Flind the product of two matrices
`A =[[cos^(2) theta , cos theta sin theta],[cos theta sin theta ,sin^(2)theta]] B= [[cos^(2) phi,cos phi sin phi],[cos phisin phi,sin^(2)phi]]`
Show that, AB is the zero matrix if `theta and phi` differ by an
odd multipl of `pi/2`.

Options

A. 5(6!)
B. 3(6!)
C. 12(6!)
D. 8(6!)

Answer

To solve the problem, we need to find the product of the two matrices \( A \) and \( B \) and show that the product \( AB \) is the zero matrix when \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \).

### Step 1: Define the Matrices
Let:
\[ A = \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{bmatrix} \]
\[ B = \begin{bmatrix} \cos^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{bmatrix} \]

### Step 2: Calculate the Product \( AB \)
The product \( AB \) is calculated as follows:

\[ AB = \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{bmatrix} \begin{bmatrix} \cos^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin^2 \phi \end{bmatrix} \]

Calculating the elements of the resulting matrix:

1. **First Element**:
\[ (AB)_{11} = \cos^2 \theta \cdot \cos^2 \phi + \cos \theta \sin \theta \cdot \cos \phi \sin \phi \]
\[ = \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi \]

2. **Second Element**:
\[ (AB)_{12} = \cos^2 \theta \cdot \cos \phi \sin \phi + \cos \theta \sin \theta \cdot \sin^2 \phi \]
\[ = \cos^2 \theta \cos \phi \sin \phi + \cos \theta \sin \theta \sin^2 \phi \]

3. **Third Element**:
\[ (AB)_{21} = \cos \theta \sin \theta \cdot \cos^2 \phi + \sin^2 \theta \cdot \cos \phi \sin \phi \]
\[ = \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi \]

4. **Fourth Element**:
\[ (AB)_{22} = \cos \theta \sin \theta \cdot \cos \phi \sin \phi + \sin^2 \theta \cdot \sin^2 \phi \]
\[ = \cos \theta \sin \theta \cos \phi \sin \phi + \sin^2 \theta \sin^2 \phi \]

### Step 3: Combine the Results
Now, we can write the product matrix \( AB \):
\[ AB = \begin{bmatrix} \cos^2 \theta \cos^2 \phi + \cos \theta \sin \theta \cos \phi \sin \phi & \cos^2 \theta \cos \phi \sin \phi + \cos \theta \sin \theta \sin^2 \phi \\ \cos \theta \sin \theta \cos^2 \phi + \sin^2 \theta \cos \phi \sin \phi & \cos \theta \sin \theta \cos \phi \sin \phi + \sin^2 \theta \sin^2 \phi \end{bmatrix} \]

### Step 4: Factor Out Common Terms
We can factor out common terms in each element:
1. **First Element**:
\[ = \cos^2 \phi (\cos^2 \theta + \sin^2 \theta) + \cos \phi \sin \phi (\cos \theta \sin \theta) \]
Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ = \cos^2 \phi + \cos \phi \sin \phi \cos \theta \sin \theta \]

2. **Second Element**:
\[ = \cos \phi \sin \phi (\cos^2 \theta + \sin^2 \theta) + \sin^2 \phi \cos \theta \sin \theta \]
\[ = \cos \phi \sin \phi + \sin^2 \phi \cos \theta \sin \theta \]

3. **Third Element**:
\[ = \cos^2 \phi \cos \theta \sin \theta + \sin \phi \cos \phi \sin^2 \theta \]

4. **Fourth Element**:
\[ = \cos \phi \sin \phi \sin^2 \theta + \sin^2 \phi \cos \theta \sin \theta \]

### Step 5: Show \( AB = 0 \) under Conditions
Now, we need to show that \( AB \) is the zero matrix when \( \theta - \phi = (2n + 1) \frac{\pi}{2} \).

Using the identity:
\[ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \]
If \( \theta - \phi = (2n + 1) \frac{\pi}{2} \), then:
\[ \cos(\theta - \phi) = 0 \]
This implies:
\[ \cos \theta \cos \phi + \sin \theta \sin \phi = 0 \]

Thus, each term in the product matrix \( AB \) will become zero, leading to:
\[ AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

### Conclusion
Therefore, we have shown that \( AB \) is the zero matrix when \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \).