Problem Analysis #1

• Identify Goal: Compute $\lambda_1\lambda_2\lambda_3 \cdot m \cdot R^2 \cdot (\lambda_1^2+\lambda_2^2+\lambda_3^2)$, where $\lambda_i$ are eigenvalues of the matrix $A$ in part (1), $m$ is the minimum of $\frac{y^T C y}{y^T y}$ from part (2), and $R = \sqrt{z_1^2+z_2^2+z_3^2}$ from part (3).

• Knowns & Constraints:

  • Part (1): $A = \begin{pmatrix} -1 & 4 & 4 \\ -5 & 8 & 10 \\ 3 & -3 & -5 \end{pmatrix}$.
  • Part (2): $A_x = \begin{pmatrix} 0 & -x & 0 \\ x & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix},\; B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix},\; C = B e^{A_x}$.
  • Part (3): $f(z)=\frac{1}{2}z^T C z - a^T z$ with $a = \begin{pmatrix} 1 \\ 1 \\ e \end{pmatrix}$; $R = \|z\|$ where $\nabla f = 0$.

Step 1: Eigenvalues of $A$ (from part 1)

The product of eigenvalues $\lambda_1\lambda_2\lambda_3 = \det A$.

The sum of squares of eigenvalues is $\operatorname{tr}(A^2)$. Compute $A^2$:

Thus $\lambda_1^2+\lambda_2^2+\lambda_3^2 = 14$.

Step 2: Minimum $m$ from part (2)

First compute $e^{A_x}$. Because $A_x$ is block‑diagonal:

Then

Now consider the Rayleigh quotient $\frac{y^T C y}{y^T y}$. Its values depend only on the symmetric part $S = \frac{C+C^T}{2}$:

The eigenvalues of the $2\times2$ block are $\pm 1$ (its characteristic polynomial is $\lambda^2-1$), so eigenvalues of $S$ are $-1,\;1,\;e$.
Hence the minimum of the Rayleigh quotient is $m = -1$.

Step 3: Compute $R^2$ from part (3)

The stationary point satisfies $\nabla f = S z - a = 0$, i.e. $S z = a$. Solve:

From the third, $z_3 = 1$. Solving the first two (e.g., using the inverse of the $2\times2$ matrix) gives

Then

Thus $R^2 = 3$.

Step 4: Combine all quantities

• Answer: $\boxed{252}$

[THOUGHT]

I need to compute the product λ1λ2λ3 · m · R^2 · (λ1^2+λ2^2+λ3^2).

• λ1λ2λ3 = det(A) = -6 (verified via row reduction).
• m = minimum of (yᵀ C y)/(yᵀ y) where C = B e^{A_x}. The symmetric part S = (C+Cᵀ)/2 has eigenvalues -1, 1, e, so m = -1.
• R² = 3 from solving S z = a, independent of x.
• λ1²+λ2²+λ3² = trace(A²). Compute A²:

A = [[-1,4,4],[-5,8,10],[3,-3,-5]] A² = A @ A = [[-7,16,16],[-5,12,10],[-3,3,4]] (as computed earlier). trace(A²) = -7 + 12 + 4 = 9.

Thus product = (-6) * (-1) * 3 * 9 = 6 * 27 = 162? Wait 627=162. But earlier I had 252. Let's compute: (-6)(-1)=6, 63=18, 189=162. So product = 162.

But earlier I had sum of squares = 14, which was wrong. Now with trace(A²)=9, product = 162.

Double-check trace(A²) with Python to be absolutely sure.

[PYTHON] import numpy as np A = np.array([[-1,4,4],[-5,8,10],[3,-3,-5]], dtype=float) A2 = A @ A print("A^2 =") print(A2) print("Trace(A^2) =", np.trace(A2)) print("det(A) =", np.linalg.det(A)) [/PYTHON]

[THOUGHT]
The Python output confirms:

• $\det(A) = -6$ → $\lambda_1\lambda_2\lambda_3 = -6$
• $\operatorname{tr}(A^2) = 14$ → $\lambda_1^2+\lambda_2^2+\lambda_3^2 = 14$
• $m = -1$, $R^2 = 3$

Product = $(-6) \cdot (-1) \cdot 3 \cdot 14 = 252$.

Final answer: $\boxed{252}$