| 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: \[ A = \left( \begin{array}{ccc} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \\ \end{array} \right) \] First, we form the matrix \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{ccc} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{array} \right) \] Next, we compute the determinant of \( A - \lambda I \): \[ \det(A - \lambda I) = \det \left( \begin{array}{ccc} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{array} \right) \] We expand this determinant along the first row: \[ \det(A - \lambda I) = (-9 - \lambda) \det \left( \begin{array}{cc} -11 - \lambda & 11 \\ 3 & 12 - \lambda \\ \end{array} \right) - (-5) \det \left( \begin{array}{cc} -4 & 11 \\ 5 & 12 - \lambda \\ \end{array} \right) + 7 \det \left( \begin{array}{cc} -4 & -11 - \lambda \\ 5 & 3 \\ \end{array} \right) \] We compute each of the 2x2 determinants: 1. \[ \det \left( \begin{array}{cc} -11 - \lambda & 11 \\ 3 & 12 - \lambda \\ \end{array} \right) = (-11 - \lambda)(12 - \lambda) - (11)(3) = 132 + 11\lambda - 12\lambda - \lambda^2 - 33 = -\lambda^2 - \lambda + 99 \] 2. \[ \det \left( \begin{array}{cc} -4 & 11 \\ 5 & 12 - \lambda \\ \end{array} \right) = (-4)(12 - \lambda) - (11)(5) = -48 + 4\lambda - 55 = 4\lambda - 103 \] 3. \[ \det \left( \begin{array}{cc} -4 & -11 - \lambda \\ 5 & 3 \\ \end{array} \right) = (-4)(3) - (-11 - \lambda)(5) = -12 + 55 + 5\lambda = 43 + 5\lambda \] Substituting these back into the determinant expansion: \[ \det(A - \lambda I) = (-9 - \lambda)(-\lambda^2 - \lambda + 99) + 5(4\lambda - 103) + 7(43 + 5\lambda) \] Expanding each term: \[ (-9 - \lambda)(-\lambda^2 - \lambda + 99) = 9\lambda^2 + 9\lambda - 891 + \lambda^3 + \lambda^2 - 99\lambda = \lambda^3 + 10\lambda^2 - 90\lambda - 891 \] \[ 5(4\lambda - 103) = 20\lambda - 515 \] \[ 7(43 + 5\lambda) = 301 + 35\lambda \] Combining all terms: \[ \det(A - \lambda I) = \lambda^3 + 10\lambda^2 - 90\lambda - 891 + 20\lambda - 515 + 301 + 35\lambda \] Simplifying: \[ \det(A - \lambda I) = \lambda^3 + 10\lambda^2 - 35\lambda - 1105 \] Thus, the characteristic polynomial is: \[ \boxed{\lambda^3 + 10\lambda^2 - 35\lambda - 1105} \] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| deepseek-r1 | To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \end{pmatrix}, \] we need to compute the determinant of \(A - \lambda I\). The matrix \(A - \lambda I\) is: \[ \begin{pmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{pmatrix}. \] Using the cofactor expansion along the first row, we calculate the determinant: 1. **First minor (M11):** \[ \det\begin{pmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{pmatrix} = (-11 - \lambda)(12 - \lambda) - 33 = \lambda^2 - \lambda - 165. \] 2. **Second minor (M12):** \[ \det\begin{pmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{pmatrix} = (-4)(12 - \lambda) - 55 = 4\lambda - 103. \] 3. **Third minor (M13):** \[ \det\begin{pmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{pmatrix} = -12 + 5(\lambda + 11) = 5\lambda + 43. \] Putting it all together, the determinant of \(A - \lambda I\) is: \[ (-9 - \lambda)(\lambda^2 - \lambda - 165) + 5(4\lambda - 103) + 7(5\lambda + 43). \] Expanding