If `{:A=[(cos theta,sintheta),(-sintheta,costheta)]:}," then "A^2=I`is true for

Options

A. `theta=0`
B. `theta=pi/4`
C. `theta=pi/2`
D. none of these

Explanation

To prove that \( A^2 = I \) for the matrix

\[ A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}, \]

we will calculate \( A^2 \) and show under what conditions it equals the identity matrix \( I \).

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

We start by multiplying 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 the matrix multiplication rules, we calculate each element of the resulting matrix:

- **First Row, First Column:**
\[ \cos \theta \cdot \cos \theta + \sin \theta \cdot (-\sin \theta) = \cos^2 \theta - \sin^2 \theta \]

- **First Row, Second Column:**
\[ \cos \theta \cdot \sin \theta + \sin \theta \cdot \cos \theta = \sin \theta \cos \theta + \sin \theta \cos \theta = 2 \sin \theta \cos \theta \]

- **Second Row, First Column:**
\[ -\sin \theta \cdot \cos \theta + \cos \theta \cdot (-\sin \theta) = -\sin \theta \cos \theta - \sin \theta \cos \theta = -2 \sin \theta \cos \theta \]

- **Second Row, Second Column:**
\[ -\sin \theta \cdot \sin \theta + \cos \theta \cdot \cos \theta = \cos^2 \theta - \sin^2 \theta \]

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 \( I \)

The identity matrix \( I \) is given by:

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

For \( A^2 \) to equal \( I \), we need:

1. \( \cos^2 \theta - \sin^2 \theta = 1 \)
2. \( 2 \sin \theta \cos \theta = 0 \)

### Step 4: Solve the equations

1. From \( \cos^2 \theta - \sin^2 \theta = 1 \):
\[ \cos^2 \theta = 1 \implies \sin^2 \theta = 0 \implies \sin \theta = 0 \]
This occurs when \( \theta = n\pi \) for \( n \in \mathbb{Z} \).

2. From \( 2 \sin \theta \cos \theta = 0 \):
\[ \sin \theta = 0 \quad \text{or} \quad \cos \theta = 0 \]
Since we already found \( \sin \theta = 0 \), this condition is satisfied.

### Conclusion

Thus, \( A^2 = I \) is true when \( \theta = n\pi \) for \( n \in \mathbb{Z} \).

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