If A=≤ft[{array}{cc}1 & θ/2 \\ - θ/2 & 1{array}] then A^{-1}=

Correct Answer is : (1−cos⁡θ2)AT≤ft(1- θ/2) A^{T}(21−cosθ)AT

Correct Answer is : (1−cos⁡θ2)AT≤ft(1- θ/2) A^{T}(21−cosθ)AT

Test Index

MHTCET 22 Apr 2025 Shift 1 Paper

Question : 102 of 150

Marks: +1, -0

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

A^{-1}=

Solution:

Concept:

The inverse of a matrix is given by

A^{-1} = \frac{1}{|A|}\operatorname{adj}(A)

Explanation:

First, compute the determinant of

A

|A| = 1 \cdot 1 - \cot\frac{\theta}{2}\left(-\cot\frac{\theta}{2}\right) = 1 + \cot^2\frac{\theta}{2}

Using

1 + \cot^2 x = \csc^2 x

, we get

|A| = \csc^2\frac{\theta}{2} \neq 0

A

is invertible.

For a

2 \times 2

matrix

\begin{bmatrix} a & b \\ c & d \end{bmatrix}

, we have

\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}

Here

a = 1

b = \cot\frac{\theta}{2}

c = -\cot\frac{\theta}{2}

d = 1

Thus

\operatorname{adj}(A) = \begin{bmatrix} 1 & -\cot\frac{\theta}{2} \\ \cot\frac{\theta}{2} & 1 \end{bmatrix} = A^{T}

Therefore:

A^{-1} = \frac{1}{\csc^2\frac{\theta}{2}} \begin{bmatrix} 1 & -\cot\frac{\theta}{2} \\ \cot\frac{\theta}{2} & 1 \end{bmatrix} = \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}

Answer:

Option D:

\left(\frac{1-\cos\theta}{2}\right)A^{T}

Go to Question:

Prev Question

Next Question