| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if a matrix is positive definite, we need to check if all its eigenvalues are positive. The given matrix is A=[[1, 4], [4, 1]]. First, we find the characteristic equation of the matrix, which is given by: det(A - λI) = 0 where λ represents the eigenvalues and I is the identity matrix. For the given matrix, the characteristic equation is: det([[1-λ, 4], [4, 1-λ]]) = (1-λ)^2 - 16 = λ^2 - 2λ - 15 = 0 Solving this quadratic equation, we find the eigenvalues to be: λ1 = 5 λ2 = -3 Since one of the eigenvalues (λ2 = -3) is negative, the given matrix A=[[1, 4], [4, 1]] is not a positive definite matrix. |
False |
| deepseekr1 | The matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \) is **not** positive definite. Here's the step-by-step reasoning: 1. **Check Symmetry**: \( A \) is symmetric since \( A = A^T \). 2. **Eigenvalues**: - Solve \( \det(A - \lambda I) = 0 \): \[ \det\begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} = (1 - \lambda)^2 - 16 = \lambda^2 - 2\lambda - 15 = 0 \] - Roots: \( \lambda = 5 \) and \( \lambda = -3 \). - **Conclusion**: Negative eigenvalue (\( -3 \)) exists, so \( A \) is not positive definite. 3. **Leading Principal Minors**: - First minor: \( 1 > 0 \). - Second minor: \( \det(A) = 1 \cdot 1 - 4 \cdot 4 = -15 < 0 \). - **Conclusion**: Second minor is negative, violating positive definiteness. 4. **Quadratic Form**: - \( x^T A x = x_1^2 + 8x_1x_2 + x_2^2 \). - Completing the square: \( (x_1 + 4x_2)^2 - 15x_2^2 \). - For \( x = [1, -1] \), \( x^T A x = -6 < 0 \), proving non-positive definiteness. **Final Answer**: No, the matrix \( A \) is not positive definite. |
False |
| gpto1 | To determine whether the matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \) is positive definite, we can use several methods. The most common approaches involve checking the eigenvalues of the matrix or applying Sylvester's Criterion, which involves evaluating the leading principal minors (determinants of the top-left \( k \times k \) submatrices). **Method 1: Eigenvalue Analysis** A symmetric matrix is positive definite if and only if all its eigenvalues are positive. Let's find the eigenvalues of matrix \( A \). **Step 1: Find the Characteristic Polynomial** The characteristic polynomial of \( A \) is given by: \[ \det(A - \lambda I) = 0 \] Compute \( A - \lambda I \): \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} \] **Step 2: Compute the Determinant** Calculate the determinant of \( A - \lambda I \): \[ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (4)(4) = (1 - \lambda)^2 - 16 \] Set the determinant equal to zero: \[ (1 - \lambda)^2 - 16 = 0 \] **Step 3: Solve for \( \lambda \)** Simplify the equation: \[ (1 - \lambda)^2 = 16 \\ 1 - \lambda = \pm 4 \] Solve for \( \lambda \): 1. \( 1 - \lambda = 4 \) \(\Rightarrow \lambda = 1 - 4 = -3 \) 2. \( 1 - \lambda = -4 \) \(\Rightarrow \lambda = 1 + 4 = 5 \) **Step 4: Analyze the Eigenvalues** The eigenvalues of \( A \) are \( \lambda_1 = -3 \) and \( \lambda_2 = 5 \). Since one of the eigenvalues (\( \lambda_1 = -3 \)) is negative, \( A \) is **not** positive definite. **Method 2: Sylvester's Criterion** Sylvester's Criterion states that a symmetric matrix is positive definite if and only if all its leading principal minors are positive. **Step 1: Compute the Leading Principal Minors** 1. **First leading principal minor (\( D_1 \)):** \[ D_1 = \det\left( \begin{bmatrix} 1 \end{bmatrix} \right) = 1 \] \( D_1 > 0 \), so the first condition is satisfied. 2. **Second leading principal minor (\( D_2 \)):** \[ D_2 = \det\left( \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \right) = (1)(1) - (4)(4) = 1 - 16 = -15 \] \( D_2 = -15 < 0 \), so the second condition is not satisfied. **Step 2: Conclude Based on Sylvester's Criterion** Since the second leading principal minor is negative (\( D_2 < 0 \)), Sylvester's Criterion tells us that \( A \) is **not** positive definite. **Conclusion** Based on both eigenvalue analysis and Sylvester's Criterion, matrix \( A \) is **not positive definite** because it has a negative eigenvalue and a negative leading principal minor. **Answer:** **No, the matrix A is not positive definite because it has a negative eigenvalue and fails Sylvester's Criterion.** |
