| 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 start by calculating the determinant of the matrix minus \( \lambda \) times the identity matrix. The given matrix is: \[ A = \left( \begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \\ \end{array} \right) \] We need to find the determinant of \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \\ \end{array} \right) \] The determinant of this matrix is: \[ \det(A - \lambda I) = \left( -\frac{5}{2} - \lambda \right) \left( \frac{27}{2} - \lambda \right) - \left( -\frac{5}{4} \right) \left( 8 \right) \] First, we expand the terms: \[ \left( -\frac{5}{2} - \lambda \right) \left( \frac{27}{2} - \lambda \right) = \left( -\frac{5}{2} \right) \left( \frac{27}{2} \right) - \left( -\frac{5}{2} \right) \lambda - \lambda \left( \frac{27}{2} \right) + \lambda^2 \] \[ = \left( -\frac{5}{2} \right) \left( \frac{27}{2} \right) + \frac{5}{2} \lambda - \frac{27}{2} \lambda + \lambda^2 \] \[ = -\frac{135}{4} + \frac{5}{2} \lambda - \frac{27}{2} \lambda + \lambda^2 \] \[ = -\frac{135}{4} - \frac{22}{2} \lambda + \lambda^2 \] \[ = -\frac{135}{4} - 11 \lambda + \lambda^2 \] Next, we calculate the second term: \[ \left( -\frac{5}{4} \right) \left( 8 \right) = -10 \] Putting it all together: \[ \det(A - \lambda I) = \left( -\frac{135}{4} - 11 \lambda + \lambda^2 \right) - (-10) \] \[ = -\frac{135}{4} - 11 \lambda + \lambda^2 + 10 \] \[ = \lambda^2 - 11 \lambda - \frac{135}{4} + 10 \] \[ = \lambda^2 - 11 \lambda - \frac{135}{4} + \frac{40}{4} \] \[ = \lambda^2 - 11 \lambda - \frac{95}{4} \] Thus, the characteristic polynomial is: \[ \boxed{\lambda^2 - 11 \lambda - \frac{95}{4}} \] |
\lambda ^2-11 \lambda -\frac{95}{4} |
| deepseek-r1 | To find the characteristic polynomial of the matrix \[ \begin{pmatrix} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \end{pmatrix}, \] we start by computing the determinant of \(\lambda I - A\), where \(I\) is the identity matrix and \(A\) is the given matrix. First, we construct the matrix \(\lambda I - A\): \[ \lambda I - A = \begin{pmatrix} \lambda + \frac{5}{2} & \frac{5}{4} \\ -8 & \lambda - \frac{27}{2} \end{pmatrix}. \] Next, we calculate the determinant of this matrix. For a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\), the determinant is \(ad - bc\). Applying this formula: \[ \det(\lambda I - A) = \left(\lambda + \frac{5}{2}\right)\left(\lambda - \frac{27}{2}\right) - \left(\frac{5}{4}\right)(-8). \] Expanding the first term: \[ \left(\lambda + \frac{5}{2}\right)\left(\lambda - \frac{27}{2}\right) = \lambda^2 - \frac{27}{2}\lambda + \frac{5}{2}\lambda - \frac{135}{4} = \lambda^2 - 11\lambda - \frac{135}{4}. \] Calculating the second term: \[ \left(\frac{5}{4}\right)(-8) = -10. \] Since this term is subtracted, we have: \[ \det(\lambda I - A) = \left(\lambda^2 - 11\lambda - \frac{135}{4}\right) - (-10) = \lambda^2 - 11\lambda - \frac{135}{4} + 10. \] Converting 10 to a fraction with denominator 4: \[ 10 = \frac{40}{4}, \] so, \[ \lambda^2 - 11\lambda - \frac{135}{4} + \frac{40}{4} = \lambda^2 - 11\lambda - \frac{95}{4}. \] Thus, the characteristic polynomial of the matrix is: \[ \boxed{x^2 - 11x - \dfrac{95}{4}}. \] |
\lambda ^2-11 \lambda -\frac{95}{4} |
