If the system of equations
`x + (sqrt 2 sin alpha) y + (sqrt 2 cos alpha) z = 0`
`x + (cos alpha)y + (sin alpha)z =0`
`x + (sin alpha)y – (cos alpha)z = 0`
has a non-trivial solution, then `alpha in (0, pi/2)` is equal to

Options

A. `(7 pi)/(24)`
B. `(3 pi)/4`
C. `(5 pi)/(24)`
D. `(11 pi)/(24)`

Answer

To solve the given system of equations for the value of \( \alpha \) such that it has a non-trivial solution, we will follow these steps:

### Step 1: Write the system of equations in matrix form
The system of equations is:
1. \( x + \sqrt{2} \sin \alpha \cdot y + \sqrt{2} \cos \alpha \cdot z = 0 \)
2. \( x + \cos \alpha \cdot y + \sin \alpha \cdot z = 0 \)
3. \( x + \sin \alpha \cdot y - \cos \alpha \cdot z = 0 \)

We can express this in matrix form as:
\[ \begin{bmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Find the determinant of the coefficient matrix
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero. Let \( A \) be the coefficient matrix:
\[ A = \begin{bmatrix} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{bmatrix} \]

### Step 3: Calculate the determinant of matrix \( A \)
Using the determinant formula for a \( 3 \times 3 \) matrix, we have:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} - \sqrt{2} \sin \alpha \cdot \begin{vmatrix} 1 & \sin \alpha \\ 1 & -\cos \alpha \end{vmatrix} + \sqrt{2} \cos \alpha \cdot \begin{vmatrix} 1 & \cos \alpha \\ 1 & \sin \alpha \end{vmatrix} \]

Calculating these determinants:
1. \( \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -\cos^2 \alpha - \sin^2 \alpha = -1 \)

2. \( \begin{vmatrix} 1 & \sin \alpha \\ 1 & -\cos \alpha \end{vmatrix} = -\cos \alpha - \sin \alpha = -(\cos \alpha + \sin \alpha) \)

3. \( \begin{vmatrix} 1 & \cos \alpha \\ 1 & \sin \alpha \end{vmatrix} = \sin \alpha - \cos \alpha \)

Substituting these back into the determinant expression:
\[ \text{det}(A) = 1 \cdot (-1) - \sqrt{2} \sin \alpha \cdot (-(\cos \alpha + \sin \alpha)) + \sqrt{2} \cos \alpha \cdot (\sin \alpha - \cos \alpha) \]
\[ = -1 + \sqrt{2} \sin \alpha (\cos \alpha + \sin \alpha) + \sqrt{2} \cos \alpha (\sin \alpha - \cos \alpha) \]

### Step 4: Set the determinant to zero
Setting \( \text{det}(A) = 0 \):
\[ -1 + \sqrt{2} \sin \alpha (\cos \alpha + \sin \alpha) + \sqrt{2} \cos \alpha (\sin \alpha - \cos \alpha) = 0 \]

### Step 5: Simplify and solve for \( \alpha \)
After simplifying and rearranging, we can express this in terms of trigonometric identities. Eventually, we will find:
\[ \sqrt{2} \sin 2\alpha - \sqrt{2} \cos 2\alpha = 1 \]
This leads to:
\[ \tan(2\alpha) = 1 \quad \Rightarrow \quad 2\alpha = \frac{\pi}{4} + n\pi \]
For \( n = 0 \):
\[ 2\alpha = \frac{\pi}{4} \quad \Rightarrow \quad \alpha = \frac{\pi}{8} \]

### Step 6: Verify the range of \( \alpha \)
Since \( \alpha \) must be in the interval \( (0, \frac{\pi}{2}) \), \( \alpha = \frac{\pi}{8} \) is valid.

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{\pi}{8} \]