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

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

We will compute the determinant \( \text{det}(A) \) and then analyze its values for \( \theta \) in the interval \( \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \).

### Step 1: Calculate the determinant of matrix \( A \)

Using the formula for the determinant of a \( 3 \times 3 \) matrix, we have:

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

where the matrix is represented as:

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

For our matrix \( A \):

- \( a = 1, b = \sin \theta, c = 1 \)
- \( d = -\sin \theta, e = 1, f = \sin \theta \)
- \( g = -1, h = -\sin \theta, i = 1 \)

Substituting these values into the determinant formula:

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

### Step 2: Simplify the determinant expression

Calculating each term:

1. The first term:
\[ 1(1 + \sin^2 \theta) = 1 + \sin^2 \theta \]

2. The second term:
\[ -\sin \theta(-\sin \theta + \sin \theta) = 0 \]

3. The third term:
\[ 1(\sin^2 \theta - 1) = \sin^2 \theta - 1 \]

Combining these results:

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

### Step 3: Analyze \( \text{det}(A) \) for \( \theta \in \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \)

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

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

In the interval \( \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \), \( \sin \theta \) varies from \( -\frac{1}{\sqrt{2}} \) to \( -\frac{1}{\sqrt{2}} \) and achieves its maximum at \( \theta = \pi \) where \( \sin \theta = 0 \).

Thus, \( \sin^2 \theta \) varies from \( 0 \) to \( \frac{1}{2} \).

### Step 4: Determine the range of \( \text{det}(A) \)

Since \( \text{det}(A) = 2\sin^2 \theta \):

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

Thus, the determinant \( \text{det}(A) \) lies in the interval \( [0, 1] \).

### Final Answer

The determinant \( \text{det}(A) \) lies in the interval \( [0, 1] \).

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