金声玉亮2.0模型评测报告

Find the characteristic polynomial of the following matrix:\n$\left(\n\begin{array}{ccc}\n -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\\n\end{array}\n\right)$. Please put your final answer in a $\\boxed{}$.

答案

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 minus \( \lambda \) times the identity matrix, where \( \lambda \) is the eigenvalue. The matrix is: \[ A = \left( \begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\ \end{array} \right) \] We form the matrix \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{array} \right) \] The characteristic polynomial is given by the determinant of \( A - \lambda I \): \[ \det(A - \lambda I) = \det \left( \begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{array} \right) \] We expand this determinant along the first row: \[ \det(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \det \left( \begin{array}{cc} \frac{93}{7} - \lambda & -6 \\ -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{array} \right) \] \[ - \left( -\frac{41}{7} \right) \det \left( \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \\ \end{array} \right) \] \[ + \left( -\frac{87}{7} \right) \det \left( \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \\ \end{array} \right) \] We compute each of these 2x2 determinants: 1. \[ \det \left( \begin{array}{cc} \frac{93}{7} - \lambda & -6 \\ -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{array} \right) = \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( -\frac{26}{7} \right) \] \[ = \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \frac{156}{7} \] \[ = \frac{8835}{49} - \frac{188}{7} \lambda + \lambda^2 - \frac{156}{7} \] \[ = \lambda^2 - \frac{188}{7} \lambda + \frac{8679}{49} \] 2. \[ \det \left( \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \\ \end{array} \right) = \left( -\frac{83}{7} \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( \frac{23}{7} \right) \] \[ = -\frac{7885}{49} + \frac{83}{7} \lambda + \frac{138}{7} \] \[ = \frac{83}{7} \lambda - \frac{7747}{49} \] 3. \[ \det \left( \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \\ \end{array} \right) = \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - \lambda \right) \left( \frac{23}{7} \right) \] \[ = \frac{2158}{49} - \frac{2139}{49} + \frac{23}{7} \lambda \] \[ = \frac{23}{7} \lambda - \frac{21}{49} \] Combining these results, we get: \[ \det(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left( \lambda^2 - \frac{188}{7} \lambda + \frac{8679}{49} \right) \] \[ + \frac{41}{7} \left( \frac{83}{7} \lambda - \frac{7747}{49} \right) \] \[ - \frac{87}{7} \left( \frac{23}{7} \lambda - \frac{21}{49} \right) \] After expanding and simplifying, we find the characteristic polynomial to be: \[ \boxed{\lambda^3 - 10 \lambda^2 - 11 \lambda - 12} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
deepseek-r1	To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \end{pmatrix}, \] we need to compute the determinant of \(\lambda I - A\). The matrix \(\lambda I - A\) is: \[ \begin{pmatrix} \lambda + \frac{61}{7} & \frac{41}{7} & \frac{87}{7} \\ \frac{83}{7} & \lambda - \frac{93}{7} & 6 \\ -\frac{23}{7} & \frac{26}{7} & \lambda - \frac{95}{7} \end{pmatrix}. \] We compute the determinant by expanding along the first row: \[ \det(\lambda I - A) = \left(\lambda + \frac{61}{7}\right) \det \begin{pmatrix} \lambda - \frac{93}{7} & 6 \\ \frac{26}{7} & \lambda - \frac{95}{7} \end{pmatrix} - \frac{41}{7} \det \begin{pmatrix} \frac{83}{7} & 6 \\ -\frac{23}{7} & \lambda - \frac{95}{7} \end{pmatrix} + \frac{87}{7} \det \begin{pmatrix} \frac{83}{7} & \lambda - \frac{93}{7} \\ -\frac{23}{7} & \frac{26}{7} \end{pmatrix}. \] Calculating each minor determinant: 1. For the first minor: \[ \det \begin{pmatrix} \lambda - \frac{93}{7} & 6 \\ \frac{26}{7} & \lambda - \frac{95}{7} \end{pmatrix} = \left(\lambda - \frac{93}{7}\right)\left(\lambda - \frac{95}{7}\right) - 6 \cdot \frac{26}{7} = \lambda^2 - \frac{188}{7}\lambda + \frac{7743}{49}. \] 2. For the second minor: \[ \det \begin{pmatrix} \frac{83}{7} & 6 \\ -\frac{23}{7} & \lambda - \frac{95}{7} \end{pmatrix} = \frac{83}{7}\left(\lambda - \frac{95}{7}\right) - 6 \cdot \left(-\frac{23}{7}\right) = \frac{83}{7}\lambda - \frac{6919}{49}. \] 3. For the third minor: \[ \det \begin{pmatrix} \frac{83}{7} & \lambda - \frac{93}{7} \\ -\frac{23}{7} & \frac{26}{7} \end{pmatrix} = \frac{83}{7} \cdot \frac{26}{7} - \left(\lambda - \frac{93}{7}\right) \cdot \left(-\frac{23}{7}\right) = \frac{2158}{49} + \frac{23}{7}\lambda - \frac{2139}{49} = \frac{23}{7}\lambda + \frac{19}{49}. \] Substituting these minors back into the expansion and combining terms, we get: \[ \det(\lambda I - A) = \lambda^3 - \frac{127}{7}\lambda^2 - \frac{5127}{49}\lambda + \frac{757655}{343}. \] Multiplying through by 343 to clear the denominators, we obtain the characteristic polynomial with integer coefficients: \[ \boxed{343\lambda^3 - 6223\lambda^2 - 35889\lambda + 757655}. \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
gpto1	To find the characteristic polynomial \( \det (A - \lambda I) \) of the given matrix \( A \), we first consider \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{pmatrix} \] To simplify calculations, we multiply the entire matrix by 7 to eliminate denominators: \[ B = 7(A - \lambda I) = \begin{pmatrix} -61 - 7\lambda & -41 & -87 \\ -83 & 93 - 7\lambda & -42 \\ 23 & -26 & 95 - 7\lambda \\ \end{pmatrix} \] We note that \( \det(B) = 7^3 \det(A - \lambda I) \), so \( \det(A - \lambda I) = \frac{1}{343} \det(B) \). Calculating \( \det(B) \) using the cofactor expansion along the first row, we obtain: \[ \begin{align} \det(B) &= (-61 - 7\lambda) \cdot \det\begin{pmatrix} 93 - 7\lambda & -42 \\ -26 & 95 - 7\lambda \end{pmatrix} - (-41) \cdot \det\begin{pmatrix} -83 & -42 \\ 23 & 95 - 7\lambda \end{pmatrix} + (-87) \cdot \det\begin{pmatrix} -83 & 93 - 7\lambda \\ 23 & -26 \end{pmatrix} \\ &= (-61 - 7\lambda)(7743 - 188\lambda + \lambda^2) - (-41)(-6919 + 83\lambda) + (-87)(19 + 23\lambda) \\ &= -190,297 - 1,679\lambda + 127\lambda^2 - \lambda^3 \end{align} \] Substituting back \( 7\lambda = x \) and simplifying, we find: \[ \det(A - \lambda I) = \frac{1}{343} \det(B) = \lambda^3 - \frac{127}{7} \lambda^2 + \frac{1679}{49} \lambda + \frac{190,297}{343} \] Therefore, the characteristic polynomial of the matrix \( A \) is: \[ \boxed{\lambda^3 - \dfrac{127}{7} \lambda^{2} + \dfrac{1679}{49} \lambda + \dfrac{190,297}{343}} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
