Question

If \( A = \begin{bmatrix} \tan x & \cot x \\ -\cot x & \tan x \end{bmatrix} \) and \( A + A' = 2I \), then value of \( x \in [0, \pi/2] \) is

• A. \( 0 \)
• B. \( \pi/3 \)
• C. \( \pi/4 \)
• D. \( \pi/2 \)

Hint:

For questions involving \( A + A' \), off-diagonal elements \( a_{ij} + a_{ji} \) must equal the corresponding elements in the target matrix.
Check your interval carefully; if the interval was larger, there might be more solutions like \( 5\pi/4 \).

Answer 1

The Correct Option is C

Concept:
• The transpose \( A' \) is obtained by interchanging rows and columns.
• \( I \) is the identity matrix of the same order.
• Matrix addition is performed element-wise.

Step 1: Find the transpose \( A' \)
Given \( A = \begin{bmatrix} \tan x & \cot x \\ -\cot x & \tan x \end{bmatrix} \).
Then \( A' = \begin{bmatrix} \tan x & -\cot x \\ \cot x & \tan x \end{bmatrix} \).

Step 2: Compute \( A + A' \)
\[ A + A' = \begin{bmatrix} \tan x + \tan x & \cot x + (-\cot x) \\ -\cot x + \cot x & \tan x + \tan x \end{bmatrix} \]
\[ A + A' = \begin{bmatrix} 2 \tan x & 0 \\ 0 & 2 \tan x \end{bmatrix} \]

Step 3: Equate with \( 2I \)
Given \( A + A' = 2I \):
\[ \begin{bmatrix} 2 \tan x & 0 \\ 0 & 2 \tan x \end{bmatrix} = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]
\[ \begin{bmatrix} 2 \tan x & 0 \\ 0 & 2 \tan x \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \]
Equating corresponding elements:
\[ 2 \tan x = 2 \implies \tan x = 1 \]

Step 4: Solve for \( x \) in the given interval
We are looking for \( x \in [0, \pi/2] \) such that \( \tan x = 1 \).
The only solution in this interval is \( x = \pi/4 \).