| gpto1 | To find the characteristic polynomial of the given matrix: \[ A = \begin{pmatrix} -\dfrac{5}{2} & -\dfrac{5}{4} \\ 8 & \dfrac{27}{2} \end{pmatrix}, \] we compute the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar (the eigenvalue). First, compute \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -\dfrac{5}{2} - \lambda & -\dfrac{5}{4} \\ 8 & \dfrac{27}{2} - \lambda \end{pmatrix}. \] Next, compute its determinant: \[ \begin{aligned} P(\lambda) &= \det(A - \lambda I) \\ &= \left( -\dfrac{5}{2} - \lambda \right) \left( \dfrac{27}{2} - \lambda \right) - \left( -\dfrac{5}{4} \cdot 8 \right) \\ &= -\left( \lambda + \dfrac{5}{2} \right) \left( \dfrac{27}{2} - \lambda \right) + 10. \\ \end{aligned} \] Now expand the product: \[ \begin{aligned} -\left( \lambda + \dfrac{5}{2} \right) \left( \dfrac{27}{2} - \lambda \right) &= -\left( \lambda \cdot \dfrac{27}{2} - \lambda^2 + \dfrac{5}{2} \cdot \dfrac{27}{2} - \dfrac{5}{2} \lambda \right) \\ &= -\left( \dfrac{27}{2} \lambda - \lambda^2 + \dfrac{135}{4} - \dfrac{5}{2} \lambda \right) \\ &= -\left( \left( \dfrac{27}{2} - \dfrac{5}{2} \right) \lambda - \lambda^2 + \dfrac{135}{4} \right) \\ &= -\left( 11\lambda - \lambda^2 + \dfrac{135}{4} \right) \\ &= -11\lambda + \lambda^2 - \dfrac{135}{4}. \end{aligned} \] Add 10 to this result to get the characteristic polynomial: \[ \begin{aligned} P(\lambda) &= \left( -11\lambda + \lambda^2 - \dfrac{135}{4} \right) + 10 \\ &= \lambda^2 - 11\lambda - \dfrac{95}{4}. \end{aligned} \] Thus, the characteristic polynomial is: \(\boxed{\lambda^2\ -\ 11\lambda\ -\ \dfrac{95}{4}}\) |
\lambda ^2-11 \lambda -\frac{95}{4} |
| gpt-4o-2024-08-06 | To find the characteristic polynomial of the matrix \[ A = \begin{pmatrix} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \end{pmatrix}, \] 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 given by \[ A - \lambda I = \begin{pmatrix} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \end{pmatrix}. \] The characteristic polynomial is the determinant of this matrix: \[ \det(A - \lambda I) = \det\begin{pmatrix} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \end{pmatrix}. \] The determinant of a \( 2 \times 2 \) matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is calculated as \( ad - bc \). Applying this formula, we have: \[ \det(A - \lambda I) = \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) - \left(-\frac{5}{4}\right)(8). \] First, calculate the product \(\left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right)\): \[ \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) = \left(-\frac{5}{2}\right)\left(\frac{27}{2}\right) - \left(-\frac{5}{2}\right)\lambda - \lambda\left(\frac{27}{2}\right) + \lambda^2. \] Calculate each term: \[ \left(-\frac{5}{2}\right)\left(\frac{27}{2}\right) = -\frac{135}{4}, \] \[ \left(-\frac{5}{2}\right)\lambda = \frac{5}{2}\lambda, \] \[ -\lambda\left(\frac{27}{2}\right) = -\frac{27}{2}\lambda, \] \[ \lambda^2 = \lambda^2. \] Combine these results: \[ -\frac{135}{4} + \frac{5}{2}\lambda - \frac{27}{2}\lambda + \lambda^2 = \lambda^2 - \frac{22}{2}\lambda - \frac{135}{4}. \] Simplify the linear term: \[ \frac{5}{2}\lambda - \frac{27}{2}\lambda = -\frac{22}{2}\lambda = -11\lambda. \] Thus, the expression becomes: \[ \lambda^2 - 11\lambda - \frac{135}{4}. \] Now, calculate the second part of the determinant: \[ \left(-\frac{5}{4}\right)(8) = -\frac{40}{4} = -10. \] Combine both parts to find the determinant: \[ \det(A - \lambda I) = \lambda^2 - 11\lambda - \frac{135}{4} + 10. \] Convert 10 to a fraction with denominator 4: \[ 10 = \frac{40}{4}. \] Thus, the expression becomes: \[ \lambda^2 - 11\lambda - \frac{135}{4} + \frac{40}{4} = \lambda^2 - 11\lambda - \frac{95}{4}. \] Therefore, the characteristic polynomial is: \[ \boxed{\lambda^2 - 11\lambda - \frac{95}{4}}. \] |