gpt-4o-2024-08-06	To find the characteristic polynomial of a matrix \( A \), we need to compute the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar. The matrix \( A \) is given by: \[ A = \begin{pmatrix} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \end{pmatrix} \] The identity matrix \( I \) is: \[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] Thus, \( A - \lambda I \) is: \[ A - \lambda I = \begin{pmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{pmatrix} \] The characteristic polynomial is the determinant of this matrix: \[ \det(A - \lambda I) = \begin{vmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix} \] We will use the cofactor expansion along the first row to compute this determinant: \[ = \left(-\frac{61}{7} - \lambda\right) \begin{vmatrix} \frac{93}{7} - \lambda & -6 \\ -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix} - \left(-\frac{41}{7}\right) \begin{vmatrix} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{vmatrix} + \left(-\frac{87}{7}\right) \begin{vmatrix} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{vmatrix} \] Now, we compute each of these 2x2 determinants: 1. \(\begin{vmatrix} \frac{93}{7} - \lambda & -6 \\ -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix} = \left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - (-6)\left(-\frac{26}{7}\right)\) \[ = \left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - \frac{156}{7} \] 2. \(\begin{vmatrix} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{vmatrix} = \left(-\frac{83}{7}\right)\left(\frac{95}{7} - \lambda\right) - (-6)\left(\frac{23}{7}\right)\) \[ = -\frac{83}{7}\left(\frac{95}{7} - \lambda\right) + \frac{138}{7} \] 3. \(\begin{vmatrix} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{vmatrix} = \left(-\frac{83}{7}\right)\left(-\frac{26}{7}\right) - \left(\frac{93}{7} - \lambda\right)\left(\frac{23}{7}\right)\) \[ = \frac{2158}{49} - \frac{23}{7}\left(\frac{93}{7} - \lambda\right) \] Substitute these back into the determinant expression: \[ = \left(-\frac{61}{7} - \lambda\right)\left(\left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - \frac{156}{7}\right) + \frac{41}{7}\left(-\frac{83}{7}\left(\frac{95}{7} - \lambda\right) + \frac{138}{7}\right) - \frac{87}{7}\left(\frac{2158}{49} - \frac{23}{7}\left(\frac{93}{7} - \lambda\right)\right) \] This expression simplifies to a cubic polynomial in \(\lambda\). After simplification, the characteristic polynomial is: \[ \boxed{\lambda^3 - 9\lambda^2 + 10\lambda - 4} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
DeepSeek-Coder-V2-Lite-Instruct	To find the characteristic polynomial of the given matrix, we need to compute the determinant of the matrix \( A - xI \), where \( A \) is the given matrix and \( I \) is the identity matrix of the same size. The characteristic polynomial is given by \( \det(A - xI) \). Given matrix \( A \): \[ A = \left( \begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\ \end{array} \right) \] First, we form the matrix \( A - xI \): \[ A - xI = \left( \begin{array}{ccc} -\frac{61}{7} - x & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - x & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - x \\ \end{array} \right) \] Next, we compute the determinant of \( A - xI \). We will use the cofactor expansion along the first row for simplicity: \[ \det(A - xI) = \left( -\frac{61}{7} - x \right) \left\| \begin{array}{cc} \frac{93}{7} - x & -6 \\ -\frac{26}{7} & \frac{95}{7} - x \\ \end{array} \right\| - \left( -\frac{41}{7} \right) \left\| \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - x \\ \end{array} \right\| - \left( -\frac{87}{7} \right) \left\| \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - x \\ \frac{23}{7} & -\frac{26}{7} \\ \end{array} \right\| \] We now compute each of the 2x2 determinants: \[ \left\| \begin{array}{cc} \frac{93}{7} - x & -6 \\ -\frac{26}{7} & \frac{95}{7} - x \\ \end{array} \right\| = \left( \frac{93}{7} - x \right) \left( \frac{95}{7} - x \right) - (-6) \left( -\frac{26}{7} \right) = \left( \frac{93}{7} - x \right) \left( \frac{95}{7} - x \right) - \frac{156}{7} = \frac{8835}{49} - \frac{188x}{7} + x^2 - \frac{156}{7} = x^2 - \frac{188x}{7} + \frac{8835}{49} - \frac{1092}{49} = x^2 - \frac{188x}{7} + \frac{7743}{49} \] \[ \left\| \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - x \\ \end{array} \right\| = \left( -\frac{83}{7} \right) \left( \frac{95}{7} - x \right) - (-6) \left( \frac{23}{7} \right) = -\frac{7885}{49} + \frac{83x}{7} + \frac{138}{7} = -\frac{7885}{49} + \frac{83x}{7} + \frac{966}{49} = \frac{83x}{7} - \frac{6919}{49} \] \[ \left\| \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - x \\ \frac{23}{7} & -\frac{26}{7} \\ \end{array} \right\| = \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - x \right) \left( \frac{23}{7} \right) = \frac{2158}{49} - \frac{2139}{49} + \frac{23x}{7} = \frac{19}{49} + \frac{23x}{7} \] Substituting these back into the determinant expression: \[ \det(A - xI) = \left( -\frac{61}{7} - x \right) \left( x^2 - \frac{188x}{7} + \frac{7743}{49} \right) - \left( -\frac{41}{7} \right) \left( \frac{83x}{7} - \frac{6919}{49} \right) - \left( -\frac{87}{7} \right) \left( \frac{19}{49} + \frac{23x}{7} \right) \] Expanding and simplifying this expression will yield the characteristic polynomial. After performing the multiplications and simplifications, we get: \[ \det(A - xI) = -x^3 + \frac{24x^2}{7} + \frac{16190x}{49} - \frac{1011644}{343} \] Thus, the characteristic polynomial is: \[ \boxed{-x^3 + \frac{24x^2}{7} + \frac{16190x}{49} - \frac{1011644}{343}} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
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 $A - \lambda I$, where $A$ is the given matrix, $\lambda$ is the eigenvalue, and $I$ is the identity matrix. ## Step 2: The given matrix is a $3 \times 3$ matrix. We will subtract $\lambda$ from the diagonal elements of the matrix and then find the determinant of the resulting matrix. ## Step 3: The matrix $A - \lambda I$ will be: \[\begin{pmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{pmatrix}\] ## Step 4: To find the determinant of this matrix, we will expand along the first row. The determinant is given by: \[\begin{vmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix}\] ## Step 5: Expanding the determinant, we get a polynomial in terms of $\lambda$, which is the characteristic polynomial of the matrix. ## Step 6: After calculating the determinant, we get the characteristic polynomial as a cubic polynomial in terms of $\lambda$. ## Step 7: The characteristic polynomial is $-\lambda^3 + 9\lambda^2 - 27\lambda + 27$. ## Step 8: Therefore, the characteristic polynomial of the given matrix is $\boxed{-\lambda^3 + 9\lambda^2 - 27\lambda + 27}$. The final answer is: $\boxed{-\lambda^3 + 9\lambda^2 - 27\lambda + 27}$	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
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: $\left(\begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{array}\right)$. ## Step 3: Calculate the determinant of $A - \lambda I$ Using the formula for the determinant of a 3x3 matrix, we get: $\det(A - \lambda I) = \left(-\frac{61}{7} - \lambda\right)\left(\left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - (-6)\left(-\frac{26}{7}\right)\right) - \left(-\frac{41}{7}\right)\left((-6)\left(\frac{23}{7}\right) - \left(-\frac{83}{7}\right)\left(\frac{95}{7} - \lambda\right)\right) + \left(-\frac{87}{7}\right)\left((-6)\left(-\frac{83}{7}\right) - \left(\frac{23}{7}\right)\left(\frac{93}{7} - \lambda\right)\right)$. ## Step 4: Simplify the determinant Expanding and simplifying the determinant, we get: $\det(A - \lambda I) = -\frac{1}{7}\left(\lambda^3 - 187\lambda^2 + 703\lambda - 451\right)$. ## Step 5: Write down the characteristic polynomial The characteristic polynomial is given by: $\left(\lambda^3 - 187\lambda^2 + 703\lambda - 451\right)$. The final answer is: $\boxed{\lambda^3 - 187\lambda^2 + 703\lambda - 451}$	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
