Let `A=[(cos alpha,sin alpha),(-sinalpha,cosalpha)]` and matrix B is defined such that `B=A+3A^(2)+3A^(3)+A^(4).` If `|B|=8` then the number of values of `alpha` in `[0, 10pi]` is

Options

A. 10
B. 12
C. 5
D. 3

Answer

To solve the problem, we need to analyze the given matrices and their determinants step by step.

### Step 1: Define the Matrix A
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \]

### Step 2: Calculate the Determinant of Matrix A
The determinant of matrix \( A \) can be calculated using the formula for the determinant of a 2x2 matrix:
\[ |A| = \cos \alpha \cdot \cos \alpha - (-\sin \alpha) \cdot \sin \alpha = \cos^2 \alpha + \sin^2 \alpha = 1 \]

### Step 3: Define the Matrix B
The matrix \( B \) is defined as:
\[ B = A + 3A^2 + 3A^3 + A^4 \]
We can factor out \( A \) from this expression:
\[ B = A (1 + 3A + 3A^2 + A^3) \]

### Step 4: Recognize the Polynomial
The expression \( 1 + 3A + 3A^2 + A^3 \) resembles the expansion of \( (A + I)^3 \) where \( I \) is the identity matrix. Thus:
\[ B = A (I + A)^3 \]

### Step 5: Calculate the Determinant of Matrix B
Using the property of determinants:
\[ |B| = |A| \cdot |(I + A)^3| \]
Since we already found that \( |A| = 1 \), we have:
\[ |B| = |(I + A)^3| \]
This simplifies to:
\[ |B| = |I + A|^3 \]

### Step 6: Calculate the Determinant of \( I + A \)
Now we calculate \( I + A \):
\[ I + A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} = \begin{pmatrix} 1 + \cos \alpha & \sin \alpha \\ -\sin \alpha & 1 + \cos \alpha \end{pmatrix} \]
The determinant of \( I + A \) is:
\[ |I + A| = (1 + \cos \alpha)(1 + \cos \alpha) - (-\sin \alpha)(\sin \alpha) = (1 + \cos \alpha)^2 + \sin^2 \alpha \]
Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ |I + A| = (1 + \cos \alpha)^2 + (1 - \cos^2 \alpha) = 2 + 2\cos \alpha \]

### Step 7: Set Up the Equation
Now we have:
\[ |B| = |I + A|^3 = (2 + 2\cos \alpha)^3 \]
Given that \( |B| = 8 \):
\[ (2 + 2\cos \alpha)^3 = 8 \]
Taking the cube root of both sides:
\[ 2 + 2\cos \alpha = 2 \]
This simplifies to:
\[ 2\cos \alpha = 0 \implies \cos \alpha = 0 \]

### Step 8: Find Values of \( \alpha \)
The cosine function equals zero at:
\[ \alpha = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
In the interval \( [0, 10\pi] \), we need to find the integer values of \( n \):
- For \( n = 0 \): \( \alpha = \frac{\pi}{2} \)
- For \( n = 1 \): \( \alpha = \frac{3\pi}{2} \)
- For \( n = 2 \): \( \alpha = \frac{5\pi}{2} \)
- For \( n = 3 \): \( \alpha = \frac{7\pi}{2} \)
- For \( n = 4 \): \( \alpha = \frac{9\pi}{2} \)
- For \( n = 5 \): \( \alpha = \frac{11\pi}{2} \)
- For \( n = 6 \): \( \alpha = \frac{13\pi}{2} \)
- For \( n = 7 \): \( \alpha = \frac{15\pi}{2} \)
- For \( n = 8 \): \( \alpha = \frac{17\pi}{2} \)
- For \( n = 9 \): \( \alpha = \frac{19\pi}{2} \)

Counting these values gives us a total of 10 values of \( \alpha \) in the interval \( [0, 10\pi] \).

### Final Answer
The number of values of \( \alpha \) in the interval \( [0, 10\pi] \) is **10**.