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 A
The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as \( ad - bc \). For our matrix \( A \):
\[ |A| = \cos \alpha \cdot \cos \alpha - (-\sin \alpha) \cdot \sin \alpha = \cos^2 \alpha + \sin^2 \alpha = 1 \]

### Step 3: Define the Matrix B
The matrix \( B \) is defined as:
\[ B = A + 3A^2 + 3A^3 + A^4 \]

### Step 4: Factor the Expression for B
Notice that we can factor the expression:
\[ B = A + 3A^2 + 3A^3 + A^4 = (I + A)^4 - I \]
where \( I \) is the identity matrix.

### Step 5: Calculate the Determinant of B
Using the property of determinants:
\[ |B| = |(I + A)^4 - I| = |(I + A)^4| \]
Thus, we can express the determinant as:
\[ |B| = |I + A|^4 \]

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

### Step 7: Set up the Equation
From the problem, we know:
\[ |B| = 8 \]
Thus:
\[ |I + A|^4 = 8 \]
Taking the fourth root:
\[ |I + A| = 2 \]
So:
\[ 2 + 2\cos \alpha = 2 \]
This simplifies to:
\[ 2\cos \alpha = 0 \implies \cos \alpha = 0 \]

### Step 8: Find Values of \( \alpha \)
The cosine function is zero at:
\[ \alpha = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]
We need to find the values of \( \alpha \) in the interval \( [0, 10\pi] \).

### Step 9: Determine the Number of Solutions
The general solutions for \( \alpha \) can be expressed as:
\[ \alpha = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}, \frac{13\pi}{2}, \frac{15\pi}{2}, \frac{17\pi}{2}, \frac{19\pi}{2} \]
Counting these, we find:
- \( n = 0 \): \( \frac{\pi}{2} \)
- \( n = 1 \): \( \frac{3\pi}{2} \)
- \( n = 2 \): \( \frac{5\pi}{2} \)
- \( n = 3 \): \( \frac{7\pi}{2} \)
- \( n = 4 \): \( \frac{9\pi}{2} \)
- \( n = 5 \): \( \frac{11\pi}{2} \)
- \( n = 6 \): \( \frac{13\pi}{2} \)
- \( n = 7 \): \( \frac{15\pi}{2} \)
- \( n = 8 \): \( \frac{17\pi}{2} \)
- \( n = 9 \): \( \frac{19\pi}{2} \)

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