Qwen2-72B-Instruct	To find the characteristic polynomial of a matrix, we need to compute its determinant after subtracting $\lambda$ from each diagonal element. The given matrix is: \[ A = \left(\begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \end{array}\right) \] The characteristic polynomial $p(\lambda)$ is defined as: \[ p(\lambda) = \det(A - \lambda I) \] where $I$ is the identity matrix of the same size as $A$. Thus, \[ A - \lambda I = \left(\begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{array}\right) \] Now, we calculate the determinant of this matrix: \[ \begin{align} p(\lambda) &= \left\| \begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{array} \right\| \\ &= \left( -\frac{61}{7} - \lambda \right) \left\| \begin{array}{cc} \frac{93}{7} - \lambda & -6 \\ - \frac{26}{7} & \frac{95}{7} - \lambda \end{array} \right\| \\ &\quad - \left( -\frac{41}{7} \right) \left\| \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{array} \right\| \\ &\quad + \left( -\frac{87}{7} \right) \left\| \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{array} \right\| \\ &= \left( -\frac{61}{7} - \lambda \right) \left( \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) + 6 \cdot \frac{26}{7} \right) \\ &\quad + \frac{41}{7} \left( -\frac{83}{7} \left( \frac{95}{7} - \lambda \right) + 6 \cdot \frac{23}{7} \right) \\ &\quad - \frac{87}{7} \left( -\frac{83}{7} \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - \lambda \right) \frac{23}{7} \right) \\ &= \left( -\frac{61}{7} - \lambda \right) \left( \frac{8835}{49} - \frac{188}{7} \lambda + \lambda^2 + \frac{156}{7} \right) \\ &\quad + \frac{41}{7} \left( -\frac{7885}{49} + \frac{83}{7} \lambda + \frac{138}{7} \right) \\ &\quad - \frac{87}{7} \left( \frac{2158}{49} - \frac{2139}{49} \lambda \right) \\ &= \left( -\frac{61}{7} - \lambda \right) \left( \lambda^2 - \frac{188}{7} \lambda + \frac{9039}{49} \right) \\ &\quad + \frac{41}{7} \left( \frac{83}{7} \lambda - \frac{7747}{49} \right) \\ &\quad - \frac{87}{7} \left( \frac{2158}{49} - \frac{2139}{49} \lambda \right) \\ &= \lambda^3 - \frac{127}{7} \lambda^2 + \frac{10202}{49} \lambda - \frac{542340}{343} \end{align} \] Thus, the characteristic polynomial of the given matrix is: \[ p(\lambda) = \lambda^3 - \frac{127}{7} \lambda^2 + \frac{10202}{49} \lambda - \frac{542340}{343} \] So, the final answer is: \[ \boxed{\lambda^3 - \frac{127}{7} \lambda^2 + \frac{10202}{49} \lambda - \frac{542340}{343}} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
baichuan4	To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix minus $x$ times the identity matrix. Given the matrix $A$: $$ A = \left( \begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\ \end{array} \right) $$ We first construct the matrix $A - xI$, where $I$ is the identity matrix: $$ A - xI = \left( \begin{array}{ccc} -\frac{61}{7} - x & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - x & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - x \\ \end{array} \right) $$ Now, we compute the determinant of this matrix: $$ \begin{aligned} \det(A - xI) &= \left\| \begin{array}{ccc} -\frac{61}{7} - x & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - x & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - x \\ \end{array} \right\| \\ &= (-\frac{61}{7} - x) \left\| \begin{array}{cc} \frac{93}{7} - x & -6 \\ -\frac{26}{7} & \frac{95}{7} - x \\ \end{array} \right\| - (-\frac{41}{7}) \left\| \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - x \\ \end{array} \right\| + (-\frac{87}{7}) \left\| \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - x \\ \frac{23}{7} & -\frac{26}{7} \\ \end{array} \right\| \\ &= (-\frac{61}{7} - x) \left( (\frac{93}{7} - x)(\frac{95}{7} - x) - (-6)(-\frac{26}{7}) \right) - (-\frac{41}{7}) \left( (-\frac{83}{7})(\frac{95}{7} - x) - (-6)(\frac{23}{7}) \right) \\ &\quad + (-\frac{87}{7}) \left( (-\frac{83}{7})(-\frac{26}{7}) - (\frac{93}{7} - x)(\frac{23}{7}) \right) \\ &= (-\frac{61}{7} - x) \left( \frac{8855}{49} - \frac{93}{7}x - \frac{95}{7}x + x^2 - \frac{156}{7} \right) + \frac{41}{7} \left( \frac{7855}{49} - \frac{83}{7}x + \frac{138}{7} \right) \\ &\quad - \frac{87}{7} \left( \frac{2158}{49} - \frac{23}{7}x \right) \\ &= (-\frac{61}{7} - x) \left( x^2 - \frac{188}{7}x + \frac{8709}{49} \right) + \frac{41}{7} \left( \frac{8031}{49} - \frac{83}{7}x \right) - \frac{87}{7} \left( \frac{2158}{49} - \frac{23}{7}x \right) \\ &= (-\frac{61}{7} - x) \left( x^2 - \frac{188}{7}x + \frac{8709}{49} \right) + \frac{41}{7} \left( \frac{8031}{49} - \frac{83}{7}x \right) - \frac{87}{7} \left( \frac{2158}{49} - \frac{23}{7}x \right). \end{aligned} $$ This expression can be further simplified by expanding and combining like terms. However, the computation is quite lengthy and can be simplified using a computer algebra system or by carefully expanding and simplifying each term manually. The final result will be a cubic polynomial in $x$ which is the characteristic polynomial of the matrix $A$. The coefficients of the polynomial will depend on the exact computation of the determinant above.	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
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 $\det(A - \lambda I) = \begin{vmatrix}a - \lambda & b & c\\d & e - \lambda & f\\g & h & i - \lambda\end{vmatrix}$. For the given matrix $A = \begin{pmatrix}-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7}\\-\frac{83}{7} & \frac{93}{7} & -6\\\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}\end{pmatrix}$, the characteristic polynomial is: \[ \begin{align} &\begin{vmatrix}-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7}\\-\frac{83}{7} & \frac{93}{7} - \lambda & -6\\\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda\end{vmatrix}\\ =&(-\frac{61}{7} - \lambda)\times\begin{vmatrix}\frac{93}{7} - \lambda & -6\\-\frac{26}{7} & \frac{95}{7} - \lambda\end{vmatrix} - (-\frac{41}{7})\times\begin{vmatrix}-\frac{83}{7} & -6\\\frac{23}{7} & \frac{95}{7} - \lambda\end{vmatrix} - (-\frac{87}{7})\times\begin{vmatrix}-\frac{83}{7} & \frac{93}{7} - \lambda\\\frac{23}{7} & -\frac{26}{7}\end{vmatrix}\\ =&(-\frac{61}{7} - \lambda)[(\frac{93}{7} - \lambda)(\frac{95}{7} - \lambda) - (-6)\times(-\frac{26}{7})] + \frac{41}{7}[(-\frac{83}{7})(\frac{95}{7} - \lambda) - (-6)\times\frac{23}{7}] + \frac{87}{7}[(-\frac{83}{7})(-\frac{26}{7}) - (\frac{93}{7} - \lambda)\times\frac{23}{7}] \end{align} \] After expanding and simplifying, we get the characteristic polynomial as: $\boxed{-\lambda^3 + \frac{249}{7}\lambda^2 - \frac{21259}{49}\lambda + \frac{37337}{343}}$.	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
gpto1	To find the characteristic polynomial \( \det (A - \lambda I) \) of the given matrix \( A \), we first consider \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{pmatrix} \] To simplify calculations, we multiply the entire matrix by 7 to eliminate denominators: \[ B = 7(A - \lambda I) = \begin{pmatrix} -61 - 7\lambda & -41 & -87 \\ -83 & 93 - 7\lambda & -42 \\ 23 & -26 & 95 - 7\lambda \\ \end{pmatrix} \] We note that \( \det(B) = 7^3 \det(A - \lambda I) \), so \( \det(A - \lambda I) = \frac{1}{343} \det(B) \). Calculating \( \det(B) \) using the cofactor expansion along the first row, we obtain: \[ \begin{align} \det(B) &= (-61 - 7\lambda) \cdot \det\begin{pmatrix} 93 - 7\lambda & -42 \\ -26 & 95 - 7\lambda \end{pmatrix} - (-41) \cdot \det\begin{pmatrix} -83 & -42 \\ 23 & 95 - 7\lambda \end{pmatrix} + (-87) \cdot \det\begin{pmatrix} -83 & 93 - 7\lambda \\ 23 & -26 \end{pmatrix} \\ &= (-61 - 7\lambda)(7743 - 188\lambda + \lambda^2) - (-41)(-6919 + 83\lambda) + (-87)(19 + 23\lambda) \\ &= -190,297 - 1,679\lambda + 127\lambda^2 - \lambda^3 \end{align} \] Substituting back \( 7\lambda = x \) and simplifying, we find: \[ \det(A - \lambda I) = \frac{1}{343} \det(B) = \lambda^3 - \frac{127}{7} \lambda^2 + \frac{1679}{49} \lambda + \frac{190,297}{343} \] Therefore, the characteristic polynomial of the matrix \( A \) is: \[ \boxed{\lambda^3 - \dfrac{127}{7} \lambda^{2} + \dfrac{1679}{49} \lambda + \dfrac{190,297}{343}} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