False |
| GPT-4o | To determine if the matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \) is positive definite, we need to check if it satisfies the following conditions: 1. \(A\) must be symmetric. 2. All eigenvalues of \(A\) must be positive. Let's check these conditions step by step. ### Step 1: Check if \(A\) is symmetric A matrix \(A\) is symmetric if \(A = A^T\). For the given matrix: \[ A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \] \[ A^T = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \] Since \(A = A^T\), the matrix is symmetric. ### Step 2: Find the eigenvalues of \(A\) To find the eigenvalues, we solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(\lambda\) is an eigenvalue and \(I\) is the identity matrix. \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} \] The determinant of this matrix is given by: \[ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - 4 \cdot 4 \] \[ \det(A - \lambda I) = (1 - \lambda)^2 - 16 \] \[ \det(A - \lambda I) = \lambda^2 - 2\lambda + 1 - 16 \] \[ \det(A - \lambda I) = \lambda^2 - 2\lambda - 15 \] Setting the determinant to zero to find the eigenvalues: \[ \lambda^2 - 2\lambda - 15 = 0 \] We can solve this quadratic equation using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -15\). \[ \lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \] \[ \lambda = \frac{2 \pm \sqrt{4 + 60}}{2} \] \[ \lambda = \frac{2 \pm \sqrt{64}}{2} \] \[ \lambda = \frac{2 \pm 8}{2} \] This gives us two solutions: \[ \lambda = \frac{2 + 8}{2} = 5 \] \[ \lambda = \frac{2 - 8}{2} = -3 \] The eigenvalues of the matrix \(A\) are \(\lambda_1 = 5\) and \(\lambda_2 = -3\). ### Step 3: Check if all eigenvalues are positive For the matrix \(A\) to be positive definite, all its eigenvalues must be positive. In this case, we have one positive eigenvalue (\(\lambda_1 = 5\)) and one negative eigenvalue (\(\lambda_2 = -3\)). Since not all eigenvalues are positive, the matrix \(A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix}\) is **not positive definite**. |
False |
| llama3.1-70B-Instruct | To determine if a matrix is positive definite, we can use the following methods: 1. Check if all eigenvalues are positive. 2. Check if the matrix is symmetric and all its leading principal minors are positive. Let's apply these methods to the given matrix A = [[1, 4], [4, 1]]. Method 1: Eigenvalues To find the eigenvalues, we can solve the characteristic equation det(A - λI) = 0, where I is the identity matrix and λ is the eigenvalue. det(A - λI) = det([[1 - λ, 4], [4, 1 - λ]]) = (1 - λ)^2 - 16 = 0 Solving for λ, we get: λ = 1 ± √17 Since one of the eigenvalues (λ = 1 - √17) is negative, the matrix A is not positive definite. Method 2: Symmetry and leading principal minors The matrix A is symmetric since A = A^T. The leading principal minors are: 1. det([[1]]) = 1 > 0 2. det([[1, 4], [4, 1]]) = -15 < 0 Since the second leading principal minor is negative, the matrix A is not positive definite. Therefore, using both methods, we conclude that the matrix A = [[1, 4], [4, 1]] is not positive definite. |
False |
| llama3.1-405B-Instruct | To determine if a matrix is positive definite, we can use the following criteria: 1. The matrix must be symmetric. 2. All the eigenvalues of the matrix must be positive. Let's check these criteria for the given matrix A = [[1, 4], [4, 1]]: 1. Symmetry: The matrix A is symmetric because A = A^T, where A^T is the transpose of A. 2. Eigenvalues: To find the eigenvalues of A, we need to solve the characteristic equation det(A - λI) = 0, where λ is the eigenvalue and I is the identity matrix. The characteristic equation for A is: (-λ + 1)^2 - 16 = 0 Solving for λ, we get: λ = 5 or λ = -3 Since one of the eigenvalues (λ = -3) is negative, the matrix A is not positive definite. Therefore, the matrix A = [[1, 4], [4, 1]] is not a positive definite matrix. |
False |
