To solve this problem, we need to address each part of the problem step-by-step.

Part (1): Compute the matrix exponential ( e^{A} )

First, we find the eigenvalues of matrix ( A ):

[ A = \begin{pmatrix} -1 & 4 & 4 \ -5 & 8 & 10 \ 3 & -3 & -5 \end{pmatrix}. ]

We use the characteristic equation ( \det(A - \lambda E) = 0 ) to find the eigenvalues.

[THOUGHT] I need to calculate the eigenvalues of matrix ( A ). I will use the sympy library to find the eigenvalues. [/THOUGHT] [PYTHON] from sympy import Matrix, symbols, solve

Define the matrix A

A = Matrix([ [-1, 4, 4], [-5, 8, 10], [3, -3, -5] ])

Define the symbol for eigenvalues

lambda_ = symbols('lambda')

Calculate the characteristic polynomial

char_poly = A.charpoly(lambda_)

Solve for the eigenvalues

eigenvalues = solve(char_poly, lambda_) print(eigenvalues)

[THOUGHT] The eigenvalues of matrix ( A ) are ( \lambda_1 = -2 ), ( \lambda_2 = 1 ), and ( \lambda_3 = 3 ). We will use these eigenvalues in subsequent parts of the problem. [/THOUGHT]

Part (2): Determine the minimum and maximum values of ( \frac{y^{T} C y}{y^{T}y} )

First, we compute the matrix ( e^{A_x} ).

The matrix ( A_x ) is:

[ A_x = \begin{pmatrix} 0 & -x & 0 \ x & 0 & 0 \ 0 & 0 & 1 \end{pmatrix}. ]

This matrix can be split into two parts:

[ A_x = \begin{pmatrix} 0 & -x & 0 \ x & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix} = X + E_{33}. ]

The exponential of ( X ) (the upper-left 2x2 block) is a rotation matrix:

[ e^X = \begin{pmatrix} \cos(x) & -\sin(x) \ \sin(x) & \cos(x) \end{pmatrix}. ]

Thus, the matrix exponential ( e^{A_x} ) is:

[ e^{A_x} = \begin{pmatrix} \cos(x) & -\sin(x) & 0 \ \sin(x) & \cos(x) & 0 \ 0 & 0 & e^1 \end{pmatrix} = \begin{pmatrix} \cos(x) & -\sin(x) & 0 \ \sin(x) & \cos(x) & 0 \ 0 & 0 & e \end{pmatrix}. ]

Now, we compute ( C = B e^{A_x} ):

[ B = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix}. ]

[ C = B e^{A_x} = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos(x) & -\sin(x) & 0 \ \sin(x) & \cos(x) & 0 \ 0 & 0 & e \end{pmatrix} = \begin{pmatrix} \cos(x) & \sin(x) & 0 \ -\sin(x) & -\cos(x) & 0 \ 0 & 0 & e \end{pmatrix}. ]

The Rayleigh quotient ( \frac{y^{T} C y}{y^{T} y} ) for matrix ( C ) is maximized and minimized by the eigenvalues of ( C ). The eigenvalues of ( C ) are ( \cos(x) \pm i\sin(x) ) and ( e ).

The real parts of ( \cos(x) \pm i\sin(x) ) are ( \cos(x) ), and the eigenvalue ( e ) is real. Therefore, the minimum and maximum values are:

[ m = \min(\cos(x), e) = -1 \quad \text{(when ( x = \pi ))} ] [ M = \max(\cos(x), e) = e \quad \text{(when ( x = 0 ))} ]

Part (3): Find ( R ) for the vector ( z ) satisfying the partial derivatives of ( f(z) )

The function ( f(z) ) is given by:

[ f(z) = \frac{1}{2} z^T C z - a^T z. ]

[ C = \begin{pmatrix} \cos(x) & \sin(x) & 0 \ -\sin(x) & -\cos(x) & 0 \ 0 & 0 & e \end{pmatrix}, \quad a = \begin{pmatrix} 1 \ 1 \ e \end{pmatrix}. ]

The gradient of ( f ) is:

[ \nabla f(z) = Cz - a. ]

Setting the gradient to zero:

[ Cz = a. ]

We solve this system for ( z ). Let's assume ( x = 0 ) (which gives the maximum eigenvalue ( e )) for simplicity:

[ C = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & e \end{pmatrix}. ]

[ Cz = \begin{pmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & e \end{pmatrix} \begin{pmatrix} z_1 \ z_2 \ z_3 \end{pmatrix} = \begin{pmatrix} z_1 \ -z_2 \ ez_3 \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ e \end{pmatrix}. ]

Solving this system:

[ z_1 = 1, \quad -z_2 = 1 \implies z_2 = -1, \quad ez_3 = e \implies z_3 = 1. ]

Thus, the vector ( z ) is:

[ z = \begin{pmatrix} 1 \ -1 \ 1 \end{pmatrix}. ]

The norm ( R ) is:

[ R = \sqrt{z_1^2 + z_2^2 + z_3^2} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}. ]

Final Calculation

Now, we compute the final expression:

[ \lambda_1 \lambda_2 \lambda_3 \cdot m \cdot R^2 \cdot (\lambda_1^2 + \lambda_2^2 + \lambda_3^2). ]

We have:

[ \lambda_1 = -2, \quad \lambda_2 = 1, \quad \lambda_3 = 3, ] [ m = -1, ] [ R^2 = 3, ] [ \lambda_1^2 + \lambda_2^2 + \lambda_3^2 = (-2)^2 + 1^2 + 3^2 = 4 + 1 + 9 = 14. ]

Thus:

[ \lambda_1 \lambda_2 \lambda_3 \cdot m \cdot R^2 \cdot (\lambda_1^2 + \lambda_2^2 + \lambda_3^2) = (-2) \cdot 1 \cdot 3 \cdot (-1) \cdot 3 \cdot 14 = 252. ]

The final answer is:

[ \boxed{252} ]