kimi	To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix A - λI, where A is the given matrix, λ is a scalar, and I is the identity matrix of the same size as A. Given matrix A: $\left(\n\begin{array}{ccc}\n -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\\n\end{array}\n\right)$ The identity matrix I of size 3x3 is: $\left(\n\begin{array}{ccc}\n 1 & 0 & 0 \\\n 0 & 1 & 0 \\\n 0 & 0 & 1 \\\n\end{array}\n\right)$ Now, we need to compute A - λI: $\left(\n\begin{array}{ccc}\n -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\n\right)$ Now, we need to compute the determinant of this matrix: $\det(A - \lambda I) = \left\|\begin{array}{ccc}\n -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right\|$ To compute the determinant, we can use the cofactor expansion along the first row: $\det(A - \lambda I) = (-\frac{61}{7} - \lambda)\left\|\begin{array}{cc}\n \frac{93}{7} - \lambda & -6 \\\n -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right\| - (-\frac{41}{7})\left\|\begin{array}{cc}\n -\frac{83}{7} & -6 \\\n \frac{23}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right\| + (-\frac{87}{7})\left\|\begin{array}{cc}\n -\frac{83}{7} & \frac{93}{7} - \lambda \\\n \frac{23}{7} & -\frac{26}{7} \\\n\end{array}\right\|$ Now, we need to compute the determinants of the 2x2 matrices: $\left\|\begin{array}{cc}\n \frac{93}{7} - \lambda & -6 \\\n -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right\| = (\frac{93}{7} - \lambda)(\frac{95}{7} - \lambda) - (-6)(-\frac{26}{7})$ $\left\|\begin{array}{cc}\n -\frac{83}{7} & -6 \\\n \frac{23}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right\| = (-\frac{83}{7})(\frac{95}{7} - \lambda) - (-6)(\frac{23}{7})$ $\left\|\begin{array}{cc}\n -\frac{83}{7} & \frac{93}{7} - \lambda \\\n \frac{23}{7} & -\frac{26}{7} \\\n\end{array}\right\| = (-\frac{83}{7})(-\frac{26}{7}) - (\frac{93}{7} - \lambda	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
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, and $\lambda$ represents the eigenvalues. The matrix $A$ is: $$ A = \left(\begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \end{array}\right) $$ The identity matrix $I$ of size $3 \times 3$ is: $$ I = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) $$ We subtract $\lambda I$ from $A$ to get: $$ A - \lambda I = \left(\begin{array}{ccc} -\frac{61}{	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
wenxinyiyan	To find the characteristic polynomial of the given matrix, we first write down the matrix and its characteristic equation: $A = \left( \begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\ \end{array} \right)$ The characteristic equation of $A$ is given by: $\det(A - \lambda I) = 0$ where $I$ is the identity matrix and $\lambda$ is the eigenvalue. Now, let's compute the determinant of $A - \lambda I$: $A - \lambda I = \left( \begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\ \end{array} \right)$ Computing the determinant, we get: $\det(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) + \left( -\frac{41}{7} \right) \left( -6 \right) \left( \frac{23}{7} \right) + \left( -\frac{87}{7} \right) \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( -\frac{87}{7} \right) \left( \frac{93}{7} - \lambda \right) \left( \frac{23}{7} \right) - \left( -\frac{41}{7} \right) \left( -6 \right) \left( \frac{95}{7} - \lambda \right) - \left( -\frac{61}{7} - \lambda \right) \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right)$ After simplifying, we obtain the characteristic polynomial: $\boxed{-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{1515 \lambda }{7}-\frac{11110}{7}}$	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
zhipu	To find the characteristic polynomial of a matrix \( A \), we need to compute the determinant of \( A - \lambda I \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix of the same size as \( A \). Given the matrix: \[ A = \left( \begin{array}{ccc} -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \end{array} \right) \] We form the matrix \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{array} \right) \] Next, we compute the determinant of \( A - \lambda I \): \[ \text{det}(A - \lambda I) = \left\| \begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{array} \right\| \] We can use the cofactor expansion along the first row: \[ \text{det}(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left\| \begin{array}{cc} \frac{93}{7} - \lambda & -6 \\ - \frac{26}{7} & \frac{95}{7} - \lambda \end{array} \right\| - \left( -\frac{41}{7} \right) \left\| \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{array} \right\| + \left( -\frac{87}{7} \right) \left\| \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{array} \right\| \] We compute each of these 2x2 determinants: \[ \left\| \begin{array}{cc} \frac{93}{7} - \lambda & -6 \\ - \frac{26}{7} & \frac{95}{7} - \lambda \end{array} \right\| = \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( -\frac{26}{7} \right) \] \[ = \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \frac{156}{7} \] \[ = \frac{93 \cdot 95}{49} - \frac{93 \lambda}{7} - \frac{95 \lambda}{7} + \lambda^2 - \frac{156}{7} \] \[ = \frac{8835}{49} - \frac{188 \lambda}{7} + \lambda^2 - \frac{156}{7} \] \[ = \frac{8835 - 1092}{49} + \lambda^2 - \frac{188 \lambda}{7} \] \[ = \frac{7743}{49} + \lambda^2 - \frac{188 \lambda}{7} \] \[ = \lambda^2 - \frac{188 \lambda}{7} + \frac{7743}{49} \] Next determinant: \[ \left\| \begin{array}{cc} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{array} \right\| = \left( -\frac{83}{7} \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( \frac{23}{7} \right) \] \[ = -\frac{83 \cdot 95}{49} + \frac{83 \lambda}{7} + \frac{138}{7} \] \[ = -\frac{7885}{49} + \frac{83 \lambda}{7} + \frac{138}{7} \] \[ = -\frac{7885}{49} + \frac{83 \lambda + 138}{7} \] \[ = -\frac{7885}{49} + \frac{83 \lambda + 138}{7} \] \[ = -\frac{7885}{49} + \frac{83 \lambda + 138}{7} \] \[ = -\frac{7885}{49} + \frac{83 \lambda + 138}{7} \] Third determinant: \[ \left\| \begin{array}{cc} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{array} \right\| = \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - \lambda \right) \left( \frac{23}{7} \right) \] \[ = \frac{83 \cdot 26}{49} - \left( \frac{93 \cdot 23}{49} - \frac{23 \lambda}{7} \right) \] \[ = \frac{2158}{49} - \frac{2139}{49} + \frac{23 \lambda}{7} \] \[ = \frac{19}{49} + \frac{23 \lambda}{7} \] Putting it all together: \[ \text{det}(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left( \lambda^2 - \frac{188 \lambda}{7} + \frac{7743}{49} \right) + \frac{41}{7} \left( -\frac{7885}{49} + \frac{83 \lambda + 138}{7} \right) - \frac{87}{7} \left( \frac{19}{49} + \frac{23 \lambda}{7} \right) \] After simplifying the above expression, we get the characteristic polynomial: \[ \boxed{\lambda^3 - \frac{188 \lambda^2}{7} + \frac{7743 \lambda}{49} - \frac{7885}{49}} \] This can be further simplified to: \[ \boxed{\lambda^3 - \frac{188}{7} \lambda^2 + \frac{7743}{49} \lambda - \frac{7885}{49}} \] Or in a more compact form: \[ \boxed{\lambda^3 - \frac{188}{7} \lambda^2 + \frac{7743 - 7885}{49} \lambda} \] \[ \boxed{\lambda^3 - \frac{188}{7} \lambda^2 - \frac{142}{49} \lambda} \] Thus, the characteristic polynomial is: \[ \boxed{\lambda^3 - \frac{188}{7} \lambda^2 - \frac{142}{49} \lambda} \]	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}
deepseek-v3	Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-3391690d-7719-9a1f-a166-88685a2bc16d', 'request_id': '3391690d-7719-9a1f-a166-88685a2bc16d'}	-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

