Let A be an involutary matrix and S be the set containing solution of `A^(x)=1` where `A = [[0,1,-1],[4,-3,4],[3,-3,4]]` Then minimum value of `sum_(x in S )^(oo) (sin^(x)theta+cos^(x)theta)`

Answer

To solve the problem step by step, we will first confirm that the matrix \( A \) is involutory, then find the set \( S \) containing solutions of \( A^x = I \), and finally compute the minimum value of the summation \( \sum_{x \in S} (\sin^x \theta + \cos^x \theta) \).

### Step 1: Verify that \( A \) is an involutory matrix
An involutory matrix satisfies the condition \( A^2 = I \), where \( I \) is the identity matrix.

Given:
\[ A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \]

We need to calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\ 3 & -3 & 4 \end{bmatrix} \]

Calculating each element of \( A^2 \):
- First row:
- \( (0 \cdot 0 + 1 \cdot 4 + (-1) \cdot 3) = 4 - 3 = 1 \)
- \( (0 \cdot 1 + 1 \cdot (-3) + (-1) \cdot (-3)) = -3 + 3 = 0 \)
- \( (0 \cdot (-1) + 1 \cdot 4 + (-1) \cdot 4) = 4 - 4 = 0 \)

- Second row:
- \( (4 \cdot 0 + (-3) \cdot 4 + 4 \cdot 3) = -12 + 12 = 0 \)
- \( (4 \cdot 1 + (-3) \cdot (-3) + 4 \cdot (-3)) = 4 + 9 - 12 = 1 \)
- \( (4 \cdot (-1) + (-3) \cdot 4 + 4 \cdot 4) = -4 - 12 + 16 = 0 \)

- Third row:
- \( (3 \cdot 0 + (-3) \cdot 4 + 4 \cdot 3) = -12 + 12 = 0 \)
- \( (3 \cdot 1 + (-3) \cdot (-3) + 4 \cdot (-3)) = 3 + 9 - 12 = 0 \)
- \( (3 \cdot (-1) + (-3) \cdot 4 + 4 \cdot 4) = -3 - 12 + 16 = 1 \)

Thus, we have:
\[ A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \]

### Step 2: Identify the set \( S \)
Since \( A^2 = I \), it follows that \( A^x = I \) for even integers \( x \). Thus, the set \( S \) is:
\[ S = \{2, 4, 6, 8, \ldots\} \]

### Step 3: Calculate the minimum value of the summation
We need to find:
\[ \sum_{x \in S} (\sin^x \theta + \cos^x \theta) \]

This can be split into two separate series:
\[ \sum_{n=1}^{\infty} \sin^{2n} \theta + \sum_{n=1}^{\infty} \cos^{2n} \theta \]

Using the formula for the sum of a geometric series:
\[ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} \quad \text{for } |r| < 1 \]

For \( \sin^{2n} \theta \):
\[ \sum_{n=1}^{\infty} \sin^{2n} \theta = \sin^2 \theta \sum_{n=0}^{\infty} (\sin^2 \theta)^n = \frac{\sin^2 \theta}{1 - \sin^2 \theta} = \frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta \]

For \( \cos^{2n} \theta \):
\[ \sum_{n=1}^{\infty} \cos^{2n} \theta = \cos^2 \theta \sum_{n=0}^{\infty} (\cos^2 \theta)^n = \frac{\cos^2 \theta}{1 - \cos^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} = \cot^2 \theta \]

Thus, the total summation is:
\[ \tan^2 \theta + \cot^2 \theta \]

Using the identity:
\[ \tan^2 \theta + \cot^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 \theta \cos^2 \theta} \]

Using the identity \( \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \):
\[ \tan^2 \theta + \cot^2 \theta = \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} \]

The minimum value occurs when \( \sin^2 \theta = \cos^2 \theta = \frac{1}{2} \):
\[ \tan^2 \theta + \cot^2 \theta = 2 \]

### Final Result
Thus, the minimum value of the summation is:
\[ \boxed{2} \]