Let `A(alpha)=[(cos alpha, 0,sin alpha),(0,1,0),(sin alpha, 0, cos alpha)] and [(x,y,z)]=[(0,1,0)]`. If the system of equations has infinite solutions and sum of all the possible value of `alpha` in `[0, 2pi]` is `kpi`, then the value of k is equal to

Options

A. `0`
B. `2`
C. `4`
D. `8`

Answer

To solve the problem step by step, we need to analyze the matrix \( A(\alpha) \) and the condition for the system of equations to have infinitely many solutions.

### Step 1: Set up the matrix and the condition for infinite solutions
Given the matrix:
\[ A(\alpha) = \begin{pmatrix} \cos \alpha & 0 & \sin \alpha \\ 0 & 1 & 0 \\ \sin \alpha & 0 & \cos \alpha \end{pmatrix} \]
For the system of equations to have infinitely many solutions, the determinant of \( A(\alpha) \) must be zero:
\[ \text{det}(A(\alpha)) = 0 \]

### Step 2: Calculate the determinant
We can calculate the determinant of the matrix \( A(\alpha) \):
\[ \text{det}(A(\alpha)) = \cos \alpha \cdot \begin{vmatrix} 1 & 0 \\ 0 & \cos \alpha \end{vmatrix} - 0 + \sin \alpha \cdot \begin{vmatrix} 0 & 1 \\ \sin \alpha & 0 \end{vmatrix} \]
Calculating the 2x2 determinants:
\[ = \cos \alpha (1 \cdot \cos \alpha - 0) + \sin \alpha (0 - \sin \alpha) \]
\[ = \cos^2 \alpha - \sin^2 \alpha \]
Using the identity \( \cos^2 \alpha - \sin^2 \alpha = \cos(2\alpha) \):
\[ \text{det}(A(\alpha)) = \cos(2\alpha) \]

### Step 3: Set the determinant to zero
To find the values of \( \alpha \) for which the determinant is zero:
\[ \cos(2\alpha) = 0 \]
The solutions to this equation occur at:
\[ 2\alpha = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ \alpha = \frac{\pi}{4} + \frac{n\pi}{2} \]

### Step 4: Find all possible values of \( \alpha \) in the interval \([0, 2\pi]\)
We can find the values of \( \alpha \) by substituting integer values for \( n \):
- For \( n = 0 \):
\[ \alpha = \frac{\pi}{4} \]
- For \( n = 1 \):
\[ \alpha = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \]
- For \( n = 2 \):
\[ \alpha = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]
- For \( n = 3 \):
\[ \alpha = \frac{\pi}{4} + \frac{3\pi}{2} = \frac{7\pi}{4} \]
- For \( n = 4 \):
\[ \alpha = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} \quad (\text{not in } [0, 2\pi]) \]

Thus, the valid values of \( \alpha \) in the interval \([0, 2\pi]\) are:
\[ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \]

### Step 5: Sum all possible values of \( \alpha \)
Now we sum these values:
\[ \text{Sum} = \frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi \]

### Step 6: Relate the sum to \( k\pi \)
We are given that this sum equals \( k\pi \):
\[ 4\pi = k\pi \implies k = 4 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{4} \]