If A=aij is a square matrix of even order such that aij=i2−j2, then

We have,aij=i2−j2∴ aji=j2−i2⇒aij=−aji.Thus, A is a skew-symmetric matrix of even order. We know that the determinant of every skew-symmetric matrix of even order is a perfect square and that of odd order is zero. Hence, option (d) is correct