If `A=[[costheta, sintheta], [-sintheta, costheta]]`, then

Options

A. A is singular
B. `A^(-1)` exists
C. `A^(-1)` does not exists
D. `A^(-1)` is singular

Answer

To solve the problem, we need to analyze the matrix \( A \) given by:

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

We will determine whether the matrix is singular or not by calculating its determinant.

### Step 1: Calculate the Determinant of Matrix \( A \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated using the formula:

\[ \text{det}(A) = ad - bc \]

For our matrix \( A \):

- \( a = \cos \theta \)
- \( b = \sin \theta \)
- \( c = -\sin \theta \)
- \( d = \cos \theta \)

Substituting these values into the determinant formula gives:

\[ \text{det}(A) = (\cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta) \]

This simplifies to:

\[ \text{det}(A) = \cos^2 \theta + \sin^2 \theta \]

### Step 2: Use the Pythagorean Identity

From trigonometry, we know that:

\[ \cos^2 \theta + \sin^2 \theta = 1 \]

Thus, we have:

\[ \text{det}(A) = 1 \]

### Step 3: Determine if the Matrix is Singular

A matrix is singular if its determinant is equal to zero. Since we found that:

\[ \text{det}(A) = 1 \neq 0 \]

This means that the matrix \( A \) is **not singular**.

### Step 4: Determine if the Inverse Exists

Since the matrix is not singular, it is invertible. Therefore, the inverse of matrix \( A \) exists.

### Step 5: Conclusion

Based on the analysis, we conclude:

- The matrix \( A \) is **not singular**.
- The inverse of matrix \( A \) **exists**.

### Final Answer

The correct option is that **the inverse of \( A \) exists**.

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