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

Step 1: Understanding the Concept:
To find the inverse of a \(2 \times 2\) matrix \(A = \begin{bmatrix} a & b
c & d \end{bmatrix}\), we use the formula \(A^{-1} = \frac{1}{|A|} \text{adj}(A)\).
Step 2: Key Formula or Approach:
For a matrix \(A = \begin{bmatrix} a & b
c & d \end{bmatrix}\), \(|A| = ad - bc\) and \(\text{adj}(A) = \begin{bmatrix} d & -b
-c & a \end{bmatrix}\).
Trigonometric identity: \(\sin^2 \frac{\theta}{2} = \frac{1-\cos \theta}{2}\).
Step 3: Detailed Explanation:
Determinant \(|A| = (1)(1) - (\cot \frac{\theta}{2})(-\cot \frac{\theta}{2}) = 1 + \cot^2 \frac{\theta}{2} = \text{cosec}^2 \frac{\theta}{2}\).
Adjoint \(A = \begin{bmatrix} 1 & -\cot \frac{\theta}{2}
\cot \frac{\theta}{2} & 1 \end{bmatrix}\).
Observe that \(A^{\text{T}} = \begin{bmatrix} 1 & -\cot \frac{\theta}{2}
\cot \frac{\theta}{2} & 1 \end{bmatrix}\). Thus, \(\text{adj}(A) = A^{\text{T}}\).
Inverse \(A^{-1} = \frac{1}{\text{cosec}^2 \frac{\theta}{2}} A^{\text{T}} = \sin^2 \frac{\theta}{2} A^{\text{T}}\).
Using the identity \(\sin^2 \frac{\theta}{2} = \frac{1-\cos \theta}{2}\), we get:
\[ A^{-1} = \left( \frac{1-\cos \theta}{2} \right) A^{\text{T}} \] Step 4: Final Answer:
The inverse is \(\left( \frac{1-\cos \theta}{2} \right) A^{\text{T}}\).