Statement I The system of linear equations
`x+(sin alpha )y+(cos alpha )z=0`
`x+(cos alpha ) y+(sin alpha )z=0`
`-x+(sin alpha )y-(cos alpha )z=0`
has a not trivial solution for only one value of ` alpha ` lying between ` 0 and pi`.
Statement II `|{:(sin x, cos x , cos x),( cos x , sin x , cos x) , (cos x , cos x , sin x ):}|=0`
has no solution in the interval `-pi//4 lt x lt pi//4 ` .

Options

A. Statement I is true , Statement II is true , Statement II is a correct explanation for Statement I.
B. Statement I is true , Statement II is true , Statement II is not a correct explanation for Statement II.
C. Statement I is true , Statement II is false
D. Statement I is false , Statement II true .

Answer

To solve the given problem, we need to analyze both statements step by step.

### Statement I
We have the system of linear equations:
1. \( x + (\sin \alpha) y + (\cos \alpha) z = 0 \)
2. \( x + (\cos \alpha) y + (\sin \alpha) z = 0 \)
3. \( -x + (\sin \alpha) y - (\cos \alpha) z = 0 \)

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

For this system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

### Step 1: Calculate the Determinant
The determinant of the matrix is given by:
\[ D = \begin{vmatrix} 1 & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ -1 & \sin \alpha & -\cos \alpha \end{vmatrix} \]

Calculating the determinant using the first row:
\[ D = 1 \cdot \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} - \sin \alpha \cdot \begin{vmatrix} 1 & \sin \alpha \\ -1 & -\cos \alpha \end{vmatrix} + \cos \alpha \cdot \begin{vmatrix} 1 & \cos \alpha \\ -1 & \sin \alpha \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} \cos \alpha & \sin \alpha \\ \sin \alpha & -\cos \alpha \end{vmatrix} = -\cos^2 \alpha - \sin^2 \alpha = -1 \)
2. \( \begin{vmatrix} 1 & \sin \alpha \\ -1 & -\cos \alpha \end{vmatrix} = -\cos \alpha + \sin \alpha = \sin \alpha - \cos \alpha \)
3. \( \begin{vmatrix} 1 & \cos \alpha \\ -1 & \sin \alpha \end{vmatrix} = \sin \alpha + \cos \alpha \)

Putting it all together:
\[ D = 1(-1) - \sin \alpha (\sin \alpha - \cos \alpha) + \cos \alpha (\sin \alpha + \cos \alpha) \]
\[ D = -1 - \sin^2 \alpha + \sin \alpha \cos \alpha + \cos \alpha \sin \alpha + \cos^2 \alpha \]
\[ D = -1 + \cos^2 \alpha + \sin^2 \alpha + 2\sin \alpha \cos \alpha \]
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ D = -1 + 1 + 2\sin \alpha \cos \alpha = 2\sin \alpha \cos \alpha = \sin(2\alpha) \]

### Step 2: Set the Determinant to Zero
For a non-trivial solution, we need:
\[ \sin(2\alpha) = 0 \]
This implies:
\[ 2\alpha = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ \alpha = \frac{n\pi}{2} \]

### Step 3: Determine Valid Values of \( \alpha \)
We need \( \alpha \) to be in the interval \( (0, \pi) \):
- For \( n = 1 \): \( \alpha = \frac{\pi}{2} \) (valid)
- For \( n = 0 \): \( \alpha = 0 \) (not valid)
- For \( n = 2 \): \( \alpha = \pi \) (not valid)

Thus, the only value of \( \alpha \) that gives a non-trivial solution in the interval \( (0, \pi) \) is \( \frac{\pi}{2} \).

### Conclusion for Statement I
Statement I is true.

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### Statement II
We need to analyze the determinant:
\[ \begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0 \]

### Step 1: Calculate the Determinant
Using similar steps as before:
\[ D = \sin x \begin{vmatrix} \sin x & \cos x \\ \cos x & \sin x \end{vmatrix} - \cos x \begin{vmatrix} \cos x & \cos x \\ \cos x & \sin x \end{vmatrix} + \cos x \begin{vmatrix} \cos x & \sin x \\ \cos x & \cos x \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} \sin x & \cos x \\ \cos x & \sin x \end{vmatrix} = \sin^2 x - \cos^2 x = \sin^2 x - \cos^2 x \)
2. \( \begin{vmatrix} \cos x & \cos x \\ \cos x & \sin x \end{vmatrix} = \cos x (\sin x - \cos x) = \cos x (\sin x - \cos x) \)
3. \( \begin{vmatrix} \cos x & \sin x \\ \cos x & \cos x \end{vmatrix} = \cos^2 x - \sin x \cos x = \cos^2 x - \sin x \cos x \)

Putting it all together:
\[ D = \sin x (\sin^2 x - \cos^2 x) - \cos x (\cos x (\sin x - \cos x)) + \cos x (\cos^2 x - \sin x \cos x) \]

### Step 2: Set the Determinant to Zero
We need to analyze when this determinant equals zero. However, we know from the problem statement that there are no solutions in the interval \( -\frac{\pi}{4} < x < \frac{\pi}{4} \).

### Conclusion for Statement II
Statement II is also true.

### Final Conclusion
Both statements are true, but Statement II does not correctly explain Statement I.

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