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

Step 1: Write the entries as a product of signs.
Notice that \( a_{ij} = (-1)^{i+j} \), which can be split as \( (-1)^i \cdot (-1)^j \). If we define a column vector \( v \) with \( v_i = (-1)^i \), every entry of the matrix is just \( v_i v_j \).

Step 2: Recognise the matrix as an outer product.
This means the whole matrix can be written as \( A = v v^{T} \), where \( v \) is a single non-zero column vector of size \( n \). An outer product of a vector with itself always has rank at most 1, since every column of \( A \) is just a multiple of \( v \).

Step 3: Confirm the rank.
Since \( v \) is not the zero vector, at least one column is non-zero, so the rank cannot be 0. Combined with the rank being at most 1, we get \[ \boxed{\text{rank}(A) = 1} \]