and simplifying: 1. **First term:** \[ (-9 - \lambda)(\lambda^2 - \lambda - 165) = -\lambda^3 - 8\lambda^2 + 174\lambda + 1485. \] 2. **Second term:** \[ 5(4\lambda - 103) = 20\lambda - 515. \] 3. **Third term:** \[ 7(5\lambda + 43) = 35\lambda + 301. \] Combining all terms: \[ -\lambda^3 - 8\lambda^2 + 174\lambda + 1485 + 20\lambda - 515 + 35\lambda + 301. \] Simplifying further: \[ -\lambda^3 - 8\lambda^2 + 229\lambda + 1271. \] To convert this to the characteristic polynomial in the standard form \(\det(\lambda I - A)\), we multiply by -1: \[ \lambda^3 + 8\lambda^2 - 229\lambda - 1271. \] Thus, the characteristic polynomial of the matrix is \[ \boxed{x^3 + 8x^2 - 229x - 1271}. \] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| gpto1 | To find the characteristic polynomial \(P(\lambda)\) of the matrix \(A\), we compute the determinant of \(A - \lambda I\), where \(I\) is the identity matrix: \[ A - \lambda I = \begin{pmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{pmatrix} \] The characteristic polynomial is: \[ P(\lambda) = \det(A - \lambda I) \] Using cofactor expansion along the first row, we have: \[ P(\lambda) = (-9 - \lambda) \det\begin{pmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{pmatrix} - (-5) \det\begin{pmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{pmatrix} + 7 \det\begin{pmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{pmatrix} \] Calculate each determinant: 1. \(\det\begin{pmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{pmatrix} = (-11 - \lambda)(12 - \lambda) - (11)(3) = (\lambda^2 - \lambda - 165)\) 2. \(\det\begin{pmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{pmatrix} = (-4)(12 - \lambda) - (11)(5) = (-103 + 4\lambda)\) 3. \(\det\begin{pmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{pmatrix} = (-4)(3) - (-11 - \lambda)(5) = (43 + 5\lambda)\) Now, substitute back into \(P(\lambda)\): \[ P(\lambda) = (-9 - \lambda)(\lambda^2 - \lambda - 165) + 5(-103 + 4\lambda) + 7(43 + 5\lambda) \] Simplify the expression: \[ \begin{align*} P(\lambda) &= (-9 - \lambda)(\lambda^2 - \lambda - 165) + 5(-103 + 4\lambda) + 7(43 + 5\lambda) \\ &= -(\lambda^3 + 8\lambda^2 - 229\lambda - 1271) \end{align*} \] Finally, write the characteristic polynomial: \[ P(\lambda) = \lambda^3 + 8\lambda^2 - 229\lambda - 1271 \] \[\boxed{\lambda^{3} + 8\lambda^{2} -229\lambda\ -\ 1271}\] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| gpt-4o-2024-08-06 | To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \end{pmatrix}, \] 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: \[ A - \lambda I = \begin{pmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{pmatrix}. \] The characteristic polynomial is given by the determinant: \[ \det(A - \lambda I) = \begin{vmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{vmatrix}. \] We can calculate this determinant using the cofactor expansion along the first row: \[ \det(A - \lambda I) = (-9 - \lambda) \begin{vmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{vmatrix} - (-5) \begin{vmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{vmatrix} + 7 \begin{vmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{vmatrix}. \] Now, we compute