| DeepSeek-Coder-V2 | To determine if the matrix A = [[1, 4], [4, 1]] is positive definite, we need to check if all its eigenvalues are positive. Here is the step-by-step solution: 1. **Find the characteristic polynomial of the matrix A:** The characteristic polynomial is given by the determinant of (A - λI), where λ is an eigenvalue and I is the identity matrix. \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} \] The determinant of this matrix is: \[ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - 4 \cdot 4 \] Simplifying this, we get: \[ \det(A - \lambda I) = (1 - \lambda)^2 - 16 \] \[ \det(A - \lambda I) = 1 - 2\lambda + \lambda^2 - 16 \] \[ \det(A - \lambda I) = \lambda^2 - 2\lambda - 15 \] 2. **Solve the characteristic polynomial to find the eigenvalues:** We need to solve the quadratic equation: \[ \lambda^2 - 2\lambda - 15 = 0 \] Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -15 \): \[ \lambda = \frac{2 \pm \sqrt{4 + 60}}{2} \] \[ \lambda = \frac{2 \pm \sqrt{64}}{2} \] \[ \lambda = \frac{2 \pm 8}{2} \] This gives us two solutions: \[ \lambda_1 = \frac{2 + 8}{2} = 5 \] \[ \lambda_2 = \frac{2 - 8}{2} = -3 \] 3. **Determine if the eigenvalues are positive:** The eigenvalues of the matrix A are 5 and -3. Since one of the eigenvalues (-3) is not positive, the matrix A is not positive definite. Therefore, the matrix A = [[1, 4], [4, 1]] is not a positive definite matrix. |
False |
| Qwen2-72B-Instruct | To determine if a matrix is positive definite, we need to check if all its eigenvalues are positive. The given matrix is A=[[1, 4], [4, 1]]. First, we find the characteristic equation of the matrix, which is given by: det(A - λI) = 0 where λ represents the eigenvalues and I is the identity matrix. For the given matrix, the characteristic equation is: det([[1-λ, 4], [4, 1-λ]]) = (1-λ)^2 - 16 = λ^2 - 2λ - 15 = 0 Solving this quadratic equation, we find the eigenvalues to be: λ1 = 5 λ2 = -3 Since one of the eigenvalues (λ2 = -3) is negative, the given matrix A=[[1, 4], [4, 1]] is not a positive definite matrix. |
False |
| baichuan4 | To determine if matrix A is positive definite, we can use one of several methods. One way is to check if all the eigenvalues of A are positive. Another way is to verify if the quadratic form associated with A is always positive for any non-zero vector x. We will use the second method here. The quadratic form associated with matrix A is given by: x^T * A * x where x is any non-zero column vector. If this quadratic form is always positive for any non-zero x, then A is positive definite. Let's calculate the quadratic form for A: A = [[1, 4], [4, 1]] x^T * A * x = [x1 x2] * [[1, 4], [4, 1]] * [x1] [x2] Expanding the multiplication, we get: x^T * A * x = (1*x1 + 4*x2) * x1 + (4*x1 + 1*x2) * x2 = x1^2 + 4*x1*x2 + 4*x1*x2 + x2^2 = x1^2 + 8*x1*x2 + x2^2 Now, we need to determine if this quadratic form can be negative for any values of x1 and x2. Since x1^2 and x2^2 are always positive, the only term that could potentially make the quadratic form negative is the 8*x1*x2 term. However, since this term is squared (8^2), it will always be positive, regardless of the signs of x1 and x2. Therefore, the quadratic form x^T * A * x = x1^2 + 8*x1*x2 + x2^2 is always positive for any non-zero vector [x1 x2], which means that matrix A is positive definite. |
False |
| doubao | To determine if a matrix is positive definite, we can check if all its eigenvalues are positive. The characteristic equation of A is given by: det(A - λI) = 0, where I is the identity matrix. So, det([[1 - λ, 4], [4, 1 - λ]]) = 0 (1 - λ)(1 - λ) - 16 = 0 1 - 2λ + λ² - 16 = 0 λ² - 2λ - 15 = 0 (λ - 5)(λ + 3) = 0 So, the eigenvalues are λ = 5 and λ = -3. Since one eigenvalue is negative (-3), the matrix A is not a positive definite matrix. |
False |
| minimax | To determine if a matrix is positive definite, we can use several methods. One common method is to check if all the leading principal minors are positive. A leading principal minor is the determinant of the top-left k x k submatrix for k = 1, 2, ..., n, where n is the size of the matrix. Let's apply this method to the given matrix A: A = [[1, 4], [4, 1]] The matrix A is a 2x2 matrix, so we need to check the following: 1. The determinant of the top-left 1x1 submatrix (which is just the element in the first row and first column): det(A1) = 1. This is positive. 2. The determinant of the entire 2x2 matrix A itself: det(A) = (1)(1) - (4)(4) = 1 - 16 = -15. This is negative. Since the determinant of the entire matrix is negative, the matrix A is not positive definite. In fact, because the determinant is negative, we can conclude that the matrix is indefinite. |
