5 & -14-x \\ \end{array} \right) $$ Now, we compute the determinant of this matrix: $$ \begin{aligned} \det(A - xI) &= \det\left( \begin{array}{ccc} -4-x & -13 & 10 \\ -9 & 10-x & 4 \\ 12 & -5 & -14-x \\ \end{array} \right) \\ &= (-4-x)\det\left( \begin{array}{cc} 10-x & 4 \\ -5 & -14-x \\ \end{array} \right) - (-13)\det\left( \begin{array}{cc} -9 & 4 \\ 12 & -14-x \\ \end{array} \right) + 10\det\left( \begin{array}{cc} -9 & 10-x \\ 12 & -5 \\ \end{array} \right) \\ &= (-4-x)((10-x)(-14-x) - (-5)(4)) - (-13)((-9)(-14-x) 4)(4)(12)) + 10((-9)(-5) - (10-x)(12)) \\ &= (-4-x)(x^2 + 4x - 140 + 20) + 13(126 + 9x - 48) - 10(45 - 120 + 12x) \\ &= (-4-x)(x^2 + 4x - 120) + 13(9x + 78) - 10(12x - 75) \\ &= (-4-x)(x^2 + 4x - 120) + 117x + 1014 - 120x + 750 \\ &= -x^3 - 4x^2 + 120x + 4x^2 + 16x - 480 - 3x + 1764 \\ &= -x^3 + 133x + 1284. \end{aligned} $$ Therefore, the characteristic polynomial of the given matrix is: $$ \boxed{-x^3 + 133x + 1284}. $$ |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| doubao | The characteristic polynomial of a 3x3 matrix $A = \begin{pmatrix}a & b & c\\d & e & f\\g & h & i\end{pmatrix}$ is given by $|A - \lambda I| = \begin{vmatrix}a - \lambda & b & c\\d & e - \lambda & f\\g & h & i - \lambda\end{vmatrix}$. For the given matrix $\begin{pmatrix}-4 & -13 & 10\\-9 & 10 & 4\\12 & -5 & -14\end{pmatrix}$, the characteristic polynomial is: \[ \begin{align*} &\begin{vmatrix}-4 - \lambda & -13 & 10\\-9 & 10 - \lambda & 4\\12 & -5 & -14 - \lambda\end{vmatrix}\\ =& (-4 - \lambda)\times\begin{vmatrix}10 - \lambda & 4\\-5 & -14 - \lambda\end{vmatrix} - (-13)\times\begin{vmatrix}-9 & 4\\12 & -14 - \lambda\end{vmatrix} + 10\times\begin{vmatrix}-9 & 10 - \lambda\\12 & -5\end{vmatrix}\\ =& (-4 - \lambda)[(10 - \lambda)(-14 - \lambda) - (-5)\times4] - (-13)[(-9)(-14 - \lambda) - 4\times12] + 10[(-9)(-5) - 12\times(10 - \lambda)]\\ =& (-4 - \lambda)[-140 + 14\lambda - 10\lambda + \lambda^2 + 20] - (-13)[126 + 9\lambda - 48] + 10[45 - 120 + 12\lambda]\\ =& (-4 - \lambda)(\lambda^2 + 4\lambda - 120) - (-13)(78 + 9\lambda) + 10(-75 + 12\lambda)\\ =& (-4\lambda^2 - 16\lambda + 480 - \lambda^3 - 4\lambda^2 + 120\lambda) - (-1014 - 117\lambda) + (-750 + 120\lambda)\\ =& -\lambda^3 - 8\lambda^2 + 104\lambda + 480 + 1014 + 117\lambda - 750 + 120\lambda\\ =& -\lambda^3 - 8\lambda^2 + 341\lambda + 744 \end{align*} \] So the answer is $\boxed{-\lambda^3 - 8\lambda^2 + 341\lambda + 744}$. |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| gpto1 | To find the characteristic polynomial \( p(\lambda) \) of the matrix \( A \): \[ A = \begin{pmatrix} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{pmatrix} \] We compute \( \det(A - \lambda I) \), where \( I \) is the identity matrix. First, compute \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{pmatrix} \] Next, compute the determinant: \[ \begin{align*} \det(A - \lambda I) &= (-4 - \lambda)\left( (10 - \lambda)(-14 - \lambda) - (4)(-5) \right) \\ &\quad{} - (-13)\left( (-9)(-14 - \lambda) - (4)(12) \right) \\ &\quad{} + 10\left( (-9)(-5) - (10 - \lambda)(12) \right) \end{align*} \] Simplify each term: 1. Compute the minors: \[ \begin{align*} M_1 &= (10 - \lambda)(-14 - \lambda) - (4)(-5) = (-140 + 4\lambda + \lambda^2) + 20 = -120 + 4\lambda + \lambda^2 \\ M_2 &= (-9)(-14 - \lambda) - (4)(12) = (126 + 