If `A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)]` and `B=[(cos^(2)betas,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. `alpha=beta`
B. `cos (alpha-beta)=0`
C. `sin(alpha-beta)=0`
D. None of these

Answer

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.

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

The product of two matrices \( A \) and \( B \) is given by:
\[ AB = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \]

Calculating each element of the resulting matrix \( AB \):

1. **First Row, First Column**:
\[ a_{11}b_{11} + a_{12}b_{21} = \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta \]

2. **First Row, Second Column**:
\[ a_{11}b_{12} + a_{12}b_{22} = \cos^2 \alpha \cos \beta \sin \beta + \cos \alpha \sin \alpha \sin^2 \beta \]

3. **Second Row, First Column**:
\[ a_{21}b_{11} + a_{22}b_{21} = \cos \alpha \sin \alpha \cos^2 \beta + \sin^2 \alpha \cos \beta \sin \beta \]

4. **Second Row, Second Column**:
\[ a_{21}b_{12} + a_{22}b_{22} = \cos \alpha \sin \alpha \cos \beta \sin \beta + \sin^2 \alpha \sin^2 \beta \]

Thus, the product \( AB \) can be written as:
\[ AB = \begin{pmatrix} \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta & \cos^2 \alpha \cos \beta \sin \beta + \cos \alpha \sin \alpha \sin^2 \beta \\ \cos \alpha \sin \alpha \cos^2 \beta + \sin^2 \alpha \cos \beta \sin \beta & \cos \alpha \sin \alpha \cos \beta \sin \beta + \sin^2 \alpha \sin^2 \beta \end{pmatrix} \]

### Step 2: Set \( AB \) equal to the null matrix

For \( AB \) to be a null matrix, all elements must equal zero. We can start with the first element:
\[ \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta = 0 \]

### Step 3: Factor the equation

We can factor out \( \cos \alpha \cos \beta \):
\[ \cos \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) = 0 \]

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

### Step 4: Analyze the cases

1. **Case 1**: If \( \cos \alpha = 0 \), then \( \alpha = \frac{\pi}{2} + n\pi \).
2. **Case 2**: If \( \cos \beta = 0 \), then \( \beta = \frac{\pi}{2} + m\pi \).
3. **Case 3**: The equation \( \cos \alpha \cos \beta + \sin \alpha \sin \beta = 0 \) simplifies to:
\[ \cos(\alpha - \beta) = 0 \]
This implies:
\[ \alpha - \beta = \frac{\pi}{2} + k\pi \]

### Conclusion

The simplest case is when \( \alpha - \beta = \frac{\pi}{2} \). Thus, the answer is:
\[ \alpha - \beta = \frac{\pi}{2} \]