Question

If \[ \left[ \begin{array}{cc} 1 & -\tan(\theta) \\ \tan(\theta) & 1 \end{array} \right] \left[ \begin{array}{cc} 1 & \tan(\theta) \\ -\tan(\theta) & 1 \end{array} \right]^{-1} = \left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right], \] then:

• A. \(a = 1, b = 1 \)
• B. \(a = \sin 2\theta, b = \cos 2\theta \)
• C. \(a = \cos 2\theta, b = \sin 2\theta \)
• D. None of these

Hint:

Matrix inverse and multiplication can be streamlined using trigonometric identities, especially in transformations involving rotation matrices.

Answer 1

The Correct Option is C

Step 1: Calculate the inverse of the second matrix. The inverse of a matrix \[ \left[ \begin{array}{cc} x & y \\ z & w \end{array} \right] \] where \(xw - yz \neq 0\) is given by: \[ \frac{1}{xw - yz}\left[ \begin{array}{cc} w & -y \\ -z & x \end{array} \right] \] For \[ \left[ \begin{array}{cc} 1 & \tan(\theta) \\ -\tan(\theta) & 1 \end{array} \right]: \] \[ \text{Determinant} = 1 - (-\tan^2(\theta)) = 1 + \tan^2(\theta) \] The inverse is: \[ \frac{1}{1+\tan^2(\theta)}\left[ \begin{array}{cc} 1 & -\tan(\theta) \\ \tan(\theta) & 1 \end{array} \right] \] Step 2: Multiply the matrices. Multiply \[ \left[ \begin{array}{cc} 1 & -\tan(\theta) \\ \tan(\theta) & 1 \end{array} \right] \] with the inverse calculated: \[ \left[ \begin{array}{cc} 1 & -\tan(\theta) \\ \tan(\theta) & 1 \end{array} \right] \frac{1}{1+\tan^2(\theta)} \left[ \begin{array}{cc} 1 & -\tan(\theta) \\ \tan(\theta) & 1 \end{array} \right] = \frac{1}{\sec^2(\theta)} \left[ \begin{array}{cc} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array} \right] \] Step 3: Identify \(a\) and \(b\). From the product matrix, we identify: \[ a = \cos 2\theta, \quad b = \sin 2\theta \]