\lambda ^2-11 \lambda -\frac{95}{4} |
| DeepSeek-Coder-V2-Lite-Instruct | To find the characteristic polynomial of the given matrix, we need to compute the determinant of the matrix \( A - \lambda I \), where \( A \) is the given matrix and \( \lambda \) is a scalar. The characteristic polynomial is given by \( \det(A - \lambda I) \). Given matrix: \[ A = \left( \begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \\ \end{array} \right) \] First, we form the matrix \( A - \lambda I \): \[ A - \lambda I = \left( \begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \\ \end{array} \right) \] Next, we compute the determinant of \( A - \lambda I \): \[ \det(A - \lambda I) = \left( -\frac{5}{2} - \lambda \right) \left( \frac{27}{2} - \lambda \right) - \left( -\frac{5}{4} \right) \cdot 8 \] We simplify each term separately: 1. Compute the product of the diagonal elements: \[ \left( -\frac{5}{2} - \lambda \right) \left( \frac{27}{2} - \lambda \right) \] 2. Compute the product of the off-diagonal elements: \[ -\frac{5}{4} \cdot 8 = -10 \] Now, combine these results: \[ \det(A - \lambda I) = \left( -\frac{5}{2} - \lambda \right) \left( \frac{27}{2} - \lambda \right) + 10 \] Expand the product of the diagonal elements: \[ \left( -\frac{5}{2} - \lambda \right) \left( \frac{27}{2} - \lambda \right) = \left( -\frac{5}{2} \cdot \frac{27}{2} \right) - \left( \frac{5}{2} \cdot \lambda \right) - \left( \lambda \cdot \frac{27}{2} \right) + \lambda^2 \] \[ = -\frac{135}{4} - \frac{5}{2} \lambda - \frac{27}{2} \lambda + \lambda^2 \] \[ = \lambda^2 - \frac{37}{2} \lambda - \frac{135}{4} \] Add the constant term: \[ \det(A - \lambda I) = \lambda^2 - \frac{37}{2} \lambda - \frac{135}{4} + 10 \] \[ = \lambda^2 - \frac{37}{2} \lambda - \frac{135}{4} + \frac{40}{4} \] \[ = \lambda^2 - \frac{37}{2} \lambda - \frac{95}{4} \] Thus, the characteristic polynomial is: \[ \boxed{\lambda^2 - \frac{37}{2} \lambda - \frac{95}{4}} \] |
\lambda ^2-11 \lambda -\frac{95}{4} |
| Meta-Llama-3.1-405B-Instruct-Turbo | ## Step 1: Recall the formula for the characteristic polynomial of a 2x2 matrix. The characteristic polynomial of a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by $(\lambda - a)(\lambda - d) - bc$, where $\lambda$ is the eigenvalue. ## Step 2: Identify the elements of the given matrix. The given matrix is $\begin{pmatrix} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \end{pmatrix}$, so $a = -\frac{5}{2}$, $b = -\frac{5}{4}$, $c = 8$, and $d = \frac{27}{2}$. ## Step 3: Substitute the elements into the characteristic polynomial formula. Substituting the elements into the formula, we get $(\lambda - (-\frac{5}{2}))(\lambda - \frac{27}{2}) - (-\frac{5}{4})(8)$. ## Step 4: Simplify the expression. Simplifying the expression, we get $(\lambda + \frac{5}{2})(\lambda - \frac{27}{2}) + 10$. ## Step 5: Expand the expression. Expanding the expression, we get $\lambda^2 - \frac{27}{2}\lambda + \frac{5}{2}\lambda - \frac{135}{4} + 10$. ## Step 6: Combine like terms. Combining like terms, we get $\lambda^2 - \frac{22}{2}\lambda - \frac{135}{4} + \frac{40}{4}$. ## Step 7: Simplify further. Simplifying further, we get $\lambda^2 - 11\lambda - \frac{95}{4}$. ## Step 8: Rewrite the expression in the required format. Rewriting the expression in the required format, we get $\lambda^2 - 11\lambda - \frac{95}{4}$. The final answer is: $\boxed{\lambda^2 - 11\lambda - \frac{95}{4}}$ |
\lambda ^2-11 \lambda -\frac{95}{4} |
| Meta-Llama-3.1-70B-Instruct-Turbo | ## Step 1: Understand the problem To find the characteristic polynomial of the given matrix, we need to follow a specific process. The characteristic polynomial is obtained by detaching the diagonal elements of the matrix, setting them to lambda, and then finding the determinant of the resulting matrix. ## Step 2: Write down the given matrix The given matrix is: $\left(\begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \end{array}\right)$ ## Step 3: Set up the characteristic polynomial equation The characteristic polynomial is given by: $|A - \lambda I| = 0$ where A is the given matrix, $\lambda$ is the eigenvalue, and I is the identity matrix. ## Step 4: Substitute the given matrix into the characteristic polynomial equation Substituting the given matrix into the equation, we get: $\left|\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \end{array}\right| = 0$ ## Step 5: Expand the determinant Expanding the determinant, we get: $(-\frac{5}{2} - \lambda)(\frac{27}{2} - \lambda) - (-\frac{5}{4})(8) = 0$ ## Step 6: Simplify the equation Simplifying the equation, we get: $\frac{135}{4} - \frac{5}{2}\lambda - \frac{27}{2}\lambda + \lambda^2 + 10 = 0$ ## Step 7: Combine like terms Combining like terms, we get: $\lambda^2 - 16\lambda + \frac{155}{4} = 0$ ## Step 8: Write the characteristic polynomial in the required format The characteristic polynomial is: $\boxed{\lambda^2 - 16\lambda + \frac{155}{4}}$ The final answer is: $\boxed{\lambda^2 - 16\lambda + \frac{155}{4}}$ |
\lambda ^2-11 \lambda -\frac{95}{4} |
| Qwen2-72B-Instruct | To find the characteristic polynomial of a 2x2 matrix, we use the formula: $$det(A - \lambda I) = 0$$ where A is the given matrix, $\lambda$ represents the eigenvalues, and I is the identity matrix. Given matrix A: $$A = \left(\begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \end{array}\right)$$ The characteristic equation is: $$det\left(\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \end{array}\right) = 0$$ Calculating the determinant: $$(-\frac{5}{2} - \lambda)(\frac{27}{2} - \lambda) - (-\frac{5}{4})(8) = 0$$ Simplifying: $$\lambda^2 - (\frac{27}{2} - \frac{5}{2})\lambda + (-\frac{5}{2})(\frac{27}{2}) + 10 = 0$$ $$\lambda^2 - 11\lambda + (-\frac{135}{4} + \frac{40}{4}) = 0$$ $$\lambda^2 - 11\lambda - \frac{95}{4} = 0$$ Thus, the characteristic polynomial is: $$\boxed{\lambda^2 - 11\lambda - \frac{95}{4}}$$ |
\lambda ^2-11 \lambda -\frac{95}{4} |