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金声玉亮2.0

To find the characteristic polynomial of the given matrix, we need to compute the determinant of the matrix minus $ \lambda $ times the identity matrix, where $ \lambda $ is the eigenvalue. The matrix is:

\[
A = \left(
\begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\
\end{array}
\right)
\]

We form the matrix $ A - \lambda I $:

\[
A - \lambda I = \left(
\begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right)
\]

The characteristic polynomial is given by the determinant of $ A - \lambda I $:

\[
\det(A - \lambda I) = \det \left(
\begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right)
\]

We expand this determinant along the first row:

\[
\det(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \det \left(
\begin{array}{cc}
\frac{93}{7} - \lambda & -6 \\
-\frac{26}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right)
\]
\[
- \left( -\frac{41}{7} \right) \det \left(
\begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right)
\]
\[
+ \left( -\frac{87}{7} \right) \det \left(
\begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - \lambda \\
\frac{23}{7} & -\frac{26}{7} \\
\end{array}
\right)
\]

We compute each of these 2x2 determinants:

1.
\[
\det \left(
\begin{array}{cc}
\frac{93}{7} - \lambda & -6 \\
-\frac{26}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right) = \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( -\frac{26}{7} \right)
\]
\[
= \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \frac{156}{7}
\]
\[
= \frac{8835}{49} - \frac{188}{7} \lambda + \lambda^2 - \frac{156}{7}
\]
\[
= \lambda^2 - \frac{188}{7} \lambda + \frac{8679}{49}
\]

2.
\[
\det \left(
\begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right) = \left( -\frac{83}{7} \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( \frac{23}{7} \right)
\]
\[
= -\frac{7885}{49} + \frac{83}{7} \lambda + \frac{138}{7}
\]
\[
= \frac{83}{7} \lambda - \frac{7747}{49}
\]

3.
\[
\det \left(
\begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - \lambda \\
\frac{23}{7} & -\frac{26}{7} \\
\end{array}
\right) = \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - \lambda \right) \left( \frac{23}{7} \right)
\]
\[
= \frac{2158}{49} - \frac{2139}{49} + \frac{23}{7} \lambda
\]
\[
= \frac{23}{7} \lambda - \frac{21}{49}
\]

Combining these results, we get:

\[
\det(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left( \lambda^2 - \frac{188}{7} \lambda + \frac{8679}{49} \right)
\]
\[
+ \frac{41}{7} \left( \frac{83}{7} \lambda - \frac{7747}{49} \right)
\]
\[
- \frac{87}{7} \left( \frac{23}{7} \lambda - \frac{21}{49} \right)
\]

After expanding and simplifying, we find the characteristic polynomial to be:

\[
\boxed{\lambda^3 - 10 \lambda^2 - 11 \lambda - 12}
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

deepseek-r1

To find the characteristic polynomial of the matrix

\[
A = \begin{pmatrix}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}
\end{pmatrix},
\]

we need to compute the determinant of $\lambda I - A$. The matrix $\lambda I - A$ is:

\[
\begin{pmatrix}
\lambda + \frac{61}{7} & \frac{41}{7} & \frac{87}{7} \\
\frac{83}{7} & \lambda - \frac{93}{7} & 6 \\
-\frac{23}{7} & \frac{26}{7} & \lambda - \frac{95}{7}
\end{pmatrix}.
\]

We compute the determinant by expanding along the first row:

\[
\det(\lambda I - A) = \left(\lambda + \frac{61}{7}\right) \det \begin{pmatrix} \lambda - \frac{93}{7} & 6 \\ \frac{26}{7} & \lambda - \frac{95}{7} \end{pmatrix} - \frac{41}{7} \det \begin{pmatrix} \frac{83}{7} & 6 \\ -\frac{23}{7} & \lambda - \frac{95}{7} \end{pmatrix} + \frac{87}{7} \det \begin{pmatrix} \frac{83}{7} & \lambda - \frac{93}{7} \\ -\frac{23}{7} & \frac{26}{7} \end{pmatrix}.
\]

Calculating each minor determinant:

1. For the first minor:
\[
\det \begin{pmatrix} \lambda - \frac{93}{7} & 6 \\ \frac{26}{7} & \lambda - \frac{95}{7} \end{pmatrix} = \left(\lambda - \frac{93}{7}\right)\left(\lambda - \frac{95}{7}\right) - 6 \cdot \frac{26}{7} = \lambda^2 - \frac{188}{7}\lambda + \frac{7743}{49}.
\]

2. For the second minor:
\[
\det \begin{pmatrix} \frac{83}{7} & 6 \\ -\frac{23}{7} & \lambda - \frac{95}{7} \end{pmatrix} = \frac{83}{7}\left(\lambda - \frac{95}{7}\right) - 6 \cdot \left(-\frac{23}{7}\right) = \frac{83}{7}\lambda - \frac{6919}{49}.
\]

3. For the third minor:
\[
\det \begin{pmatrix} \frac{83}{7} & \lambda - \frac{93}{7} \\ -\frac{23}{7} & \frac{26}{7} \end{pmatrix} = \frac{83}{7} \cdot \frac{26}{7} - \left(\lambda - \frac{93}{7}\right) \cdot \left(-\frac{23}{7}\right) = \frac{2158}{49} + \frac{23}{7}\lambda - \frac{2139}{49} = \frac{23}{7}\lambda + \frac{19}{49}.
\]

Substituting these minors back into the expansion and combining terms, we get:

\[
\det(\lambda I - A) = \lambda^3 - \frac{127}{7}\lambda^2 - \frac{5127}{49}\lambda + \frac{757655}{343}.
\]

Multiplying through by 343 to clear the denominators, we obtain the characteristic polynomial with integer coefficients:

\[
\boxed{343\lambda^3 - 6223\lambda^2 - 35889\lambda + 757655}.
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

gpto1

To find the characteristic polynomial $ \det (A - \lambda I) $ of the given matrix $ A $, we first consider $ A - \lambda I $:

\[
A - \lambda I = \begin{pmatrix}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\
\end{pmatrix}
\]

To simplify calculations, we multiply the entire matrix by 7 to eliminate denominators:

\[
B = 7(A - \lambda I) = \begin{pmatrix}
-61 - 7\lambda & -41 & -87 \\
-83 & 93 - 7\lambda & -42 \\
23 & -26 & 95 - 7\lambda \\
\end{pmatrix}
\]

We note that $ \det(B) = 7^3 \det(A - \lambda I) $, so $ \det(A - \lambda I) = \frac{1}{343} \det(B) $.

Calculating $ \det(B) $ using the cofactor expansion along the first row, we obtain:

\[
\begin{align*}
\det(B) &= (-61 - 7\lambda) \cdot \det\begin{pmatrix} 93 - 7\lambda & -42 \\ -26 & 95 - 7\lambda \end{pmatrix} - (-41) \cdot \det\begin{pmatrix} -83 & -42 \\ 23 & 95 - 7\lambda \end{pmatrix} + (-87) \cdot \det\begin{pmatrix} -83 & 93 - 7\lambda \\ 23 & -26 \end{pmatrix} \\
&= (-61 - 7\lambda)(7743 - 188\lambda + \lambda^2) - (-41)(-6919 + 83\lambda) + (-87)(19 + 23\lambda) \\
&= -190,297 - 1,679\lambda + 127\lambda^2 - \lambda^3
\end{align*}
\]

Substituting back $ 7\lambda = x $ and simplifying, we find:

\[
\det(A - \lambda I) = \frac{1}{343} \det(B) = \lambda^3 - \frac{127}{7} \lambda^2 + \frac{1679}{49} \lambda + \frac{190,297}{343}
\]

Therefore, the characteristic polynomial of the matrix $ A $ is:

\[
\boxed{\lambda^3 - \dfrac{127}{7} \lambda^{2} + \dfrac{1679}{49} \lambda + \dfrac{190,297}{343}}
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

gpt-4o-2024-08-06

To find the characteristic polynomial of a matrix $ A $, we need to compute the determinant of $ A - \lambda I $, where $ I $ is the identity matrix and $ \lambda $ is a scalar. The matrix $ A $ is given by:

\[
A = \begin{pmatrix}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}
\end{pmatrix}
\]

The identity matrix $ I $ is:

\[
I = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\]

Thus, $ A - \lambda I $ is:

\[
A - \lambda I = \begin{pmatrix}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda
\end{pmatrix}
\]

The characteristic polynomial is the determinant of this matrix:

\[
\det(A - \lambda I) = \begin{vmatrix}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda
\end{vmatrix}
\]

We will use the cofactor expansion along the first row to compute this determinant:

\[
= \left(-\frac{61}{7} - \lambda\right) \begin{vmatrix} \frac{93}{7} - \lambda & -6 \\ -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix}
- \left(-\frac{41}{7}\right) \begin{vmatrix} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{vmatrix}
+ \left(-\frac{87}{7}\right) \begin{vmatrix} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{vmatrix}
\]

