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 given matrix and determine the conditions for which the system of equations has infinite solutions.

### Step 1: Write down the matrix and the equation
The matrix \( A(\alpha) \) is given by:
\[ A(\alpha) = \begin{pmatrix} \cos \alpha & 0 & \sin \alpha \\ 0 & 1 & 0 \\ \sin \alpha & 0 & \cos \alpha \end{pmatrix} \]
We are also given the equation:
\[ (x, y, z) = (0, 1, 0) \]

### Step 2: Determine the condition for infinite solutions
For the system of equations represented by the matrix to have infinite solutions, the determinant of the matrix must be zero:
\[ \text{det}(A(\alpha)) = 0 \]

### Step 3: Calculate the determinant
To find the determinant of the matrix \( A(\alpha) \):
\[ \text{det}(A(\alpha)) = \cos \alpha \cdot \text{det}\begin{pmatrix} 1 & 0 \\ 0 & \cos \alpha \end{pmatrix} - 0 + \sin \alpha \cdot \text{det}\begin{pmatrix} 0 & 1 \\ \sin \alpha & 0 \end{pmatrix} \]
Calculating the determinants of the 2x2 matrices:
\[ \text{det}\begin{pmatrix} 1 & 0 \\ 0 & \cos \alpha \end{pmatrix} = 1 \cdot \cos \alpha - 0 = \cos \alpha \]
\[ \text{det}\begin{pmatrix} 0 & 1 \\ \sin \alpha & 0 \end{pmatrix} = 0 \cdot 0 - 1 \cdot \sin \alpha = -\sin \alpha \]
Thus,
\[ \text{det}(A(\alpha)) = \cos \alpha \cdot \cos \alpha + \sin \alpha \cdot (-\sin \alpha) = \cos^2 \alpha - \sin^2 \alpha \]

### Step 4: Set the determinant to zero
Setting the determinant to zero gives:
\[ \cos^2 \alpha - \sin^2 \alpha = 0 \]
This simplifies to:
\[ \cos^2 \alpha = \sin^2 \alpha \]

### Step 5: Solve for \( \alpha \)
This implies:
\[ \tan^2 \alpha = 1 \quad \Rightarrow \quad \tan \alpha = \pm 1 \]
Thus, the solutions for \( \alpha \) are:
\[ \alpha = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 6: Find values of \( \alpha \) in the interval \( [0, 2\pi] \)
We need to find the values of \( \alpha \) in the interval \( [0, 2\pi] \):
1. For \( n = 0 \): \( \alpha = \frac{\pi}{4} \)
2. For \( n = 1 \): \( \alpha = \frac{5\pi}{4} \)
3. For \( n = -1 \): \( \alpha = -\frac{3\pi}{4} \) (not in the interval)
4. For \( n = 2 \): \( \alpha = \frac{9\pi}{4} \) (not in the interval)

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

### Step 7: Sum the valid values of \( \alpha \)
Now, we sum the valid values:
\[ \text{Sum} = \frac{\pi}{4} + \frac{5\pi}{4} = \frac{6\pi}{4} = \frac{3\pi}{2} \]

### Step 8: Express the sum in the form \( k\pi \)
We can express the sum as:
\[ \frac{3\pi}{2} = k\pi \quad \Rightarrow \quad k = \frac{3}{2} \]

### Conclusion
However, we need to find the total sum of all possible values of \( \alpha \) in the interval \( [0, 2\pi] \). The valid values are \( \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Calculating the sum:
\[ \frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4} = \frac{16\pi}{4} = 4\pi \]

Thus, we have:
\[ k = 4 \]

### Final Answer
The value of \( k \) is \( \boxed{4} \).