Let `A=B B^(T)+C C^(T)`, where `B=[(cos theta), (sintheta)], C=[(sin theta), (-costheta)], theta in R`. Then A is :

Options

A. `[(0, 0),(0, 0)]`
B. `[(0, 1), (1, 0)]`
C. `[(1, 0), (0,1)]`
D. `[(0,0),(0, 1)]`

Answer

To solve the problem, we need to compute the matrix \( A = B B^T + C C^T \) where:

\[ B = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}, \quad C = \begin{pmatrix} \sin \theta \\ -\cos \theta \end{pmatrix} \]

### Step 1: Calculate \( B B^T \)

First, we compute \( B B^T \):

\[ B^T = \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} \]

Now, multiply \( B \) by \( B^T \):

\[ B B^T = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta \end{pmatrix} = \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{pmatrix} \]

### Step 2: Calculate \( C C^T \)

Next, we compute \( C C^T \):

\[ C^T = \begin{pmatrix} \sin \theta & -\cos \theta \end{pmatrix} \]

Now, multiply \( C \) by \( C^T \):

\[ C C^T = \begin{pmatrix} \sin \theta \\ -\cos \theta \end{pmatrix} \begin{pmatrix} \sin \theta & -\cos \theta \end{pmatrix} = \begin{pmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{pmatrix} \]

### Step 3: Add \( B B^T \) and \( C C^T \)

Now we add the two matrices:

\[ A = B B^T + C C^T = \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{pmatrix} + \begin{pmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{pmatrix} \]

Calculating the sum:

\[ A = \begin{pmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ \sin \theta \cos \theta - \sin \theta \cos \theta & \sin^2 \theta + \cos^2 \theta \end{pmatrix} \]

### Step 4: Simplify the Matrix

Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):

\[ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Conclusion

Thus, the final result is:

\[ A = I_2 \]

where \( I_2 \) is the \( 2 \times 2 \) identity matrix.