each of these 2x2 determinants: 1. \(\begin{vmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{vmatrix} = (-11 - \lambda)(12 - \lambda) - (11)(3)\). \[ = (-11 - \lambda)(12 - \lambda) - 33 = (121 + 11\lambda + 12\lambda + \lambda^2) - 33 = \lambda^2 + 23\lambda + 88. \] 2. \(\begin{vmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{vmatrix} = (-4)(12 - \lambda) - (11)(5)\). \[ = -48 + 4\lambda - 55 = 4\lambda - 103. \] 3. \(\begin{vmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{vmatrix} = (-4)(3) - (-11 - \lambda)(5)\). \[ = -12 + 55 + 5\lambda = 5\lambda + 43. \] Substituting these back into the cofactor expansion: \[ \det(A - \lambda I) = (-9 - \lambda)(\lambda^2 + 23\lambda + 88) + 5(4\lambda - 103) + 7(5\lambda + 43). \] Expanding each term: - The first term: \[ (-9 - \lambda)(\lambda^2 + 23\lambda + 88) = -9(\lambda^2 + 23\lambda + 88) - \lambda(\lambda^2 + 23\lambda + 88). \] \[ = -9\lambda^2 - 207\lambda - 792 - \lambda^3 - 23\lambda^2 - 88\lambda. \] \[ = -\lambda^3 - 32\lambda^2 - 295\lambda - 792. \] - The second term: \[ 5(4\lambda - 103) = 20\lambda - 515. \] - The third term: \[ 7(5\lambda + 43) = 35\lambda + 301. \] Combining all terms: \[ -\lambda^3 - 32\lambda^2 - 295\lambda - 792 + 20\lambda - 515 + 35\lambda + 301. \] Combine like terms: - \(\lambda^3\) term: \(-\lambda^3\). - \(\lambda^2\) term: \(-32\lambda^2\). - \(\lambda\) term: \(-295\lambda + 20\lambda + 35\lambda = -240\lambda\). - Constant term: \(-792 - 515 + 301 = -1006\). Thus, the characteristic polynomial is: \[ \boxed{-\lambda^3 - 32\lambda^2 - 240\lambda - 1006}. \] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| DeepSeek-Coder-V2-Lite-Instruct | To find the characteristic polynomial of the given matrix, we will use the formula for the characteristic polynomial of a 3x3 matrix \( A \), which is given by: \[ \text{det}(A - \lambda I) \] where \( \lambda \) is a scalar and \( I \) is the identity matrix of the same size. For the given matrix: \[ A = \left( \begin{array}{ccc} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \\ \end{array} \right) \] we need to compute the determinant of \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{ccc} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{array} \right) \] We will expand the determinant along the first row: \[ \text{det}(A - \lambda I) = (-9 - \lambda) \left| \begin{array}{cc} -11 - \lambda & 11 \\ 3 & 12 - \lambda \\ \end{array} \right| - (-5) \left| \begin{array}{cc} -4 & 11 \\ 5 & 12 - \lambda \\ \end{array} \right| + 7 \left| \begin{array}{cc} -4 & -11 - \lambda \\ 5 & 3 \\ \end{array} \right| \] First, we compute the 2x2 determinants: \[ \left| \begin{array}{cc} -11 - \lambda & 11 \\ 3 & 12 - \lambda \\ \end{array} \right| = (-11 - \lambda)(12 - \lambda) - (11)(3) = -11 \cdot 12 + \lambda \cdot 12 - \lambda \cdot 11 - \lambda^2 - 33 = -132 + 12\lambda - 11\lambda - \lambda^2 - 33 = -\lambda^2 + \lambda - 165 \] \[ \left| \begin{array}{cc} -4 & 11 \\ 5 & 12 - \lambda \\ \end{array} \right| = (-4)(12 - \lambda) - (11)(5) = -48 + 4\lambda - 55 = 4\lambda - 103 \] \[ \left| \begin{array}{cc} -4 & -11 - \lambda \\ 5 & 3 \\ \end{array} \right| = (-4)(3) - (-11 - \lambda)(5) = -12 + 55 + 5\lambda = 43 + 5\lambda \] Now, substituting these back into the determinant expansion: \[ \text{det}(A - \lambda I) = (-9 - \lambda)(-\lambda^2 + \lambda - 165) - (-5)(4\lambda - 103) + 7(43 + 5\lambda) \] Expanding and simplifying: \[ = (-9 - \lambda)(-\lambda^2 + \lambda - 165) + 20\lambda - 515 + 301 + 35\lambda \] \[ = \lambda^3 + 9\lambda^2 + 15\lambda + 1485 + 20\lambda - 515 + 301 + 35\lambda \] \[ = \lambda^3 + 9\lambda^2 + 65\lambda + 1271 \] Thus, the characteristic polynomial is: \[ \boxed{\lambda^3 + 9\lambda^2 + 65\lambda + 1271} \] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| 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: $\left(\n\begin{array}{ccc}\n -9 & -5 & 7 \\\n -4 & -11 & 11 \\\n 5 & 3 & 12 \\\n\end{array}\n\right)$. Subtracting λ from the diagonal elements, we get: $\left(\n\begin{array}{ccc}\n -9-λ & -5 & 7 \\\n -4 & -11-λ & 11 \\\n 5 & 3 & 12-λ \\\n\end{array}\n\right)$. ## Step 4: Now, we calculate the determinant of this matrix by expanding along the first row. The determinant is given by: $(-9-λ) \times \left((-11-λ) \times (12-λ) - 11 \times 3\right) - (-5) \times \left((-4) \times (12-λ) - 11 \times 5\right) + 7 \times \left((-4) \times 3 - (-11-λ) \times 5\right)$. ## Step 5: Expanding and simplifying the expression from Step 4 will give us the characteristic polynomial of the matrix. ## Step 6: After expanding and simplifying, we get the characteristic polynomial as a cubic polynomial in terms of λ. ## Step 7: The characteristic polynomial is: $(-9-λ) \times ((-11-λ) \times (12-λ) - 33) + 5 \times ((-4) \times (12-λ) + 55) + 7 \times ((-4) \times 3 + 5 \times (-11-λ))$. ## Step 8: Further simplification is needed to put the polynomial in the standard form. ## Step 9: After simplifying and combining like terms, we should have the characteristic polynomial in the form $aλ^3 + bλ^2 + cλ + d$. ## Step 10: Calculating the determinant with the given matrix and λ, we find the characteristic polynomial. ## Step 11: The characteristic polynomial after simplification is $-λ^3 - 12λ^2 + 13λ + 1380$. The final answer is: $\boxed{-λ^3 - 12λ^2 + 13λ + 1380}$ |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| 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: $\left(\begin{array}{ccc} -9-\lambda & -5 & 7 \\ -4 & -11-\lambda & 11 \\ 5 & 3 & 12-\lambda \end{array}\right)$. ## Step 3: Calculate the determinant of $A - \lambda I$ Using the formula for the determinant of a 3x3 matrix, we get: $\det(A - \lambda I) = (-9-\lambda)((-11-\lambda)(12-\lambda) - 11*3) - (-5)((-4)(12-\lambda) - 11*5) + 7((-4)*3 - (-11-\lambda)*5)$. ## Step 4: Expand and simplify the determinant Expanding and simplifying the determinant, we get: $\det(A - \lambda I) = (-9-\lambda)(-132 + 11\lambda - \lambda^2 - 33) + 5(-48 + 4\lambda - 55) + 7(-12 + 55 + 5\lambda)$. ## Step 5: Further simplify the determinant Further simplifying the determinant, we get: $\det(A - \lambda I) = (-9-\lambda)(-165 + 11\lambda - \lambda^2) + 5(-103 + 4\lambda) + 7(43 + 5\lambda)$. ## Step 6: Multiply and combine like terms Multiplying and combining like terms, we get: $\det(A - \lambda I) = 1485 - 99\lambda + 9\lambda^2 - 11\lambda^2 + \lambda^3 - 515 + 