Let `A = [(1,sin theta,1),(- sin theta,1,sin theta),(-1,-sin theta,1)], " where " 0 le theta lt 2 pi`. then, which of the following is not correct ?

Options

A. `D = 0`
B. `D in (0, oo)`
C. `D in [2, 4]`
D. `D in [2, oo)`

Explanation

To solve the problem, we need to find the determinant of the matrix \( A \) given by:

\[ A = \begin{pmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{pmatrix} \]

### Step 1: Apply Column Operation
We will perform the column operation \( C_1 \leftarrow C_1 + C_3 \). This means we will add the first column to the third column.

After performing this operation, the matrix \( A \) becomes:

\[ A = \begin{pmatrix} 2 & \sin \theta & 1 \\ 0 & 1 & \sin \theta \\ 0 & -\sin \theta & 1 \end{pmatrix} \]

### Step 2: Calculate the Determinant
Next, we will calculate the determinant of the modified matrix \( A \) using the first column for expansion:

\[ \text{det}(A) = 2 \cdot \begin{vmatrix} 1 & \sin \theta \\ -\sin \theta & 1 \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinant
Now we compute the 2x2 determinant:

\[ \begin{vmatrix} 1 & \sin \theta \\ -\sin \theta & 1 \end{vmatrix} = (1)(1) - (-\sin \theta)(\sin \theta) = 1 + \sin^2 \theta \]

### Step 4: Substitute Back
Substituting back into the determinant expression, we have:

\[ \text{det}(A) = 2(1 + \sin^2 \theta) \]

### Step 5: Analyze the Result
Since \( \sin^2 \theta \) lies between 0 and 1 for \( 0 \leq \theta < 2\pi \), we can analyze the bounds:

- Minimum value: When \( \sin^2 \theta = 0 \), \( \text{det}(A) = 2(1 + 0) = 2 \).
- Maximum value: When \( \sin^2 \theta = 1 \), \( \text{det}(A) = 2(1 + 1) = 4 \).

Thus, we conclude that:

\[ 2 \leq \text{det}(A) \leq 4 \]

### Final Conclusion
From the analysis, we can determine which statements regarding the determinant are correct or not. The statement that is not correct would be any that suggests the determinant falls outside the range \( [2, 4] \).