Let `Delta=|(1,sinalpha,1),(-sinalpha,1,sinalpha),(-1,-sinalpha,1)|` then `Delta` lies in the interval

Options

A. [2,3]
B. [3,4]
C. [1,4]
D. [2,4]

Answer

To find the value of the determinant \( \Delta = \begin{vmatrix} 1 & \sin \alpha & 1 \\ -\sin \alpha & 1 & \sin \alpha \\ -1 & -\sin \alpha & 1 \end{vmatrix} \), we will expand it and simplify step by step.

### Step 1: Expand the Determinant
We will expand the determinant along the first row (R1):

\[ \Delta = 1 \cdot \begin{vmatrix} 1 & \sin \alpha \\ -\sin \alpha & 1 \end{vmatrix} - \sin \alpha \cdot \begin{vmatrix} -\sin \alpha & \sin \alpha \\ -1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -\sin \alpha & 1 \\ -1 & -\sin \alpha \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants
Now, we will calculate each of the 2x2 determinants:

1. For the first determinant:
\[ \begin{vmatrix} 1 & \sin \alpha \\ -\sin \alpha & 1 \end{vmatrix} = (1)(1) - (-\sin \alpha)(\sin \alpha) = 1 + \sin^2 \alpha \]

2. For the second determinant:
\[ \begin{vmatrix} -\sin \alpha & \sin \alpha \\ -1 & 1 \end{vmatrix} = (-\sin \alpha)(1) - (\sin \alpha)(-1) = -\sin \alpha + \sin \alpha = 0 \]

3. For the third determinant:
\[ \begin{vmatrix} -\sin \alpha & 1 \\ -1 & -\sin \alpha \end{vmatrix} = (-\sin \alpha)(-\sin \alpha) - (1)(-1) = \sin^2 \alpha + 1 \]

### Step 3: Substitute Back into the Determinant
Now substitute these results back into the expression for \( \Delta \):

\[ \Delta = 1 \cdot (1 + \sin^2 \alpha) - \sin \alpha \cdot 0 + 1 \cdot (\sin^2 \alpha + 1) \]

This simplifies to:

\[ \Delta = 1 + \sin^2 \alpha + \sin^2 \alpha + 1 = 2 + 2\sin^2 \alpha \]

### Step 4: Determine the Range of \( \Delta \)
Now we need to find the range of \( \Delta = 2 + 2\sin^2 \alpha \).

Since \( \sin^2 \alpha \) varies from 0 to 1, we can find the minimum and maximum values of \( \Delta \):

- **Minimum value**: When \( \sin^2 \alpha = 0 \):
\[ \Delta_{\text{min}} = 2 + 2 \cdot 0 = 2 \]

- **Maximum value**: When \( \sin^2 \alpha = 1 \):
\[ \Delta_{\text{max}} = 2 + 2 \cdot 1 = 4 \]

Thus, \( \Delta \) lies in the interval \( [2, 4] \).

### Final Answer
The determinant \( \Delta \) lies in the interval \( [2, 4] \).

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