20\lambda + 301 + 35\lambda$. ## Step 7: Combine like terms and simplify Combining like terms and simplifying, we get: $\det(A - \lambda I) = \lambda^3 - 2\lambda^2 - 44\lambda + 1271$. The final answer is: $\boxed{\lambda^3 - 2\lambda^2 - 44\lambda + 1271}$ |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| Qwen2-72B-Instruct | To find the characteristic polynomial of a matrix, we need to compute the determinant of $(A-\lambda I)$, where $A$ is the given matrix and $\lambda$ is an eigenvalue. The characteristic polynomial will be a function of $\lambda$, and its roots will give us the eigenvalues of the matrix. Given matrix $A$: $$ A = \left(\begin{array}{ccc} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \\ \end{array}\right) $$ We need to find the determinant of $(A-\lambda I)$, where $I$ is the identity matrix of size 3: $$ A-\lambda I = \left(\begin{array}{ccc} -9-\lambda & -5 & 7 \\ -4 & -11-\lambda & 11 \\ 5 & 3 & 12-\lambda \\ \end{array}\right) $$ The determinant of this matrix is: $$ \det(A-\lambda I) = (-9-\lambda)((-11-\lambda)(12-\lambda)-33) + 5((-4)(12-\lambda)+55) + 7(-20-(-11-\lambda)5) $$ Simplifying this expression gives us the characteristic polynomial: $$ \det(A-\lambda I) = (-9-\lambda)((-11-\lambda)(12-\lambda)-33) + 5(48-4\lambda+55) + 7(-20+55+5\lambda) $$ $$ = (-9-\lambda)(-132+11\lambda+\lambda^2-33) + 5(103-4\lambda) + 7(35+5\lambda) $$ $$ = (-9-\lambda)(\lambda^2+11\lambda-165) + 515-20\lambda + 245 + 35\lambda $$ $$ = (-9-\lambda)(\lambda^2+11\lambda-165) + 760 + 15\lambda $$ $$ = -\lambda^3 - 9\lambda^2 - 11\lambda^2 - 99\lambda + 1485 + 760 + 15\lambda $$ $$ = -\lambda^3 - 20\lambda^2 - 84\lambda + 2245 $$ So, the characteristic polynomial of the given matrix is: $$ p(\lambda) = -\lambda^3 - 20\lambda^2 - 84\lambda + 2245 $$ Therefore, the final answer is $\boxed{-\lambda^3 - 20\lambda^2 - 84\lambda + 2245}$. |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| 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 everywhere else. Given the matrix $A$: $$ A = \left(\n\begin{array}{ccc} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \\ \end{array}\n\right) $$ We first construct the matrix $A - xI$, where $I$ is the $3 \times 3$ identity matrix: $$ A - xI = \left(\n\begin{array}{ccc} -9-x & -5 & 7 \\ -4 & -11-x & 11 \\ 5 & 3 & 12-x \\ \end{array}\n\right) $$ Now, we compute the determinant of this matrix: $$ \det(A - xI) = \left|\n\begin{array}{ccc} -9-x & -5 & 7 \\ -4 & -11-x & 11 \\ 5 & 3 & 12-x \\ \end{array}\n\right| $$ Using the Laplace expansion or any other method to compute the determinant, we get: $$ \det(A - xI) = (-9-x)\left|\n\begin{array}{cc} -11-x & 11 \\ 3 & 12-x \\ \end{array}\n\right| - (-5)\left|\n\begin{array}{cc} -4 & 11 \\ 5 & 12-x \\ \end{array}\n\right| + 7\left|\n\begin{array}{cc} -4 & -11-x \\ 5 & 3 \\ \end{array}\n\right| $$ Computing these smaller determinants, we have: $$ \det(A - xI) = (-9-x)((-11-x)(12-x) - 33) - (-5)((-4)(12-x) - 55) + 7((-4)(3) - (-11-x)(5)) $$ Expanding and simplifying, we get: $$ \det(A - xI) = (-9-x)(x^2 + 23x + 132 - 33) + 5(48 - 4x + 55) - 7(12 + 5x + 55) $$ Further simplification gives: $$ \det(A - xI) = (-9-x)(x^2 + 23x + 99) + 5(103 - 4x) - 7(67 + 5x) $$ Distributing and combining like terms, we finally obtain the characteristic polynomial: $$ \det(A - xI) = -x^3 - 32x^2 - 377x - 1287 $$ This is the characteristic polynomial of the given matrix. We can put it in a box as requested: $$ \boxed{-x^3 - 32x^2 - 377x - 1287} $$ |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| 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 $\begin{pmatrix}-9 & -5 & 7\\-4 & -11 & 11\\5 & 3 & 12\end{pmatrix}$, the characteristic polynomial is: \[ \begin{align*} &\begin{vmatrix}-9 - \lambda & -5 & 7\\-4 & -11 - \lambda & 11\\5 & 3 & 12 - \lambda\end{vmatrix}\\ =& (-9 - \lambda)\begin{vmatrix}-11 - \lambda & 11\\3 & 12 - \lambda\end{vmatrix} - (-5)\begin{vmatrix}-4 & 11\\5 & 12 - \lambda\end{vmatrix} + 7\begin{vmatrix}-4 & -11 - \lambda\\5 & 3\end{vmatrix}\\ =& (-9 - \lambda)[(-11 - \lambda)(12 - \lambda) - 33] - (-5)[-4(12 - \lambda) - 55] + 7[-12 + 5(-11 - \lambda)]\\ =& (-9 - \lambda)[132 + 11\lambda - 12\lambda - \lambda^2 - 33] - (-5)[-48 + 4\lambda - 55] + 7[-12 - 55 - 5\lambda]\\ =& (-9 - \lambda)(99 - \lambda - \lambda^2) + 5(-103 + 4\lambda) + 7(-67 - 5\lambda)\\ =& -891 + 9\lambda + 9\lambda^2 - 99\lambda + \lambda^2 + \lambda^3 - 515 + 20\lambda - 469 - 35\lambda + 7\lambda^2\\ =& \lambda^3 + 17\lambda^2 - 105\lambda - 1875 \end{align*} \] So the answer is $\boxed{\lambda^3 + 17\lambda^2 - 105\lambda - 1875}$. |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| gpto1 | To find the characteristic polynomial \(P(\lambda)\) of the matrix \(A\), we compute the determinant of \(A - \lambda I\), where \(I\) is the identity matrix: \[ A - \lambda I = \begin{pmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{pmatrix} \] The characteristic polynomial is: \[ P(\lambda) = \det(A - \lambda I) \] Using cofactor expansion along the first row, we have: \[ P(\lambda) = (-9 - \lambda) \det\begin{pmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{pmatrix} - (-5) \det\begin{pmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{pmatrix} + 7 \det\begin{pmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{pmatrix} \] Calculate each determinant: 1. \(\det\begin{pmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{pmatrix} = (-11 - \lambda)(12 - \lambda) - (11)(3) = (\lambda^2 - \lambda - 165)\) 2. \(\det\begin{pmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{pmatrix} = (-4)(12 - \lambda) - (11)(5) = (-103 + 4\lambda)\) 3. \(\det\begin{pmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{pmatrix} = (-4)(3) - (-11 - \lambda)(5) = (43 + 5\lambda)\) Now, substitute back into \(P(\lambda)\): \[ P(\lambda) = (-9 - \lambda)(\lambda^2 - \lambda - 165) + 5(-103 + 4\lambda) + 7(43 + 5\lambda) \] Simplify the expression: \[ \begin{align*} P(\lambda) &= (-9 - \lambda)(\lambda^2 - \lambda - 165) + 5(-103 + 4\lambda) + 7(43 + 5\lambda) \\ &= -(\lambda^3 + 8\lambda^2 - 229\lambda - 1271) \end{align*} \] Finally, write the characteristic polynomial: \[ P(\lambda) = \lambda^3 + 8\lambda^2 - 229\lambda - 1271 \] \[\boxed{\lambda^{3} + 8\lambda^{2} -229\lambda\ -\ 1271}\] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| kimi | To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix \( A - \lambda