False |
| kimi | To determine if a matrix is positive definite, we can use the following criteria: 1. The matrix must be symmetric, meaning \( A = A^T \). 2. All eigenvalues of the matrix must be positive. For the given matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \), let's check these criteria: 1. The matrix is symmetric because \( A^T = A \). 2. To check the eigenvalues, we first find the characteristic equation of the matrix. The characteristic equation is given by: \[ \det(A - \lambda I) = 0 \] where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. For matrix \( A \), the characteristic equation is: \[ \det\begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} = 0 \] Expanding the determinant, we get: \[ (1 - \lambda)^2 - (4)(4) = 0 \] \[ \lambda^2 - 2\lambda + 1 - 16 = 0 \] \[ \lambda^2 - 2\lambda - 15 = 0 \] Now, we solve this quadratic equation for \( \lambda \). The solutions to the equation are the eigenvalues of the matrix. The quadratic formula is: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation \( \lambda^2 - 2\lambda - 15 = 0 \), \( a = 1 \), \( b = -2 \), and \( c = -15 \). Plugging these values into the quadratic formula gives us: \[ \lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} \] \[ \lambda = \frac{2 \pm \sqrt{4 + 60}}{2} \] \[ \lambda = \frac{2 \pm \sqrt{64}}{2} \] \[ \lambda = \frac{2 \pm 8}{2} \] This gives us two eigenvalues: \[ \lambda_1 = \frac{2 + 8}{2} = 5 \] \[ \lambda_2 = \frac{2 - 8}{2} = -3 \] Since one of the eigenvalues is negative (\( \lambda_2 = -3 \)), the matrix \( A \) is not positive definite. |
False |
| wenxinyiyan | To determine if the matrix A is positive definite, we need to check if all its eigenvalues are positive. The eigenvalues of a matrix A can be found by solving the characteristic equation det(A - λI) = 0, where λ is an eigenvalue and I is the identity matrix. For the given matrix A = [[1, 4], [4, 1]], the characteristic equation is: det(A - λI) = det([[1-λ, 4], [4, 1-λ]]) = (1-λ)^2 - 4*4 = λ^2 - 2λ - 15 = 0 Solving this quadratic equation, we get: λ = (2 ± sqrt(2^2 - 4*(-15))) / 2 = (2 ± sqrt(64)) / 2 = (2 ± 8) / 2 So, the eigenvalues are λ1 = (2+8)/2 = 5 and λ2 = (2-8)/2 = -3. Since one of the eigenvalues (λ2 = -3) is negative, the matrix A is not positive definite. |
False |
| zhipu | To determine whether a matrix is positive definite, we can use several criteria. One of the common methods is to check the eigenvalues of the matrix. A matrix is positive definite if and only if all of its eigenvalues are positive. The given matrix A is: \[ A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \] Let's find the eigenvalues of A. The eigenvalues \(\lambda\) of a matrix A are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0 \] where \(I\) is the identity matrix of the same size as A. For the matrix A, the characteristic equation is: \[ \det \begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} = 0 \] Calculating the determinant, we get: \[ (1 - \lambda)(1 - \lambda) - (4)(4) = 0 \] \[ (1 - \lambda)^2 - 16 = 0 \] \[ 1 - 2\lambda + \lambda^2 - 16 = 0 \] \[ \lambda^2 - 2\lambda - 15 = 0 \] This is a quadratic equation. Solving for \(\lambda\) using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -15\): \[ \lambda = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} \] \[ \lambda = \frac{2 \pm \sqrt{4 + 60}}{2} \] \[ \lambda = \frac{2 \pm \sqrt{64}}{2} \] \[ \lambda = \frac{2 \pm 8}{2} \] So, the eigenvalues are: \[ \lambda_1 = \frac{2 + 8}{2} = 5 \] \[ \lambda_2 = \frac{2 - 8}{2} = -3 \] Since one of the eigenvalues (\(\lambda_2 = -3\)) is negative, the matrix A is not positive definite. Another method to check for positive definiteness is to use the principal minors. For a 2x2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), the conditions for positive definiteness are: 1. \(a > 0\) 2. \(ad - bc > 0\) For the given matrix \(A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix}\): 1. \(a = 1 > 0\) 2. \(ad - bc = (1)(1) - (4)(4) = 1 - 16 = -15\) Since the second condition fails (\(ad - bc\) is not greater than 0), the matrix A is not positive definite. Therefore, the matrix \(A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix}\) is **not** a positive definite matrix. |
