Let `A=[{:(costheta,sintheta),(-sintheta,costheta):}]`, then `abs(2A)` is equal to

Options

A. `4 cos 2 theta`
B. `1`
C. `2`
D. `4`

Answer

To find the value of \( \text{abs}(2A) \) where \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the determinant of matrix \( A \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by the formula:

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

For our matrix \( A \):

\[ a = \cos \theta, \quad b = \sin \theta, \quad c = -\sin \theta, \quad d = \cos \theta \]

Substituting these values into the determinant formula:

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

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

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

### Step 2: Use the property of determinants to find \( \text{det}(2A) \)

The property of determinants states that for a scalar \( k \) and an \( n \times n \) matrix \( A \):

\[ \text{det}(kA) = k^n \cdot \text{det}(A) \]

Here, \( k = 2 \) and \( n = 2 \) (since \( A \) is a \( 2 \times 2 \) matrix):

\[ \text{det}(2A) = 2^2 \cdot \text{det}(A) = 4 \cdot 1 = 4 \]

### Conclusion

Thus, the absolute value of the determinant of \( 2A \) is:

\[ \text{abs}(2A) = 4 \]