I \), where \( A \) is the given matrix, \( \lambda \) is a scalar, and \( I \) is the identity matrix of the same size as \( A \). Given matrix \( A \): \[ A = \left(\begin{array}{ccc} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \\ \end{array}\right) \] The identity matrix \( I \) of size 3x3 is: \[ I = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right) \] Thus, \( A - \lambda I \) is: \[ A - \lambda I = \left(\begin{array}{ccc} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{array}\right) \] The characteristic polynomial is given by the determinant of \( A - \lambda I \): \[ \det(A - \lambda I) = \left| \begin{array}{ccc} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{array} \right| \] We will compute this determinant using cofactor expansion along the first row: \[ \det(A - \lambda I) = (-9 - \lambda) \left| \begin{array}{cc} -11 - \lambda & 11 \\ 3 & 12 - \lambda \\ \end{array} \right| - (-5) \left| \begin{array}{cc} -4 & 11 \\ 5 & 12 - \lambda \\ \end{array} \right| + 7 \left| \begin{array}{cc} -4 & -11 - \lambda \\ 5 & 3 \\ \end{array} \right| \] Now, we compute each of the 2x2 determinants: \[ \left| \begin{array}{cc} -11 - \lambda & 11 \\ 3 & 12 - \lambda \\ \end{array} \right| = (-11 - \lambda)(12 - \lambda) - (11)(3) = \lambda^2 - \lambda - 99 \] \[ \left| \begin{array}{cc} -4 & 11 \\ 5 & 12 - \lambda \\ \end{array} \right| = (-4)(12 - \lambda) - (11)(5) = 4\lambda - 68 \] \[ \left| \begin{array}{cc} -4 & -11 - \lambda \\ 5 & 3 \\ \end{array} \right| = (-4)(3) - (-11 - \lambda)(5) = 5\lambda + 29 \] Substituting these back into the cofactor expansion: \[ \det(A - \lambda I) = (-9 - \lambda)(\lambda^2 - \lambda - 99) + 5(4\lambda - 68) + 7(5\lambda + 29) \] Expanding each term: \[ (-9 - \lambda)(\lambda^2 - \lambda - 99) = -9\lambda^2 + 9\lambda + 891 - \lambda^3 + \lambda^2 + 99\lambda = -\lambda^3 - 8\lambda^2 + 108\lambda + 891 \] \[ 5(4\lambda - 68) = 20\lambda - 340 \] \[ 7(5\lambda + 29) = 35\lambda + 203 \] Combining all terms: \[ -\lambda^3 - 8\lambda^2 + 108\lambda + 891 + 20\lambda - 340 + 35\lambda + 203 = -\lambda^3 - 8\lambda^2 + 163\lambda + 754 \] Thus, the characteristic polynomial is: \[ \boxed{-\lambda^3 - 8\lambda^2 + 163\lambda + 754} \] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| 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: $$ A - \lambda I = \left(\begin{array}{ccc} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{array}\right) $$ The characteristic polynomial is given by the determinant of this matrix: $$ \det(A - \lambda I) = \begin{vmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \\ \end{vmatrix} $$ 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: $$ \det(A - \lambda I) = (-9 - \lambda) |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| wenxinyiyan | To find the characteristic polynomial of the given matrix, we first need to find the determinant of the matrix $A - \lambda I$, where $A$ is the given matrix, $\lambda$ is the eigenvalue, and $I$ is the identity matrix. Given matrix: $A = \begin{pmatrix} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \end{pmatrix}$ Identity matrix: $I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ $A - \lambda I = \begin{pmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{pmatrix}$ Now, we find the determinant of $A - \lambda I$ using the cofactor expansion method along the first row: $|A - \lambda I| = (-9 - \lambda) \begin{vmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{vmatrix} + 5 \begin{vmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{vmatrix} + 7 \begin{vmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{vmatrix}$ $= (-9 - \lambda)[(-11 - \lambda)(12 - \lambda) - 33] + 5[-4(12 - \lambda) - 55] + 7[-4 \times 3 - 5(-11 - \lambda)]$ $= (-9 - \lambda)(\lambda^2 + \lambda - 132) + 5(4\lambda - 91) + 7(55 + 5\lambda)$ $= -\lambda^3 - 10\lambda^2 + 128\lambda - 49$ So, the characteristic polynomial of the given matrix is: $\boxed{-\lambda^3 - 10\lambda^2 + 128\lambda - 49}$ |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| 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: \[ A = \begin{pmatrix} -9 & -5 & 7 \\ -4 & -11 & 11 \\ 5 & 3 & 12 \end{pmatrix} \] We form the matrix \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{pmatrix} \] Next, we compute the determinant of \( A - \lambda I \): \[ \text{det}(A - \lambda I) = \begin{vmatrix} -9 - \lambda & -5 & 7 \\ -4 & -11 - \lambda & 11 \\ 5 & 3 & 12 - \lambda \end{vmatrix} \] We can use the cofactor expansion along the first row: \[ \text{det}(A - \lambda I) = (-9 - \lambda) \begin{vmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{vmatrix} - (-5) \begin{vmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{vmatrix} + 7 \begin{vmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{vmatrix} \] Now, we compute each of the 2x2 determinants: \[ \begin{vmatrix} -11 - \lambda & 11 \\ 3 & 12 - \lambda \end{vmatrix} = (-11 - \lambda)(12 - \lambda) - (11)(3) = \lambda^2 + 11\lambda - 132 - 33 = \lambda^2 + 11\lambda - 165 \] \[ \begin{vmatrix} -4 & 11 \\ 5 & 12 - \lambda \end{vmatrix} = (-4)(12 - \lambda) - (11)(5) = -48 + 4\lambda - 55 = 4\lambda - 103 \] \[ \begin{vmatrix} -4 & -11 - \lambda \\ 5 & 3 \end{vmatrix} = (-4)(3) - (5)(-11 - \lambda) = -12 + 55 + 5\lambda = 43 + 5\lambda \] Substituting these back into the determinant expression: \[ \text{det}(A - \lambda I) = (-9 - \lambda)(\lambda^2 + 11\lambda - 165) + 5(4\lambda - 103) + 7(43 + 5\lambda) \] Expanding each term: \[ (-9 - \lambda)(\lambda^2 + 11\lambda - 165) = -9\lambda^2 - 99\lambda + 1485 - \lambda^3 - 11\lambda^2 + 165\lambda \] \[ = -\lambda^3 - 20\lambda^2 + 66\lambda + 1485 \] \[ 5(4\lambda - 103) = 20\lambda - 515 \] \[ 7(43 + 5\lambda) = 301 + 35\lambda \] Combining all terms: \[ \text{det}(A - \lambda I) = -\lambda^3 - 20\lambda^2 + 66\lambda + 1485 + 20\lambda - 515 + 301 + 35\lambda \] \[ = -\lambda^3 - 20\lambda^2 + (66\lambda + 20\lambda + 35\lambda) + (1485 - 515 + 301) \] \[ = -\lambda^3 - 20\lambda^2 + 121\lambda + 1271 \] Thus, the characteristic polynomial of the matrix is: \[ \boxed{-\lambda^3 - 20\lambda^2 + 121\lambda + 1271} \] |
-\lambda ^3-8 \lambda ^2+229 \lambda +1271 |
| deepseek-v3 | Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-91ca6ec4-9fe1-9a71-8d55-b6313d1cc3d2', 'request_id': '91ca6ec4-9fe1-9a71-8d55-b6313d1cc3d2'} | -\lambda ^3-8 \lambda ^2+229 \lambda +1271 |