Question

If \( A=\begin{bmatrix}\sec\theta & -\tan\theta-\tan\theta & \sec\theta\end{bmatrix} \) and \( A+\operatorname{adj}A=4I \), then find the value of \( \theta \).

• (A) \( \frac{\pi}{6} \)
• (B) \( \frac{\pi}{4} \)
• (C) \( 0 \)
• (D) \( \frac{\pi}{3} \)

Hint:

For symmetric matrices with equal diagonal entries, adding the matrix and its adjoint often simplifies the off-diagonal terms automatically.

Answer 1

The Correct Option is D

Concept:
For a matrix \[ M=\begin{bmatrix}a & b c & d\end{bmatrix}, \] the adjoint is \[ \operatorname{adj}M= \begin{bmatrix} d & -b -c & a \end{bmatrix}. \]

Step 1: Finding the adjoint of matrix \( A \).
Given, \[ A= \begin{bmatrix} \sec\theta & -\tan\theta -\tan\theta & \sec\theta \end{bmatrix}. \] Comparing with \[ \begin{bmatrix} a & b c & d \end{bmatrix}, \] we get: \[ a=d=\sec\theta, \] and \[ b=c=-\tan\theta. \] Hence, \[ \operatorname{adj}A= \begin{bmatrix} \sec\theta & \tan\theta \tan\theta & \sec\theta \end{bmatrix}. \]

Step 2: Applying the given condition.
We are given: \[ A+\operatorname{adj}A=4I. \] Substituting the matrices: \[ \begin{bmatrix} \sec\theta & -\tan\theta -\tan\theta & \sec\theta \end{bmatrix} + \begin{bmatrix} \sec\theta & \tan\theta \tan\theta & \sec\theta \end{bmatrix} = \begin{bmatrix} 4 & 0 0 & 4 \end{bmatrix}. \] Adding the matrices: \[ \begin{bmatrix} 2\sec\theta & 0 0 & 2\sec\theta \end{bmatrix} = \begin{bmatrix} 4 & 0 0 & 4 \end{bmatrix}. \]

Step 3: Comparing corresponding elements.
Equating diagonal entries: \[ 2\sec\theta=4. \] Therefore, \[ \sec\theta=2. \] Thus, \[ \cos\theta=\frac{1}{2}. \] Hence, \[ \boxed{\theta=\frac{\pi}{3}}. \]