Question

A \( 2n \times 2n \) matrix \( A = [a_{ij}] \) has its elements defined as: \[ a_{ij} = \begin{cases} \beta(i + j), & \text{if } i + j \text{ is odd} \\ -\beta(i + j), & \text{if } i + j \text{ is even} \end{cases} \] where \( n \) is an integer greater than 2, and \( \beta \) is any non-zero real number. What is the rank of matrix \( A \)?

• A. 1
• B. 2
• C. \( n \)
• D. \( 2n \)

Hint:

When matrix elements depend only on the sum \( i + j \), look for symmetry or repetition. If every row is a scaled version of one vector, the matrix has rank 1.

Answer 1

The Correct Option is A

Step 1: Factor out the non-zero scalar \( \beta \)
Since \( \beta \) is a non-zero constant, we can factor it out of every matrix entry. The structure of matrix \( A \) is preserved, and the rank remains unchanged.

\[ A = \beta \cdot M, \quad \text{where } M = [m_{ij}], \text{ with } m_{ij} = \begin{cases} (i + j), & \text{if } i + j \text{ is odd} \\ -(i + j), & \text{if } i + j \text{ is even} \end{cases} \]
Step 2: Examine matrix pattern
Every entry in the matrix depends only on \( i + j \). Thus, all rows and columns follow a highly regular pattern.
In fact, if we examine the entire matrix, all rows are scalar multiples of one another — which implies:

\[ \text{All rows are linearly dependent.} \]
Step 3: Conclusion from structure
Since all rows and all columns are linearly dependent, the matrix has only one linearly independent row (or column).
\[ \text{So, rank of } A = 1 \]