False |
| gpto1 | To determine whether the matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \) is positive definite, we can use several methods. The most common approaches involve checking the eigenvalues of the matrix or applying Sylvester's Criterion, which involves evaluating the leading principal minors (determinants of the top-left \( k \times k \) submatrices). **Method 1: Eigenvalue Analysis** A symmetric matrix is positive definite if and only if all its eigenvalues are positive. Let's find the eigenvalues of matrix \( A \). **Step 1: Find the Characteristic Polynomial** The characteristic polynomial of \( A \) is given by: \[ \det(A - \lambda I) = 0 \] Compute \( A - \lambda I \): \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 4 \\ 4 & 1 - \lambda \end{bmatrix} \] **Step 2: Compute the Determinant** Calculate the determinant of \( A - \lambda I \): \[ \det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (4)(4) = (1 - \lambda)^2 - 16 \] Set the determinant equal to zero: \[ (1 - \lambda)^2 - 16 = 0 \] **Step 3: Solve for \( \lambda \)** Simplify the equation: \[ (1 - \lambda)^2 = 16 \\ 1 - \lambda = \pm 4 \] Solve for \( \lambda \): 1. \( 1 - \lambda = 4 \) \(\Rightarrow \lambda = 1 - 4 = -3 \) 2. \( 1 - \lambda = -4 \) \(\Rightarrow \lambda = 1 + 4 = 5 \) **Step 4: Analyze the Eigenvalues** The eigenvalues of \( A \) are \( \lambda_1 = -3 \) and \( \lambda_2 = 5 \). Since one of the eigenvalues (\( \lambda_1 = -3 \)) is negative, \( A \) is **not** positive definite. **Method 2: Sylvester's Criterion** Sylvester's Criterion states that a symmetric matrix is positive definite if and only if all its leading principal minors are positive. **Step 1: Compute the Leading Principal Minors** 1. **First leading principal minor (\( D_1 \)):** \[ D_1 = \det\left( \begin{bmatrix} 1 \end{bmatrix} \right) = 1 \] \( D_1 > 0 \), so the first condition is satisfied. 2. **Second leading principal minor (\( D_2 \)):** \[ D_2 = \det\left( \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \right) = (1)(1) - (4)(4) = 1 - 16 = -15 \] \( D_2 = -15 < 0 \), so the second condition is not satisfied. **Step 2: Conclude Based on Sylvester's Criterion** Since the second leading principal minor is negative (\( D_2 < 0 \)), Sylvester's Criterion tells us that \( A \) is **not** positive definite. **Conclusion** Based on both eigenvalue analysis and Sylvester's Criterion, matrix \( A \) is **not positive definite** because it has a negative eigenvalue and a negative leading principal minor. **Answer:** **No, the matrix A is not positive definite because it has a negative eigenvalue and fails Sylvester's Criterion.** |
False |
| deepseekv3 | To determine whether the matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \) is positive definite, we can use the following criteria for a \( 2 \times 2 \) matrix: A matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is positive definite if: 1. \( a > 0 \), 2. The determinant of \( A \) is positive (\( ad - bc > 0 \)). Let's apply these criteria to the given matrix: ### Step 1: Check the first condition The first element of \( A \) is \( a = 1 \). Since \( 1 > 0 \), the first condition is satisfied. ### Step 2: Check the determinant The determinant of \( A \) is calculated as: \[ \det(A) = ad - bc = (1)(1) - (4)(4) = 1 - 16 = -15 \] Since the determinant (\( -15 \)) is not positive, the second condition is not satisfied. ### Conclusion Since the determinant of \( A \) is not positive, the matrix \( A \) is not positive definite. **Final Answer:** No, the matrix \( A = \begin{bmatrix} 1 & 4 \\ 4 & 1 \end{bmatrix} \) is not positive definite. |
False |