If `A = [(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1)]`, then for all
`thetain ((3pi)/(4),(5pi)/(4))`, det. (A) lies in the interval

Options

A. `[(5)/(2),4)`
B. `((3)/(2),3]`
C. `(0,(3)/(2)]`
D. `(1,(5)/(2)]`

Answer

To find the determinant of the matrix \( A = \begin{pmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{pmatrix} \) for \( \theta \) in the interval \( \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right) \), we will follow these steps:

### Step 1: Calculate the Determinant of \( A \)

We will use the determinant formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

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

Let’s denote the elements as follows:
- \( a = 1, b = \sin \theta, c = 1 \)
- \( d = -\sin \theta, e = 1, f = \sin \theta \)
- \( g = -1, h = -\sin \theta, i = 1 \)

Now, substituting these values into the determinant formula:

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

Calculating each term:

1. \( 1 \cdot (1 + \sin^2 \theta) = 1 + \sin^2 \theta \)
2. \( -\sin \theta \cdot (-\sin \theta + \sin \theta) = 0 \)
3. \( 1 \cdot (\sin^2 \theta - 1) = \sin^2 \theta - 1 \)

Putting it all together:

\[ \text{det}(A) = (1 + \sin^2 \theta) + 0 + (\sin^2 \theta - 1) = 2\sin^2 \theta \]

### Step 2: Determine the Range of \( \sin^2 \theta \)

Now, we need to find the range of \( \sin^2 \theta \) for \( \theta \) in the interval \( \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right) \).

1. At \( \theta = \frac{3\pi}{4} \), \( \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \) so \( \sin^2 \frac{3\pi}{4} = \frac{1}{2} \).
2. At \( \theta = \frac{5\pi}{4} \), \( \sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2} \) so \( \sin^2 \frac{5\pi}{4} = \frac{1}{2} \).

Since \( \sin \theta \) is negative in the third quadrant, \( \sin^2 \theta \) will take values from \( 0 \) to \( \frac{1}{2} \) in this interval.

### Step 3: Calculate the Range of \( 2\sin^2 \theta \)

Now, multiplying the range of \( \sin^2 \theta \) by 2:

- Minimum: \( 2 \cdot 0 = 0 \)
- Maximum: \( 2 \cdot \frac{1}{2} = 1 \)

Thus, the determinant \( \text{det}(A) = 2\sin^2 \theta \) lies in the interval \( (0, 1) \).

### Conclusion

The determinant of the matrix \( A \) lies in the interval \( (0, 1) \).