if `A=[(costheta,sintheta),(-sintheta,costheta)],then A^(2)=l ` is true for

Options

A. `theta=0`
B. `theta=(pi)/(4)`
C. `theta=(pi)/(2)`
D. None of these

Answer

To solve the problem, we need to find the square of the matrix \( A \) and determine for which values of \( \theta \) the result equals the identity matrix.

Given the matrix:
\[ A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply the matrix \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \cdot \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

### Step 2: Perform the multiplication

Using matrix multiplication:
- The element at (1,1):
\[ \cos \theta \cdot \cos \theta + \sin \theta \cdot (-\sin \theta) = \cos^2 \theta - \sin^2 \theta \]
- The element at (1,2):
\[ \cos \theta \cdot \sin \theta + \sin \theta \cdot \cos \theta = 2 \sin \theta \cos \theta \]
- The element at (2,1):
\[ -\sin \theta \cdot \cos \theta + \cos \theta \cdot (-\sin \theta) = -2 \sin \theta \cos \theta \]
- The element at (2,2):
\[ -\sin \theta \cdot \sin \theta + \cos \theta \cdot \cos \theta = \cos^2 \theta - \sin^2 \theta \]

Thus, we have:
\[ A^2 = \begin{pmatrix} \cos^2 \theta - \sin^2 \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos^2 \theta - \sin^2 \theta \end{pmatrix} \]

### Step 3: Set \( A^2 \) equal to the identity matrix

The identity matrix \( I \) is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

We set \( A^2 = I \):
\[ \begin{pmatrix} \cos^2 \theta - \sin^2 \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos^2 \theta - \sin^2 \theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

From this, we get two equations:
1. \( \cos^2 \theta - \sin^2 \theta = 1 \)
2. \( 2 \sin \theta \cos \theta = 0 \)

### Step 4: Solve the equations

**From the second equation:**
\[ 2 \sin \theta \cos \theta = 0 \implies \sin \theta = 0 \quad \text{or} \quad \cos \theta = 0 \]

- If \( \sin \theta = 0 \):
- \( \theta = 0, \pi, 2\pi, \ldots \)

- If \( \cos \theta = 0 \):
- \( \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots \)

**From the first equation:**
\[ \cos^2 \theta - \sin^2 \theta = 1 \implies \cos^2 \theta = 1 \quad \text{and} \quad \sin^2 \theta = 0 \]
This confirms \( \theta = 0 \) or \( \theta = \pi \).

### Conclusion

The values of \( \theta \) for which \( A^2 = I \) are:
- \( \theta = 0 \)
- \( \theta = \pi \)