| baichuan4 | To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix minus $x$ times the identity matrix. For the given matrix: $$ A = \left( \begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \\ \end{array} \right) $$ We first construct the matrix $A - xI$, where $I$ is the identity matrix: $$ A - xI = \left( \begin{array}{cc} -\frac{5}{2} - x & -\frac{5}{4} \\ 8 & \frac{27}{2} - x \\ \end{array} \right) $$ Next, we compute the determinant of this matrix: $$ \text{det}(A - xI) = \left(-\frac{5}{2} - x\right)\left(\frac{27}{2} - x\right) - \left(-frac{5frac{5}{4}\right)(8) $$ Expanding the determinant, we get: $$ \text{det}(A - xI) = \left(-\frac{5}{2} - x\right)\left(\frac{27}{2} - x\right) + \frac{5}{4} \cdot 8 $$ $$ \text{det}(A - xI) = \left(-\frac{5}{2} \cdot \frac{27}{2} + \frac{5}{2}x - \frac{27}{2}x + x^2\right) + 10 $$ $$ \text{det}(A - xI) = \left(-\frac{135}{4} + \frac{5}{2}x - \frac{27}{2}x + x^2\right) + 10 $$ $$ \text{det}(A - xI) = x^2 - \left(\frac{5}{2} + \frac{27}{2}\right)x + \left(-\frac{135}{4} + 10\right) $$ $$ \text{det}(A - xI) = x^2 - 16x + \left(-\frac{135}{4} + \frac{40}{4}\right) $$ $$ \text{det}(A - xI) = x^2 - 16x - \frac{95}{4} $$ Therefore, the characteristic polynomial of the given matrix is: $$ \boxed{x^2 - 16x - \frac{95}{4}} $$ |
\lambda ^2-11 \lambda -\frac{95}{4} |
| doubao | The characteristic polynomial of a 2x2 matrix $A=\begin{pmatrix}a & b\\c & d\end{pmatrix}$ is given by $p(\lambda)=\lambda^2-(a+d)\lambda+ad-bc$. For the given matrix $\begin{pmatrix}-\frac{5}{2} & -\frac{5}{4}\\8 & \frac{27}{2}\end{pmatrix}$, $a=-\frac{5}{2}$, $d=\frac{27}{2}$, $b=-\frac{5}{4}$, $c=8$. The characteristic polynomial is: \[ \begin{align*} p(\lambda)&=\lambda^2-(-\frac{5}{2}+\frac{27}{2})\lambda+(-\frac{5}{2}\times\frac{27}{2})-(-\frac{5}{4}\times8)\\ &=\lambda^2-11\lambda+(-\frac{135}{4})-(-10)\\ &=\lambda^2-11\lambda-\frac{135}{4}+10\\ &=\lambda^2-11\lambda-\frac{95}{4}\\ \end{align*} \] So the answer is $\boxed{\lambda^2-11\lambda-\frac{95}{4}}$. |
\lambda ^2-11 \lambda -\frac{95}{4} |
| gpto1 | To find the characteristic polynomial of the given matrix: \[ A = \begin{pmatrix} -\dfrac{5}{2} & -\dfrac{5}{4} \\ 8 & \dfrac{27}{2} \end{pmatrix}, \] we compute the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar (the eigenvalue). First, compute \( A - \lambda I \): \[ A - \lambda I = \begin{pmatrix} -\dfrac{5}{2} - \lambda & -\dfrac{5}{4} \\ 8 & \dfrac{27}{2} - \lambda \end{pmatrix}. \] Next, compute its determinant: \[ \begin{aligned} P(\lambda) &= \det(A - \lambda I) \\ &= \left( -\dfrac{5}{2} - \lambda \right) \left( \dfrac{27}{2} - \lambda \right) - \left( -\dfrac{5}{4} \cdot 8 \right) \\ &= -\left( \lambda + \dfrac{5}{2} \right) \left( \dfrac{27}{2} - \lambda \right) + 10. \\ \end{aligned} \] Now expand the product: \[ \begin{aligned} -\left( \lambda + \dfrac{5}{2} \right) \left( \dfrac{27}{2} - \lambda \right) &= -\left( \lambda \cdot \dfrac{27}{2} - \lambda^2 + \dfrac{5}{2} \cdot \dfrac{27}{2} - \dfrac{5}{2} \lambda \right) \\ &= -\left( \dfrac{27}{2} \lambda - \lambda^2 + \dfrac{135}{4} - \dfrac{5}{2} \lambda \right) \\ &= -\left( \left( \dfrac{27}{2} - \dfrac{5}{2} \right) \lambda - \lambda^2 + \dfrac{135}{4} \right) \\ &= -\left( 11\lambda - \lambda^2 + \dfrac{135}{4} \right) \\ &= -11\lambda + \lambda^2 - \dfrac{135}{4}. \end{aligned} \] Add 10 to this result to get the characteristic polynomial: \[ \begin{aligned} P(\lambda) &= \left( -11\lambda + \lambda^2 - \dfrac{135}{4} \right) + 10 \\ &= \lambda^2 - 11\lambda - \dfrac{95}{4}. \end{aligned} \] Thus, the characteristic polynomial is: \(\boxed{\lambda^2\ -\ 11\lambda\ -\ \dfrac{95}{4}}\) |
