Question

Let \( A = [a_{ij}] \) for \( i, j = 1, 2, 3, \ldots, n \) be an \( n \times n \) matrix such that \( a_{ij} = \begin{cases} 1, & \text{if } (i+j) \text{ is even} \\ -1, & \text{if } (i+j) \text{ is odd} \end{cases} \) where \( n \gt 1 \). Then the rank of \( A \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

Whenever the elements of a matrix satisfy \( a_{ij} = f(i)g(j) \), the rank of the matrix will always be 1. Here, \( a_{ij} = (-1)^i \cdot (-1)^j \).

Answer 1

The Correct Option is A

Concept: The rank of a matrix corresponds to the number of linearly independent rows or columns.

• A matrix has rank 1 if every row is a scalar multiple of a single non-zero row.

• Linear dependence occurs when rows can be reduced to zero using elementary operations.

• Parity rules for indices determine the sign of the elements in the matrix.

Step 1: Constructing the matrix structure.
The element \( a_{ij} \) is defined as \( (-1)^{i+j} \). Let us write the matrix for \( n = 3 \) to observe the pattern. \[ A = \begin{bmatrix} (-1)^2 & (-1)^3 & (-1)^4 \\ (-1)^3 & (-1)^4 & (-1)^5 \\ (-1)^4 & (-1)^5 & (-1)^6 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]

Step 2: Analyzing row relationships.
Notice the relationship between the first row and subsequent rows. Row 2 is exactly \( -1 \) times Row 1. Row 3 is exactly \( 1 \) times Row 1. In general, Row \( i \) is \( (-1)^{i-1} \) times the first row.

Step 3: Determining the Rank.
Applying row operations \( R_i \to R_i - (-1)^{i-1}R_1 \) for all \( i \gt 1 \). All rows except the first become zero. The number of non-zero rows in row-echelon form is 1. Therefore, the rank of the matrix is 1.