Now, we compute each of these 2x2 determinants:

1. $\begin{vmatrix} \frac{93}{7} - \lambda & -6 \\ -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix} = \left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - (-6)\left(-\frac{26}{7}\right)$

\[
= \left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - \frac{156}{7}
\]

2. $\begin{vmatrix} -\frac{83}{7} & -6 \\ \frac{23}{7} & \frac{95}{7} - \lambda \end{vmatrix} = \left(-\frac{83}{7}\right)\left(\frac{95}{7} - \lambda\right) - (-6)\left(\frac{23}{7}\right)$

\[
= -\frac{83}{7}\left(\frac{95}{7} - \lambda\right) + \frac{138}{7}
\]

3. $\begin{vmatrix} -\frac{83}{7} & \frac{93}{7} - \lambda \\ \frac{23}{7} & -\frac{26}{7} \end{vmatrix} = \left(-\frac{83}{7}\right)\left(-\frac{26}{7}\right) - \left(\frac{93}{7} - \lambda\right)\left(\frac{23}{7}\right)$

\[
= \frac{2158}{49} - \frac{23}{7}\left(\frac{93}{7} - \lambda\right)
\]

Substitute these back into the determinant expression:

\[
= \left(-\frac{61}{7} - \lambda\right)\left(\left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - \frac{156}{7}\right)
+ \frac{41}{7}\left(-\frac{83}{7}\left(\frac{95}{7} - \lambda\right) + \frac{138}{7}\right)
- \frac{87}{7}\left(\frac{2158}{49} - \frac{23}{7}\left(\frac{93}{7} - \lambda\right)\right)
\]

This expression simplifies to a cubic polynomial in $\lambda$. After simplification, the characteristic polynomial is:

\[
\boxed{\lambda^3 - 9\lambda^2 + 10\lambda - 4}
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

DeepSeek-Coder-V2-Lite-Instruct

To find the characteristic polynomial of the given matrix, we need to compute the determinant of the matrix $ A - xI $, where $ A $ is the given matrix and $ I $ is the identity matrix of the same size. The characteristic polynomial is given by $ \det(A - xI) $.

Given matrix $ A $:
\[
A = \left(
\begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\
\end{array}
\right)
\]

First, we form the matrix $ A - xI $:
\[
A - xI = \left(
\begin{array}{ccc}
-\frac{61}{7} - x & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - x & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - x \\
\end{array}
\right)
\]

Next, we compute the determinant of $ A - xI $. We will use the cofactor expansion along the first row for simplicity:
\[
\det(A - xI) = \left( -\frac{61}{7} - x \right) \left|
\begin{array}{cc}
\frac{93}{7} - x & -6 \\
-\frac{26}{7} & \frac{95}{7} - x \\
\end{array}
\right|
- \left( -\frac{41}{7} \right) \left|
\begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - x \\
\end{array}
\right|
- \left( -\frac{87}{7} \right) \left|
\begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - x \\
\frac{23}{7} & -\frac{26}{7} \\
\end{array}
\right|
\]

We now compute each of the 2x2 determinants:
\[
\left|
\begin{array}{cc}
\frac{93}{7} - x & -6 \\
-\frac{26}{7} & \frac{95}{7} - x \\
\end{array}
\right|
= \left( \frac{93}{7} - x \right) \left( \frac{95}{7} - x \right) - (-6) \left( -\frac{26}{7} \right)
= \left( \frac{93}{7} - x \right) \left( \frac{95}{7} - x \right) - \frac{156}{7}
= \frac{8835}{49} - \frac{188x}{7} + x^2 - \frac{156}{7}
= x^2 - \frac{188x}{7} + \frac{8835}{49} - \frac{1092}{49}
= x^2 - \frac{188x}{7} + \frac{7743}{49}
\]

\[
\left|
\begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - x \\
\end{array}
\right|
= \left( -\frac{83}{7} \right) \left( \frac{95}{7} - x \right) - (-6) \left( \frac{23}{7} \right)
= -\frac{7885}{49} + \frac{83x}{7} + \frac{138}{7}
= -\frac{7885}{49} + \frac{83x}{7} + \frac{966}{49}
= \frac{83x}{7} - \frac{6919}{49}
\]

\[
\left|
\begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - x \\
\frac{23}{7} & -\frac{26}{7} \\
\end{array}
\right|
= \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - x \right) \left( \frac{23}{7} \right)
= \frac{2158}{49} - \frac{2139}{49} + \frac{23x}{7}
= \frac{19}{49} + \frac{23x}{7}
\]

Substituting these back into the determinant expression:
\[
\det(A - xI) = \left( -\frac{61}{7} - x \right) \left( x^2 - \frac{188x}{7} + \frac{7743}{49} \right) - \left( -\frac{41}{7} \right) \left( \frac{83x}{7} - \frac{6919}{49} \right) - \left( -\frac{87}{7} \right) \left( \frac{19}{49} + \frac{23x}{7} \right)
\]

Expanding and simplifying this expression will yield the characteristic polynomial. After performing the multiplications and simplifications, we get:
\[
\det(A - xI) = -x^3 + \frac{24x^2}{7} + \frac{16190x}{49} - \frac{1011644}{343}
\]

Thus, the characteristic polynomial is:
\[
\boxed{-x^3 + \frac{24x^2}{7} + \frac{16190x}{49} - \frac{1011644}{343}}
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

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 $A - \lambda I$, where $A$ is the given matrix, $\lambda$ is the eigenvalue, and $I$ is the identity matrix.
## Step 2: The given matrix is a $3 \times 3$ matrix. We will subtract $\lambda$ from the diagonal elements of the matrix and then find the determinant of the resulting matrix.
## Step 3: The matrix $A - \lambda I$ will be:
\[\begin{pmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{pmatrix}\]
## Step 4: To find the determinant of this matrix, we will expand along the first row. The determinant is given by:
\[\begin{vmatrix} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{vmatrix}\]
## Step 5: Expanding the determinant, we get a polynomial in terms of $\lambda$, which is the characteristic polynomial of the matrix.
## Step 6: After calculating the determinant, we get the characteristic polynomial as a cubic polynomial in terms of $\lambda$.
## Step 7: The characteristic polynomial is $-\lambda^3 + 9\lambda^2 - 27\lambda + 27$.
## Step 8: Therefore, the characteristic polynomial of the given matrix is $\boxed{-\lambda^3 + 9\lambda^2 - 27\lambda + 27}$.

The final answer is: $\boxed{-\lambda^3 + 9\lambda^2 - 27\lambda + 27}$

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

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:
$\left(\begin{array}{ccc} -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\ -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\ \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \end{array}\right)$.

## Step 3: Calculate the determinant of $A - \lambda I$
Using the formula for the determinant of a 3x3 matrix, we get:
$\det(A - \lambda I) = \left(-\frac{61}{7} - \lambda\right)\left(\left(\frac{93}{7} - \lambda\right)\left(\frac{95}{7} - \lambda\right) - (-6)\left(-\frac{26}{7}\right)\right) - \left(-\frac{41}{7}\right)\left((-6)\left(\frac{23}{7}\right) - \left(-\frac{83}{7}\right)\left(\frac{95}{7} - \lambda\right)\right) + \left(-\frac{87}{7}\right)\left((-6)\left(-\frac{83}{7}\right) - \left(\frac{23}{7}\right)\left(\frac{93}{7} - \lambda\right)\right)$.

## Step 4: Simplify the determinant
Expanding and simplifying the determinant, we get:
$\det(A - \lambda I) = -\frac{1}{7}\left(\lambda^3 - 187\lambda^2 + 703\lambda - 451\right)$.

## Step 5: Write down the characteristic polynomial
The characteristic polynomial is given by:
$\left(\lambda^3 - 187\lambda^2 + 703\lambda - 451\right)$.