9\lambda) - 48 = 78 + 9\lambda \\ M_3 &= (-9)(-5) - (12)(10 - \lambda) = 45 - (120 - 12\lambda) = -75 + 12\lambda \end{align*} \] 2. Multiply each minor by the corresponding element of \( A - \lambda I \): \[ \begin{align*} T_1 &= (-4 - \lambda)(-120 + 4\lambda + \lambda^2) \\ T_2 &= -(-13)(78 + 9\lambda) = 13(78 + 9\lambda) \\ T_3 &= 10(-75 + 12\lambda) \end{align*} \] 3. Expand and simplify: \[ \begin{align*} T_1 &= (-4 - \lambda)(-120 + 4\lambda + \lambda^2) = 480 + 104\lambda - 8\lambda^2 - \lambda^3 \\ T_2 &= 1014 + 117\lambda \\ T_3 &= -750 + 120\lambda \end{align*} \] 4. Sum the terms: \[ \det(A - \lambda I) = T_1 + T_2 + T_3 = (480 + 1014 - 750) + (104\lambda + 117\lambda + 120\lambda) - 8\lambda^2 - \lambda^3 \] 5. Simplify: \[ \begin{align*} \det(A - \lambda I) &= 744 + 341\lambda - 8\lambda^2 - \lambda^3 = -\lambda^3 - 8\lambda^2 + 341\lambda + 744 \end{align*} \] Therefore, the characteristic polynomial is: \[ \boxed{-\lambda^3\ -\ 8\lambda^2\ +\ 341\lambda\ +\ 744} \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| kimi | To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix minus a scalar multiple of the identity matrix. Let's denote the given matrix as A: $A = \left(\begin{array}{ccc} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \end{array}\right)$ The characteristic polynomial is given by the determinant of (A - λI), where λ is a scalar and I is the identity matrix. So, we have: $A - λI = \left(\begin{array}{ccc} -4-λ & -13 & 10 \\ -9 & 10-λ & 4 \\ 12 & -5 & -14-λ \end{array}\right)$ Now, we need to find the determinant of this matrix: $\det(A - λI) = \left| \begin{array}{ccc} -4-λ & -13 & 10 \\ -9 & 10-λ & 4 \\ 12 & -5 & -14-λ \end{array} \right|$ We can use the cofactor expansion along the first row to compute the determinant: $\det(A - λI) = (-4-λ) \left| \begin{array}{cc} 10-λ & 4 \\ -5 & -14-λ \end{array} \right| - (-13) \left| \begin{array}{cc} -9 & 4 \\ 12 & -14-λ \end{array} \right| + 10 \left| \begin{array}{cc} -9 & 10-λ \\ 12 & -5 \end{array} \right|$ Now, we compute the determinants of the 2x2 matrices: $\left| \begin{array}{cc} 10-λ & 4 \\ -5 & -14-λ \end{array} \right| = (10-λ)(-14-λ) - (4)(-5) = -140 + 14λ + λ^2 + 20 = λ^2 + 14λ - 120$ $\left| \begin{array}{cc} -9 & 4 \\ 12 & -14-λ \end{array} \right| = (-9)(-14-λ) - (4)(12) = 126 + 9λ - 48 = 9λ + 78$ $\left| \begin{array}{cc} -9 & 10-λ \\ 12 & -5 \end{array} \right| = (-9)(-5) - (10-λ)(12) = 45 - 120 + 12λ = 12λ - 75$ Now, substitute these back into the cofactor expansion: $\det(A - λI) = (-4-λ)(λ^2 + 14λ - 120) + 13(9λ + 78) + 10(12λ - 75)$ Expand and simplify: $\det(A - λI) = -λ^3 - 14λ^2 + 120λ + 4λ^2 + 56λ - 480 - 117λ - 1014 + 120λ - 750$ Combine like terms: $\det(A - λI) = -λ^3 - 10λ^2 + 79λ - 2244$ So, the characteristic polynomial is: $\boxed{-λ^3 - 10λ^2 + 79λ - 2244}$ |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| minimax | To find the characteristic polynomial of the given matrix, we need to compute the determinant of $A - \lambda I$, where $A$ is the matrix in question and $I$ is the identity matrix of the same size, and $\lambda$ represents the eigenvalues. The matrix $A - \lambda I$ is: $$ A - \lambda I = \left(\begin{array}{ccc} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{array}\right) $$ The characteristic polynomial is given by the