Question

If A = \(\begin{bmatrix}     1 & \tan\ \alpha/2 \\     -\tan\ \alpha/2 & 1 \end{bmatrix}\) and AB=I then B =

• A. cos² α/2 . I
• B. cos² α/2 . A^T
• C. sin² α/2 . A
• D. cos² α/2 . A

Hint:

To find the inverse of a matrix, use the formula for the inverse of a 2x2 matrix \( \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \). In this case, recognizing trigonometric identities like \( 1 + \tan^2 \theta = \sec^2 \theta \) helps simplify the computation. Additionally, when dealing with matrices, remember that the inverse of a matrix can be written as the transpose of the matrix multiplied by a scalar factor.

Answer 1

The Correct Option is B

The correct answer is (B) : cos² α/2 . A^T.

Answer 2

The correct answer is: (B) \( \cos^2 \left(\frac{\alpha}{2}\right) A^T \).

We are given that:

\( A = \begin{bmatrix} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{bmatrix} \)

and \( AB = I \), where \( I \) is the identity matrix.

We need to find matrix \( B \).

Step 1: Use the condition \( AB = I \)

The condition \( AB = I \) tells us that matrix \( B \) is the inverse of matrix \( A \). Therefore, \( B = A^{-1} \).

Step 2: Find the inverse of matrix \( A \)

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For matrix \( A \), we have: \[ a = 1, \, b = \tan \frac{\alpha}{2}, \, c = -\tan \frac{\alpha}{2}, \, d = 1 \] The determinant of \( A \) is: \[ \text{det}(A) = 1 \cdot 1 - (-\tan \frac{\alpha}{2}) \cdot \tan \frac{\alpha}{2} = 1 + \tan^2 \frac{\alpha}{2} \] Using the trigonometric identity \( 1 + \tan^2 \theta = \sec^2 \theta \), we get: \[ \text{det}(A) = \sec^2 \frac{\alpha}{2} \] Therefore, the inverse of \( A \) is: \[ A^{-1} = \frac{1}{\sec^2 \frac{\alpha}{2}} \begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix} \] Simplifying further: \[ A^{-1} = \cos^2 \frac{\alpha}{2} \begin{bmatrix} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{bmatrix} \] The inverse of \( A \) is \( B = \cos^2 \frac{\alpha}{2} A^T \), where \( A^T \) is the transpose of matrix \( A \).

Thus, the correct answer is (B) \( \cos^2 \left(\frac{\alpha}{2}\right) A^T \).