The final answer is: $\boxed{\lambda^3 - 187\lambda^2 + 703\lambda - 451}$

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

Qwen2-72B-Instruct

To find the characteristic polynomial of a matrix, we need to compute its determinant after subtracting $\lambda$ from each diagonal element. The given matrix is:

\[
A = \left(\begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}
\end{array}\right)
\]

The characteristic polynomial $p(\lambda)$ is defined as:

\[
p(\lambda) = \det(A - \lambda I)
\]

where $I$ is the identity matrix of the same size as $A$. Thus,

\[
A - \lambda I = \left(\begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda
\end{array}\right)
\]

Now, we calculate the determinant of this matrix:

\[
\begin{align*}
p(\lambda) &= \left| \begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda
\end{array} \right| \\
&= \left( -\frac{61}{7} - \lambda \right) \left| \begin{array}{cc}
\frac{93}{7} - \lambda & -6 \\
- \frac{26}{7} & \frac{95}{7} - \lambda
\end{array} \right| \\
&\quad - \left( -\frac{41}{7} \right) \left| \begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - \lambda
\end{array} \right| \\
&\quad + \left( -\frac{87}{7} \right) \left| \begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - \lambda \\
\frac{23}{7} & -\frac{26}{7}
\end{array} \right| \\
&= \left( -\frac{61}{7} - \lambda \right) \left( \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) + 6 \cdot \frac{26}{7} \right) \\
&\quad + \frac{41}{7} \left( -\frac{83}{7} \left( \frac{95}{7} - \lambda \right) + 6 \cdot \frac{23}{7} \right) \\
&\quad - \frac{87}{7} \left( -\frac{83}{7} \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - \lambda \right) \frac{23}{7} \right) \\
&= \left( -\frac{61}{7} - \lambda \right) \left( \frac{8835}{49} - \frac{188}{7} \lambda + \lambda^2 + \frac{156}{7} \right) \\
&\quad + \frac{41}{7} \left( -\frac{7885}{49} + \frac{83}{7} \lambda + \frac{138}{7} \right) \\
&\quad - \frac{87}{7} \left( \frac{2158}{49} - \frac{2139}{49} \lambda \right) \\
&= \left( -\frac{61}{7} - \lambda \right) \left( \lambda^2 - \frac{188}{7} \lambda + \frac{9039}{49} \right) \\
&\quad + \frac{41}{7} \left( \frac{83}{7} \lambda - \frac{7747}{49} \right) \\
&\quad - \frac{87}{7} \left( \frac{2158}{49} - \frac{2139}{49} \lambda \right) \\
&= \lambda^3 - \frac{127}{7} \lambda^2 + \frac{10202}{49} \lambda - \frac{542340}{343}
\end{align*}
\]

Thus, the characteristic polynomial of the given matrix is:

\[
p(\lambda) = \lambda^3 - \frac{127}{7} \lambda^2 + \frac{10202}{49} \lambda - \frac{542340}{343}
\]

So, the final answer is:

\[
\boxed{\lambda^3 - \frac{127}{7} \lambda^2 + \frac{10202}{49} \lambda - \frac{542340}{343}}
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

baichuan4

To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix minus $x$ times the identity matrix. Given the matrix $A$:

$$
A = \left(
\begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\
\end{array}
\right)
$$

We first construct the matrix $A - xI$, where $I$ is the identity matrix:

$$
A - xI = \left(
\begin{array}{ccc}
-\frac{61}{7} - x & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - x & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - x \\
\end{array}
\right)
$$

Now, we compute the determinant of this matrix:

$$
\begin{aligned}
\det(A - xI) &= \left|
\begin{array}{ccc}
-\frac{61}{7} - x & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - x & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - x \\
\end{array}
\right| \\
&= (-\frac{61}{7} - x) \left|
\begin{array}{cc}
\frac{93}{7} - x & -6 \\
-\frac{26}{7} & \frac{95}{7} - x \\
\end{array}
\right| - (-\frac{41}{7}) \left|
\begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - x \\
\end{array}
\right| + (-\frac{87}{7}) \left|
\begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - x \\
\frac{23}{7} & -\frac{26}{7} \\
\end{array}
\right| \\
&= (-\frac{61}{7} - x) \left( (\frac{93}{7} - x)(\frac{95}{7} - x) - (-6)(-\frac{26}{7}) \right) - (-\frac{41}{7}) \left( (-\frac{83}{7})(\frac{95}{7} - x) - (-6)(\frac{23}{7}) \right) \\
&\quad + (-\frac{87}{7}) \left( (-\frac{83}{7})(-\frac{26}{7}) - (\frac{93}{7} - x)(\frac{23}{7}) \right) \\
&= (-\frac{61}{7} - x) \left( \frac{8855}{49} - \frac{93}{7}x - \frac{95}{7}x + x^2 - \frac{156}{7} \right) + \frac{41}{7} \left( \frac{7855}{49} - \frac{83}{7}x + \frac{138}{7} \right) \\
&\quad - \frac{87}{7} \left( \frac{2158}{49} - \frac{23}{7}x \right) \\
&= (-\frac{61}{7} - x) \left( x^2 - \frac{188}{7}x + \frac{8709}{49} \right) + \frac{41}{7} \left( \frac{8031}{49} - \frac{83}{7}x \right) - \frac{87}{7} \left( \frac{2158}{49} - \frac{23}{7}x \right) \\
&= (-\frac{61}{7} - x) \left( x^2 - \frac{188}{7}x + \frac{8709}{49} \right) + \frac{41}{7} \left( \frac{8031}{49} - \frac{83}{7}x \right) - \frac{87}{7} \left( \frac{2158}{49} - \frac{23}{7}x \right).
\end{aligned}
$$

This expression can be further simplified by expanding and combining like terms. However, the computation is quite lengthy and can be simplified using a computer algebra system or by carefully expanding and simplifying each term manually.

The final result will be a cubic polynomial in $x$ which is the characteristic polynomial of the matrix $A$. The coefficients of the polynomial will depend on the exact computation of the determinant above.

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

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 $\det(A - \lambda I) = \begin{vmatrix}a - \lambda & b & c\\d & e - \lambda & f\\g & h & i - \lambda\end{vmatrix}$.

For the given matrix $A = \begin{pmatrix}-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7}\\-\frac{83}{7} & \frac{93}{7} & -6\\\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}\end{pmatrix}$, the characteristic polynomial is:

\[
\begin{align*}
&\begin{vmatrix}-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7}\\-\frac{83}{7} & \frac{93}{7} - \lambda & -6\\\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda\end{vmatrix}\\
=&(-\frac{61}{7} - \lambda)\times\begin{vmatrix}\frac{93}{7} - \lambda & -6\\-\frac{26}{7} & \frac{95}{7} - \lambda\end{vmatrix} - (-\frac{41}{7})\times\begin{vmatrix}-\frac{83}{7} & -6\\\frac{23}{7} & \frac{95}{7} - \lambda\end{vmatrix} - (-\frac{87}{7})\times\begin{vmatrix}-\frac{83}{7} & \frac{93}{7} - \lambda\\\frac{23}{7} & -\frac{26}{7}\end{vmatrix}\\
=&(-\frac{61}{7} - \lambda)[(\frac{93}{7} - \lambda)(\frac{95}{7} - \lambda) - (-6)\times(-\frac{26}{7})] + \frac{41}{7}[(-\frac{83}{7})(\frac{95}{7} - \lambda) - (-6)\times\frac{23}{7}] + \frac{87}{7}[(-\frac{83}{7})(-\frac{26}{7}) - (\frac{93}{7} - \lambda)\times\frac{23}{7}]
\end{align*}
\]

After expanding and simplifying, we get the characteristic polynomial as:

$\boxed{-\lambda^3 + \frac{249}{7}\lambda^2 - \frac{21259}{49}\lambda + \frac{37337}{343}}$.

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

gpto1

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

kimi

To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix A - λI, where A is the given matrix, λ is a scalar, and I is the identity matrix of the same size as A.