\lambda ^2-11 \lambda -\frac{95}{4} |
| kimi | To find the characteristic polynomial of a matrix, we need to compute the determinant of the matrix subtracted by $\lambda I$, where $\lambda$ is a scalar and $I$ is the identity matrix. In this case, the matrix is: $\left(\begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \\ \end{array}\right)$ The identity matrix $I$ for a 2x2 matrix is: $\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array}\right)$ Now, we subtract $\lambda I$ from the given matrix: $\left(\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \\ \end{array}\right)$ Now, we compute the determinant of this matrix: $\det\left(\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \\ \end{array}\right) = \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) - \left(-\frac{5}{4}\right)(8)$ Expanding the determinant: $= \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) + 10$ $= \left(-\frac{5}{2} \cdot \frac{27}{2} + \frac{5}{2} \cdot \lambda - \lambda \cdot \frac{27}{2} + \lambda^2\right) + 10$ $= \left(-\frac{135}{4} + \frac{5}{2} \cdot \lambda - \frac{27}{2} \cdot \lambda + \lambda^2\right) + 10$ Now, we simplify the expression: $= \lambda^2 - 6\lambda - \frac{135}{4} + 10$ $= \lambda^2 - 6\lambda - \frac{135}{4} + \frac{40}{4}$ $= \lambda^2 - 6\lambda - \frac{95}{4}$ So, the characteristic polynomial of the given matrix is: $\boxed{\lambda^2 - 6\lambda - \frac{95}{4}}$ |
\lambda ^2-11 \lambda -\frac{95}{4} |
| minimax | To find the characteristic polynomial of a 2x2 matrix, we use the formula: $$ \text{det}(A - \lambda I) = 0 $$ where $A$ is the matrix, $\lambda$ represents the eigenvalues, and $I$ is the identity matrix of the same size as $A$. For the given matrix $$ A = \left(\begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \\ \end{array}\right) $$ we have: $$ A - \lambda I = \left(\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \\ \end{array}\right) $$ The determinant of this matrix is: $$ \text{det}(A - \lambda I) = \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) - \left(-\frac{5}{4}\right)(8) $$ |
\lambda ^2-11 \lambda -\frac{95}{4} |
| wenxinyiyan | 为了找到矩阵$\begin{pmatrix} -\frac{5}{2} & -\frac{5}{4} \ 8 & \frac{27}{2} \end{pmatrix}$的特征多项式，我们需要计算行列式$\begin{vmatrix} -\frac{5}{2} - \lambda & -\frac{5}{4} \ 8 & \frac{27}{2} - \lambda \end{vmatrix}$。 这个行列式可以展开为： $(-\frac{5}{2} - \lambda)(\frac{27}{2} - \lambda) - (-\frac{5}{4} \times 8) = \lambda^2 - 11\lambda + \frac{125}{4} - \frac{100}{4} = \lambda^2 - 11\lambda + \frac{25}{4}$。因此，特征多项式为$\boxed{\lambda^2 - 11\lambda + \frac{25}{4}}$。 |
\lambda ^2-11 \lambda -\frac{95}{4} |