determinant of this matrix: $$ \text{det}(A - \lambda I) = \left| \begin{array}{ccc} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{array} \right| $$ We can expand this determinant using the rule of Sarrus or the cofactor expansion along any row or column. For simplicity, let's expand along the first column: $$ \text |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| wenxinyiyan | To find the characteristic polynomial of the given matrix, we first write down the matrix and its characteristic equation: $A = \begin{pmatrix} -4 & -13 & 10 \ -9 & 10 & 4 \ 12 & -5 & -14 \end{pmatrix}$ The characteristic equation of $A$ is given by: $f(\lambda) = \det(\lambda I - A) = \det \begin{pmatrix} \lambda + 4 & 13 & -10 \ 9 & \lambda - 10 & -4 \ -12 & 5 & \lambda + 14 \end{pmatrix}$ To compute the determinant, we expand along the first row: $f(\lambda) = (\lambda + 4) \cdot \det \begin{pmatrix} \lambda - 10 & -4 \ 5 & \lambda + 14 \end{pmatrix} - 13 \cdot \det \begin{pmatrix} 9 & -4 \ -12 & \lambda + 14 \end{pmatrix} + (-10) \cdot \det \begin{pmatrix} 9 & \lambda - 10 \ -12 & 5 \end{pmatrix}$ Next, we compute the determinants of the 2x2 matrices: $f(\lambda) = (\lambda + 4)[(\lambda - 10)(\lambda + 14) - (-4) \cdot 5] - 13[9(\lambda + 14) - (-4) \cdot (-12)] + (-10)[9 \cdot 5 - (-12)(\lambda - 10)]$ Simplifying further, we get: $f(\lambda) = (\lambda + 4)(\lambda^2 + 4\lambda - 120) - 13(9\lambda + 126) - 10(45 + 12\lambda)$ $f(\lambda) = (\lambda + 4)(\lambda^2 + 4\lambda - 120) - 117\lambda - 1638 - 450 - 120\lambda$ $f(\lambda) = (\lambda + 4)(\lambda^2 + 4\lambda - 120) - 237\lambda - 2088$ $f(\lambda) = \lambda^3 + 4\lambda^2 - 120\lambda + 4\lambda^2 + 16\lambda - 480 - 237\lambda - 2088$ $f(\lambda) = \lambda^3 + 8\lambda^2 - 341\lambda - 2568$ Therefore, the characteristic polynomial of the given matrix is: $\boxed{f(\lambda) = \lambda^3 + 8\lambda^2 - 341\lambda - 2568}$ |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| zhipu | To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \end{pmatrix}, \] we need to compute the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar. This matrix \( A - \lambda I \) is given by: \[ A - \lambda I = \begin{pmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \end{pmatrix}. \] The characteristic polynomial \( p(\lambda) \) is the determinant of this matrix: \[ p(\lambda) = \det(A - \lambda I) = \begin{vmatrix} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \end{vmatrix}. \] We will expand this determinant using the first row: \[ p(\lambda) = (-4 - \lambda) \begin{vmatrix} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{vmatrix} - (-13) \begin{vmatrix} -9 & 4 \\ 12 & -14 - \lambda \end{vmatrix} + 10 \begin{vmatrix} -9 & 10 - \lambda \\ 12 & -5 \end{vmatrix}. \] Now, we compute each of the 2x2 determinants: \[ \begin{vmatrix} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{vmatrix} = (10 - \lambda)(-14 - \lambda) - (4)(-5) = -140 - 10\lambda + \lambda^2 + 20 = \lambda^2 - 10\lambda - 120, \] \[ \begin{vmatrix} -9 & 4 \\ 12 & -14 - \lambda \end{vmatrix} = (-9)(-14 - \lambda) - (4)(12) = 126 + 9\lambda - 48 = 9\lambda + 78, \] \[ \begin{vmatrix} -9 & 10 - \lambda \\ 12 & -5 \end{vmatrix} = (-9)(-5) - (12)(10 - \lambda) = 45 - 120 + 12\lambda = 12\lambda - 75. \] Substituting these back into the