Problem Analysis #1

34.81s

9,320 toks

Problem Statement

For a square matrix $A$ , the matrix exponential $e^{A}$ is defined by $e^{A} = E + \sum_{k=1}^{\infty} \frac{1}{k!}A^{k},$ where $E$ is the identity matrix.

[(1)] Let $A= \begin{pmatrix} -1 & 4 & 4 \\ -5 & 8 & 10 \\ 3 & -3 & -5 \end{pmatrix}.$ Compute the matrix exponential $e^{A}$ .
[(2)] Consider the matrices $A_x= \begin{pmatrix} 0 & -x & 0 \\ x & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\qquad B= \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix},$ where $x$ is a real number, and define the vector $a= \begin{pmatrix} 1 \\[2mm] 1 \\[2mm] e \end{pmatrix}.$ Let $C = B e^{A_x}$ . Determine the minimum and maximum values of $\frac{y^{T} C y}{y^{T}y}$ over all nonzero real vectors $y\in\mathbb{R}^{3}$ .
[(3)] Let $f(z)=\frac{1}{2}z^{T} C z - a^{T}z, \qquad z= \begin{pmatrix} z_{1}\\ z_{2}\\ z_{3} \end{pmatrix},$ where $C$ is the matrix from part (2). Find the value of $\sqrt{z_1^2+z_2^2+z_3^2}$ for the vector $z$ satisfying $\frac{\partial f}{\partial z_{1}} = \frac{\partial f}{\partial z_{2}} = \frac{\partial f}{\partial z_{3}} =0.$ Let $\lambda_1,\lambda_2,\lambda_3$ be the eigenvalues of the matrix $A$ in part {\rm (1)}. Let $m$ and $M$ be the minimum and maximum values, respectively, obtained in part {\rm (2)}, and let $R = \sqrt{z_1^2+z_2^2+z_3^2}$ be the value obtained in part {\rm (3)}. Compute $\lambda_1\lambda_2\lambda_3 \cdot m \cdot R^2 \cdot \bigl(\lambda_1^2+\lambda_2^2+\lambda_3^2\bigr).$ \

Ground Truth

252

Execution Trace

Reasoning Process

[THOUGHT] The eigenvalues of A are needed. Use sympy to compute them exactly. Verify product and sum of squares. Confirm m=-1, R^2=3 from analytical derivation. Compute the expression λ1λ2λ3 * m * R^2 * (sum λ_i^2). [/THOUGHT] [PYTHON] import sympy as sp

A = sp.Matrix([[-1, 4, 4], [-5, 8, 10], [3, -3, -5]]) eigenvals = A.eigenvals() print("Eigenvalues:", eigenvals)

lambdas = list(eigenvals.keys()) prod = sp.prod(lambdas) sum_sq = sum(l**2 for l in lambdas) m = -1 R_sq = 3 result = prod * m * R_sq * sum_sq print("Product lambdas:", prod) print("Sum squares:", sum_sq) print("Final result:", result) [/PYTHON]

Tool Output

[PYTHON OUTPUT] Eigenvalues: {3: 1, 1: 1, -2: 1} Product lambdas: -6 Sum squares: 14 Final result: 252 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The Python output confirms eigenvalues 3,1,-2; product -6; sum squares 14. With m=-1, R^2=3, the result is (-6)*(-1)314=252. [/THOUGHT] \boxed{252}

Explanation

The reasoning process follows the problem's requirements by first computing the eigenvalues of matrix A from part (1) using SymPy, then using these to evaluate the final expression involving λ1λ2λ3, m from part (2), R^2 from part (3), and the sum of squares of the eigenvalues.

Compute eigenvalues of A = [[-1, 4, 4], [-5, 8, 10], [3, -3, -5]] via SymPy's eigenvals() method, yielding {3: 1, 1: 1, -2: 1}, so the eigenvalues are λ1=3, λ2=1, λ3=-2.
Calculate the product λ1λ2λ3 = 3 * 1 * (-2) = -6.
Compute the sum of squares λ1² + λ2² + λ3² = 3² + 1² + (-2)² = 9 + 1 + 4 = 14.
Use m = -1 (the minimum value from part (2), as derived from the Rayleigh quotient of C = B e^{A_x} at its extremal eigenvalues).
Use R² = 3 (from part (3), where z satisfies the critical point of f(z), leading to ||z||² = 3 via solving ∇f(z) = 0, i.e., C z = a).
Evaluate the full expression: λ1λ2λ3 * m * R² * (λ1² + λ2² + λ3²) = (-6) * (-1) * 3 * 14 = 6 * 3 * 14 = 18 * 14 = 252.

This matches the Python output, confirming the result as ground truth.

Final Model Output

252

Comparative Analysis (P1) (Average across runs)

Avg Time Taken

GPT-5.1 (high)

150.0

Llama-VARCO-8B-Instruct

141.2

Deepseek-V3.2

124.1

A.X-4.0 (72B)

99.6

Gemini-3-Pro-Preview

97.2

Claude-Opus-4.5

43.6

Solar-Pro-2 (31B)(high)

40.8

HCX-007(high)

39.6

Grok-4.1-fast

34.8

EXAONE-4.0.1-32B (high)

6.3

Avg Token Usage

Claude-Opus-4.5

19293.0

Gemini-3-Pro-Preview

14285.0

Llama-VARCO-8B-Instruct

13418.0

Deepseek-V3.2

9555.0

GPT-5.1 (high)

9506.0

Grok-4.1-fast

9320.0

Solar-Pro-2 (31B)(high)

8677.0

HCX-007(high)

5746.0

A.X-4.0 (72B)

5358.0

EXAONE-4.0.1-32B (high)

5236.0