If `{:E(theta)=[(cos^2 theta,costhetasintheta),(costhetasintheta,sin^2theta)]:},and thetaand phi` differ by an odd multiple of `pi//2," then "E(theta)E(phi)` is a

Options

A. null matrix
B. unit matrix
C. diagonal matrix
D. none of these

Answer

To solve the problem, we need to find the product \( E(\theta)E(\phi) \) where \( E(\theta) \) and \( E(\phi) \) are defined as follows:

\[ E(\theta) = \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{pmatrix} \]

Given that \( \theta \) and \( \phi \) differ by an odd multiple of \( \frac{\pi}{2} \), we can express this as:

\[ \phi = \theta + (2n + 1) \frac{\pi}{2} \]

for some integer \( n \).

### Step 1: Write down \( E(\phi) \)

Using the above relationship, we can find \( E(\phi) \):

\[ E(\phi) = \begin{pmatrix} \cos^2(\phi) & \cos(\phi) \sin(\phi) \\ \cos(\phi) \sin(\phi) & \sin^2(\phi) \end{pmatrix} \]

Substituting \( \phi = \theta + (2n + 1) \frac{\pi}{2} \):

1. **Calculate \( \cos(\phi) \)**:
\[ \cos(\phi) = \cos\left(\theta + (2n + 1) \frac{\pi}{2}\right) = -\sin(\theta) \]

2. **Calculate \( \sin(\phi) \)**:
\[ \sin(\phi) = \sin\left(\theta + (2n + 1) \frac{\pi}{2}\right) = \cos(\theta) \]

Now substituting these into \( E(\phi) \):

\[ E(\phi) = \begin{pmatrix} \cos^2(\phi) & \cos(\phi) \sin(\phi) \\ \cos(\phi) \sin(\phi) & \sin^2(\phi) \end{pmatrix} = \begin{pmatrix} \sin^2(\theta) & -\sin(\theta) \cos(\theta) \\ -\sin(\theta) \cos(\theta) & \cos^2(\theta) \end{pmatrix} \]

### Step 2: Multiply \( E(\theta) \) and \( E(\phi) \)

Now we need to compute the product \( E(\theta)E(\phi) \):

\[ E(\theta)E(\phi) = \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin^2 \theta \end{pmatrix} \begin{pmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{pmatrix} \]

### Step 3: Calculate the entries of the product

1. **First row, first column**:
\[ \cos^2 \theta \cdot \sin^2 \theta + \cos \theta \sin \theta \cdot (-\sin \theta \cos \theta) = \cos^2 \theta \sin^2 \theta - \cos^2 \theta \sin^2 \theta = 0 \]

2. **First row, second column**:
\[ \cos^2 \theta \cdot (-\sin \theta \cos \theta) + \cos \theta \sin \theta \cdot \cos^2 \theta = -\cos^3 \theta \sin \theta + \cos^3 \theta \sin \theta = 0 \]

3. **Second row, first column**:
\[ \cos \theta \sin \theta \cdot \sin^2 \theta + \sin^2 \theta \cdot (-\sin \theta \cos \theta) = \cos \theta \sin^3 \theta - \sin^3 \theta \cos \theta = 0 \]

4. **Second row, second column**:
\[ \cos \theta \sin \theta \cdot (-\sin \theta \cos \theta) + \sin^2 \theta \cdot \cos^2 \theta = -\cos^2 \theta \sin^2 \theta + \sin^2 \theta \cos^2 \theta = 0 \]

### Final Result

Putting it all together, we have:

\[ E(\theta)E(\phi) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

This is a null matrix.

### Conclusion

Thus, \( E(\theta)E(\phi) \) is a null matrix.

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