| zhipu | To find the characteristic polynomial of the matrix \[ A = \left(\begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \end{array}\right), \] 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 given by: \[ A - \lambda I = \left(\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \end{array}\right). \] The characteristic polynomial \(p(\lambda)\) is the determinant of this matrix: \[ p(\lambda) = \det\left(\begin{array}{cc} -\frac{5}{2} - \lambda & -\frac{5}{4} \\ 8 & \frac{27}{2} - \lambda \end{array}\right). \] The determinant of a \(2 \times 2\) matrix \(\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\) is given by \(ad - bc\). Applying this to our matrix, we get: \[ p(\lambda) = \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) - \left(8 \cdot -\frac{5}{4}\right). \] First, compute the product of the diagonal elements: \[ \left(-\frac{5}{2} - \lambda\right)\left(\frac{27}{2} - \lambda\right) = \left(-\frac{5}{2}\right)\left(\frac{27}{2}\right) + \left(-\frac{5}{2}\right)(-\lambda) + (-\lambda)\left(\frac{27}{2}\right) + \lambda^2. \] Simplifying each term: \[ \left(-\frac{5}{2}\right)\left(\frac{27}{2}\right) = -\frac{135}{4}, \] \[ \left(-\frac{5}{2}\right)(-\lambda) = \frac{5}{2}\lambda, \] \[ (-\lambda)\left(\frac{27}{2}\right) = -\frac{27}{2}\lambda, \] \[ \lambda^2 = \lambda^2. \] Combining these, we have: \[ -\frac{135}{4} + \frac{5}{2}\lambda - \frac{27}{2}\lambda + \lambda^2 = \lambda^2 - \frac{22}{2}\lambda - \frac{135}{4}. \] Simplify the coefficient of \(\lambda\): \[ -\frac{22}{2} = -11. \] So, the expression becomes: \[ \lambda^2 - 11\lambda - \frac{135}{4}. \] Next, compute the off-diagonal product: \[ 8 \cdot -\frac{5}{4} = -10. \] Thus, the characteristic polynomial is: \[ p(\lambda) = \lambda^2 - 11\lambda - \frac{135}{4} + 10. \] Convert \(10\) to a fraction with a denominator of \(4\): \[ 10 = \frac{40}{4}. \] So, the polynomial becomes: \[ p(\lambda) = \lambda^2 - 11\lambda - \frac{135}{4} + \frac{40}{4} = \lambda^2 - 11\lambda - \frac{95}{4}. \] To clear the fraction, multiply through by \(4\): \[ 4\lambda^2 - 44\lambda - 95. \] Therefore, the characteristic polynomial is: \[ \boxed{4\lambda^2 - 44\lambda - 95}. \] |
\lambda ^2-11 \lambda -\frac{95}{4} |
| deepseek-v3 | To find the **characteristic polynomial** of the matrix: \[ A = \left( \begin{array}{cc} -\frac{5}{2} & -\frac{5}{4} \\ 8 & \frac{27}{2} \\ \end{array} \right) \] we follow these steps: ### Step 1: Recall the Formula for the Characteristic Polynomial For a \(2 \times 2\) matrix \(A = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\), the characteristic polynomial \(p(\lambda)\) is given by: \[ p(\lambda) = \det(A - \lambda I) = \lambda^2 - (a + d)\lambda + (ad - bc) \] where \(I\) is the identity matrix. ### Step 2: Identify the Elements of the Matrix From matrix \(A\): \[ a = -\frac{5}{2}, \quad b = -\frac{5}{4}, \quad c = 8, \quad d = \frac{27}{2} \] ### Step 3: Compute the Trace and Determinant - **Trace** (\(a + d\)): \[ a + d = -\frac{5}{2} + \frac{27}{2} = \frac{22}{2} = 11 \] - **Determinant** (\(ad - bc\)): \[ ad = \left(-\frac{5}{2}\right) \times \frac{27}{2} = -\frac{135}{4} \] \[ bc = \left(-\frac{5}{4}\right) \times 8 = -10 \] \[ ad - bc = -\frac{135}{4} - (-10) = -\frac{135}{4} + \frac{40}{4} = -\frac{95}{4} \] ### Step 4: Formulate the Characteristic Polynomial Substituting the trace and determinant into the characteristic polynomial formula: \[ p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc) = \lambda^2 - 11\lambda - \frac{95}{4} \] ### Step 5: Present the Final Answer \[ \boxed{\lambda^2 - 11\lambda - \frac{95}{4}} \] |
\lambda ^2-11 \lambda -\frac{95}{4} |