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 correct ?

Options

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

Answer

To solve the problem, we need to calculate 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 a column operation to simplify the determinant. Specifically, we will replace the first column \( C_1 \) with \( C_1 + C_3 \).

\[ C_1 \rightarrow C_1 + C_3 \]

This gives us:

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

Simplifying this, we have:

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

### Step 2: Calculate the Determinant
Now we will calculate the determinant of the modified matrix:

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

We can expand this determinant using the first row:

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

### Step 3: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:

1. For the first determinant:

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

2. For the second determinant:

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

3. For the third determinant:

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

### Step 4: Substitute Back
Now substituting these back into the determinant expression:

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

Expanding this gives:

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

Combining like terms:

\[ \text{det}(A) = 4 + 2\sin^2 \theta + \sin \theta \]

### Step 5: Determine the Range of the Determinant
Since \( \sin^2 \theta \) lies between 0 and 1, we can find the range of \( \text{det}(A) \):

- When \( \sin^2 \theta = 0 \):
\[ \text{det}(A) = 4 + 0 + 0 = 4 \]

- When \( \sin^2 \theta = 1 \):
\[ \text{det}(A) = 4 + 2(1) + 1 = 7 \]

Thus, the determinant \( \text{det}(A) \) lies between 4 and 7.

### Conclusion
The range of the determinant is \( 4 \leq \text{det}(A) < 7 \).