Question

If \(A = \left[ \begin{array}{cc} 1 & \cot \frac{\theta}{2} \\ -\cot \frac{\theta}{2} & 1 \end{array} \right]\) then \(A^{-1} =\)

• A. \(\text{cosec}^2 \frac{\theta}{2} A^{\text{T}}\)
• B. \(\frac{-\sin^2 \theta}{2} A^{\text{T}}\)
• C. \(\left( \frac{1+\cos \theta}{2} \right) A^{\text{T}}\)
• D. \(\left( \frac{1-\cos \theta}{2} \right) A^{\text{T}}\)

Hint:

Whenever a matrix inverse option is given in terms of \(A^T\), first compute the determinant and check whether the adjoint resembles the transpose.

Answer 1

The Correct Option is D

Concept:
For a \(2\times2\) matrix \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the inverse is: \[ \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] ip

Step 1: Find the determinant of \(A\).
\[ A= \begin{bmatrix} 1 & \cot\frac{\theta}{2} \\ -\cot\frac{\theta}{2} & 1 \end{bmatrix} \] So, \[ |A| = 1\cdot1 - \left(\cot\frac{\theta}{2}\right)\left(-\cot\frac{\theta}{2}\right) \] \[ |A| = 1+\cot^2\frac{\theta}{2} \] \[ |A| = \csc^2\frac{\theta}{2} \] ip

Step 2: Find the inverse matrix.
\[ A^{-1} = \frac{1}{\csc^2\frac{\theta}{2}} \begin{bmatrix} 1 & -\cot\frac{\theta}{2} \\ \cot\frac{\theta}{2} & 1 \end{bmatrix} \] But, \[ \frac{1}{\csc^2\frac{\theta}{2}}=\sin^2\frac{\theta}{2} \] So, \[ A^{-1} = \sin^2\frac{\theta}{2} \begin{bmatrix} 1 & -\cot\frac{\theta}{2} \\ \cot\frac{\theta}{2} & 1 \end{bmatrix} \] ip

Step 3: Relate it to \(A^T\).
\[ A^T= \begin{bmatrix} 1 & -\cot\frac{\theta}{2} \\ \cot\frac{\theta}{2} & 1 \end{bmatrix} \] Thus, \[ A^{-1}=\sin^2\frac{\theta}{2}\,A^T \] Now use the identity: \[ \sin^2\frac{\theta}{2}=\frac{1-\cos\theta}{2} \] Hence, \[ A^{-1}= \left(\frac{1-\cos\theta}{2}\right)A^T \] ip Hence, the correct answer is:
\[ \boxed{(D)\ \left( \frac{1-\cos \theta}{2} \right) A^{T}} \]