If `A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)]` and `B=[(cos^(2)beta, cos beta sin beta),(cos beta sin beta, sin^(2) beta)]` are two matrices such that the product AB is null matrix, then `alpha-beta` is

Options

A. 0
B. multiple of `pi`
C. an odd multiple of `pi//2`
D. None of the above

Explanation

To solve the problem, we need to find the value of \( \alpha - \beta \) given that the product of matrices \( A \) and \( B \) is a null matrix.

Given:
\[ A = \begin{pmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{pmatrix} \]
\[ B = \begin{pmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{pmatrix} \]

We need to compute the product \( AB \) and set it equal to the null matrix \( 0 \).

### Step 1: Compute the product \( AB \)

The product \( AB \) is calculated as follows:

\[ AB = \begin{pmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{pmatrix} \begin{pmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{pmatrix} \]

Calculating the elements of the resulting matrix:

1. First element:
\[ \cos^2 \alpha \cdot \cos^2 \beta + \cos \alpha \sin \alpha \cdot \cos \beta \sin \beta \]

2. Second element:
\[ \cos^2 \alpha \cdot \cos \beta \sin \beta + \cos \alpha \sin \alpha \cdot \sin^2 \beta \]

3. Third element:
\[ \cos \alpha \sin \alpha \cdot \cos^2 \beta + \sin^2 \alpha \cdot \cos \beta \sin \beta \]

4. Fourth element:
\[ \cos \alpha \sin \alpha \cdot \cos \beta \sin \beta + \sin^2 \alpha \cdot \sin^2 \beta \]

Thus, we have:
\[ AB = \begin{pmatrix} \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta & \cos^2 \alpha \cos \beta \sin \beta + \cos \alpha \sin \alpha \sin^2 \beta \\ \cos \alpha \sin \alpha \cos^2 \beta + \sin^2 \alpha \cos \beta \sin \beta & \cos \alpha \sin \alpha \cos \beta \sin \beta + \sin^2 \alpha \sin^2 \beta \end{pmatrix} \]

### Step 2: Set \( AB = 0 \)

Since \( AB \) is a null matrix, we set each element to zero:

1. \( \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta = 0 \)
2. \( \cos^2 \alpha \cos \beta \sin \beta + \cos \alpha \sin \alpha \sin^2 \beta = 0 \)
3. \( \cos \alpha \sin \alpha \cos^2 \beta + \sin^2 \alpha \cos \beta \sin \beta = 0 \)
4. \( \cos \alpha \sin \alpha \cos \beta \sin \beta + \sin^2 \alpha \sin^2 \beta = 0 \)

### Step 3: Factor out common terms

From the first equation, we can factor out \( \cos \beta \):

\[ \cos \beta (\cos^2 \alpha \cos \beta + \sin \alpha \sin \beta) = 0 \]

This gives us two cases:
1. \( \cos \beta = 0 \)
2. \( \cos^2 \alpha \cos \beta + \sin \alpha \sin \beta = 0 \)

### Step 4: Solve for \( \alpha - \beta \)

If \( \cos \beta = 0 \), then \( \beta = \frac{\pi}{2} + n\pi \) for some integer \( n \).

Now, consider the second case:
\[ \cos^2 \alpha \cos \beta + \sin \alpha \sin \beta = 0 \]

This can be rewritten using the cosine of the difference formula:
\[ \cos(\alpha - \beta) = 0 \]

Thus, \( \alpha - \beta = \frac{\pi}{2} + k\pi \) for some integer \( k \).

### Final Result

The general solution for \( \alpha - \beta \) is:
\[ \alpha - \beta = (2n + 1)\frac{\pi}{2} \]
where \( n \) is an integer.