determinant expansion: \[ p(\lambda) = (-4 - \lambda)(\lambda^2 - 10\lambda - 120) + 13(9\lambda + 78) + 10(12\lambda - 75). \] Expanding each term: \[ (-4 - \lambda)(\lambda^2 - 10\lambda - 120) = -4\lambda^2 + 40\lambda + 480 - \lambda^3 + 10\lambda^2 + 120\lambda = -\lambda^3 + 6\lambda^2 + 160\lambda + 480, \] \[ 13(9\lambda + 78) = 117\lambda + 1014, \] \[ 10(12\lambda - 75) = 120\lambda - 750. \] Adding these together: \[ p(\lambda) = -\lambda^3 + 6\lambda^2 + 160\lambda + 480 + 117\lambda + 1014 + 120\lambda - 750, \] \[ p(\lambda) = -\lambda^3 + 6\lambda^2 + (160 + 117 + 120)\lambda + (480 + 1014 - 750), \] \[ p(\lambda) = -\lambda^3 + 6\lambda^2 + 397\lambda + 744. \] Thus, the characteristic polynomial of the matrix is: \[ \boxed{-\lambda^3 + 6\lambda^2 + 397\lambda + 744}. \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |
| deepseek-v3 | To find the **characteristic polynomial** of the given matrix, we follow these steps: ### Given Matrix: \[ A = \left(\begin{array}{ccc} -4 & -13 & 10 \\ -9 & 10 & 4 \\ 12 & -5 & -14 \\ \end{array}\right) \] ### Step 1: Set Up the Characteristic Equation The characteristic polynomial is found by calculating the determinant of \( A - \lambda I \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. \[ A - \lambda I = \left(\begin{array}{ccc} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{array}\right) \] ### Step 2: Compute the Determinant We need to find: \[ \det(A - \lambda I) = \left|\begin{array}{ccc} -4 - \lambda & -13 & 10 \\ -9 & 10 - \lambda & 4 \\ 12 & -5 & -14 - \lambda \\ \end{array}\right| \] Using the rule for 3x3 matrices: \[ \det(A - \lambda I) = (-4 - \lambda) \cdot \left| \begin{array}{cc} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{array} \right| - (-13) \cdot \left| \begin{array}{cc} -9 & 4 \\ 12 & -14 - \lambda \end{array} \right| + 10 \cdot \left| \begin{array}{cc} -9 & 10 - \lambda \\ 12 & -5 \end{array} \right| \] Calculating each minor: 1. **First Minor:** \[ \left| \begin{array}{cc} 10 - \lambda & 4 \\ -5 & -14 - \lambda \end{array} \right| = (10 - \lambda)(-14 - \lambda) - (4)(-5) = \lambda^2 + 4\lambda - 140 + 20 = \lambda^2 + 4\lambda - 120 \] 2. **Second Minor:** \[ \left| \begin{array}{cc} -9 & 4 \\ 12 & -14 - \lambda \end{array} \right| = (-9)(-14 - \lambda) - (4)(12) = 126 + 9\lambda - 48 = 9\lambda + 78 \] 3. **Third Minor:** \[ \left| \begin{array}{cc} -9 & 10 - \lambda \\ 12 & -5 \end{array} \right| = (-9)(-5) - (10 - \lambda)(12) = 45 - 120 + 12\lambda = 12\lambda - 75 \] Now, combine these results: \[ \det(A - \lambda I) = (-4 - \lambda)(\lambda^2 + 4\lambda - 120) + 13(9\lambda + 78) + 10(12\lambda - 75) \] Expanding each term: 1. **First Term:** \[ (-4 - \lambda)(\lambda^2 + 4\lambda - 120) = -4\lambda^2 - 16\lambda + 480 - \lambda^3 - 4\lambda^2 + 120\lambda = -\lambda^3 - 8\lambda^2 + 104\lambda + 480 \] 2. **Second Term:** \[ 13(9\lambda + 78) = 117\lambda + 1014 \] 3. **Third Term:** \[ 10(12\lambda - 75) = 120\lambda - 750 \] Combine all terms: \[ \det(A - \lambda I) = (-\lambda^3 - 8\lambda^2 + 104\lambda + 480) + (117\lambda + 1014) + (120\lambda - 750) \] Simplify: \[ \det(A - \lambda I) = -\lambda^3 - 8\lambda^2 + 341\lambda + 744 \] ### Final Answer: \[ \boxed{-\lambda^3 - 8\lambda^2 + 341\lambda + 744} \] |
-\lambda ^3-8 \lambda ^2+341 \lambda +744 |