📘 Question
If \(A = [a_{ij}]\) is a square matrix of even order such that:
\[
a_{ij} = i^2 – j^2
\]
Then:
(a) \(A\) is skew-symmetric and \(|A| = 0\)
(b) \(A\) is symmetric and \(|A|\) is a square
(c) \(A\) is symmetric and \(|A| = 0\)
(d) none of these
✏️ Step-by-Step Solution
Step 1: Check symmetry
\[
a_{ji} = j^2 – i^2 = -(i^2 – j^2) = -a_{ij}
\]
So,
\[
A^T = -A
\]
Hence, \(A\) is a skew-symmetric matrix.
Step 2: Determinant property
For a skew-symmetric matrix of even order:
\[
|A| = (\text{Pfaffian})^2
\]
So determinant is a perfect square (not necessarily zero).
❌ Check options
- (a) says determinant is zero → ❌ not always true
- (b) says symmetric → ❌ wrong
- (c) says symmetric → ❌ wrong
✅ Final Answer
\[
\boxed{(d)\; \text{none of these}}
\]
💡 Key Concept
If \(a_{ji} = -a_{ij}\), the matrix is skew-symmetric. For even order, determinant is a perfect square (not necessarily zero).