Given matrix A:
$\left(\n\begin{array}{ccc}\n -\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\\n\end{array}\n\right)$

The identity matrix I of size 3x3 is:
$\left(\n\begin{array}{ccc}\n 1 & 0 & 0 \\\n 0 & 1 & 0 \\\n 0 & 0 & 1 \\\n\end{array}\n\right)$

Now, we need to compute A - λI:
$\left(\n\begin{array}{ccc}\n -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\n\right)$

Now, we need to compute the determinant of this matrix:

$\det(A - \lambda I) = \left|\begin{array}{ccc}\n -\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\\n -\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\\n \frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right|$

To compute the determinant, we can use the cofactor expansion along the first row:

$\det(A - \lambda I) = (-\frac{61}{7} - \lambda)\left|\begin{array}{cc}\n \frac{93}{7} - \lambda & -6 \\\n -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right| - (-\frac{41}{7})\left|\begin{array}{cc}\n -\frac{83}{7} & -6 \\\n \frac{23}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right| + (-\frac{87}{7})\left|\begin{array}{cc}\n -\frac{83}{7} & \frac{93}{7} - \lambda \\\n \frac{23}{7} & -\frac{26}{7} \\\n\end{array}\right|$

Now, we need to compute the determinants of the 2x2 matrices:

$\left|\begin{array}{cc}\n \frac{93}{7} - \lambda & -6 \\\n -\frac{26}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right| = (\frac{93}{7} - \lambda)(\frac{95}{7} - \lambda) - (-6)(-\frac{26}{7})$

$\left|\begin{array}{cc}\n -\frac{83}{7} & -6 \\\n \frac{23}{7} & \frac{95}{7} - \lambda \\\n\end{array}\right| = (-\frac{83}{7})(\frac{95}{7} - \lambda) - (-6)(\frac{23}{7})$

$\left|\begin{array}{cc}\n -\frac{83}{7} & \frac{93}{7} - \lambda \\\n \frac{23}{7} & -\frac{26}{7} \\\n\end{array}\right| = (-\frac{83}{7})(-\frac{26}{7}) - (\frac{93}{7} - \lambda

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

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, and $\lambda$ represents the eigenvalues.

The matrix $A$ is:

$$
A = \left(\begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}
\end{array}\right)
$$

The identity matrix $I$ of size $3 \times 3$ is:

$$
I = \left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)
$$

We subtract $\lambda I$ from $A$ to get:

$$
A - \lambda I = \left(\begin{array}{ccc}
-\frac{61}{

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

wenxinyiyan

To find the characteristic polynomial of the given matrix, we first write down the matrix and its characteristic equation:

$A = \left(
\begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} \\
\end{array}
\right)$

The characteristic equation of $A$ is given by:

$\det(A - \lambda I) = 0$

where $I$ is the identity matrix and $\lambda$ is the eigenvalue.

Now, let's compute the determinant of $A - \lambda I$:

$A - \lambda I = \left(
\begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda \\
\end{array}
\right)$

Computing the determinant, we get:

$\det(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) + \left( -\frac{41}{7} \right) \left( -6 \right) \left( \frac{23}{7} \right) + \left( -\frac{87}{7} \right) \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( -\frac{87}{7} \right) \left( \frac{93}{7} - \lambda \right) \left( \frac{23}{7} \right) - \left( -\frac{41}{7} \right) \left( -6 \right) \left( \frac{95}{7} - \lambda \right) - \left( -\frac{61}{7} - \lambda \right) \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right)$

After simplifying, we obtain the characteristic polynomial:

$\boxed{-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{1515 \lambda }{7}-\frac{11110}{7}}$

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

zhipu

To find the characteristic polynomial of a matrix $ A $, we need to compute the determinant of $ A - \lambda I $, where $ \lambda $ is an eigenvalue and $ I $ is the identity matrix of the same size as $ A $.

Given the matrix:
\[ A = \left( \begin{array}{ccc}
-\frac{61}{7} & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7}
\end{array} \right) \]

We form the matrix $ A - \lambda I $:
\[ A - \lambda I = \left( \begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda
\end{array} \right) \]

Next, we compute the determinant of $ A - \lambda I $:
\[
\text{det}(A - \lambda I) = \left| \begin{array}{ccc}
-\frac{61}{7} - \lambda & -\frac{41}{7} & -\frac{87}{7} \\
-\frac{83}{7} & \frac{93}{7} - \lambda & -6 \\
\frac{23}{7} & -\frac{26}{7} & \frac{95}{7} - \lambda
\end{array} \right|
\]

We can use the cofactor expansion along the first row:
\[
\text{det}(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left| \begin{array}{cc}
\frac{93}{7} - \lambda & -6 \\
- \frac{26}{7} & \frac{95}{7} - \lambda
\end{array} \right|
- \left( -\frac{41}{7} \right) \left| \begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - \lambda
\end{array} \right|
+ \left( -\frac{87}{7} \right) \left| \begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - \lambda \\
\frac{23}{7} & -\frac{26}{7}
\end{array} \right|
\]

We compute each of these 2x2 determinants:
\[
\left| \begin{array}{cc}
\frac{93}{7} - \lambda & -6 \\
- \frac{26}{7} & \frac{95}{7} - \lambda
\end{array} \right| = \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( -\frac{26}{7} \right)
\]
\[
= \left( \frac{93}{7} - \lambda \right) \left( \frac{95}{7} - \lambda \right) - \frac{156}{7}
\]
\[
= \frac{93 \cdot 95}{49} - \frac{93 \lambda}{7} - \frac{95 \lambda}{7} + \lambda^2 - \frac{156}{7}
\]
\[
= \frac{8835}{49} - \frac{188 \lambda}{7} + \lambda^2 - \frac{156}{7}
\]
\[
= \frac{8835 - 1092}{49} + \lambda^2 - \frac{188 \lambda}{7}
\]
\[
= \frac{7743}{49} + \lambda^2 - \frac{188 \lambda}{7}
\]
\[
= \lambda^2 - \frac{188 \lambda}{7} + \frac{7743}{49}
\]

Next determinant:
\[
\left| \begin{array}{cc}
-\frac{83}{7} & -6 \\
\frac{23}{7} & \frac{95}{7} - \lambda
\end{array} \right| = \left( -\frac{83}{7} \right) \left( \frac{95}{7} - \lambda \right) - \left( -6 \right) \left( \frac{23}{7} \right)
\]
\[
= -\frac{83 \cdot 95}{49} + \frac{83 \lambda}{7} + \frac{138}{7}
\]
\[
= -\frac{7885}{49} + \frac{83 \lambda}{7} + \frac{138}{7}
\]
\[
= -\frac{7885}{49} + \frac{83 \lambda + 138}{7}
\]
\[
= -\frac{7885}{49} + \frac{83 \lambda + 138}{7}
\]
\[
= -\frac{7885}{49} + \frac{83 \lambda + 138}{7}
\]
\[
= -\frac{7885}{49} + \frac{83 \lambda + 138}{7}
\]

Third determinant:
\[
\left| \begin{array}{cc}
-\frac{83}{7} & \frac{93}{7} - \lambda \\
\frac{23}{7} & -\frac{26}{7}
\end{array} \right| = \left( -\frac{83}{7} \right) \left( -\frac{26}{7} \right) - \left( \frac{93}{7} - \lambda \right) \left( \frac{23}{7} \right)
\]
\[
= \frac{83 \cdot 26}{49} - \left( \frac{93 \cdot 23}{49} - \frac{23 \lambda}{7} \right)
\]
\[
= \frac{2158}{49} - \frac{2139}{49} + \frac{23 \lambda}{7}
\]
\[
= \frac{19}{49} + \frac{23 \lambda}{7}
\]

Putting it all together:
\[
\text{det}(A - \lambda I) = \left( -\frac{61}{7} - \lambda \right) \left( \lambda^2 - \frac{188 \lambda}{7} + \frac{7743}{49} \right) + \frac{41}{7} \left( -\frac{7885}{49} + \frac{83 \lambda + 138}{7} \right) - \frac{87}{7} \left( \frac{19}{49} + \frac{23 \lambda}{7} \right)
\]

After simplifying the above expression, we get the characteristic polynomial:
\[
\boxed{\lambda^3 - \frac{188 \lambda^2}{7} + \frac{7743 \lambda}{49} - \frac{7885}{49}}
\]

This can be further simplified to:
\[
\boxed{\lambda^3 - \frac{188}{7} \lambda^2 + \frac{7743}{49} \lambda - \frac{7885}{49}}
\]

Or in a more compact form:
\[
\boxed{\lambda^3 - \frac{188}{7} \lambda^2 + \frac{7743 - 7885}{49} \lambda}
\]
\[
\boxed{\lambda^3 - \frac{188}{7} \lambda^2 - \frac{142}{49} \lambda}
\]

Thus, the characteristic polynomial is:
\[
\boxed{\lambda^3 - \frac{188}{7} \lambda^2 - \frac{142}{49} \lambda}
\]

-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

deepseek-v3

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-\lambda ^3+\frac{127 \lambda ^2}{7}+\frac{5127 \lambda }{49}-\frac{757655}{343}

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