If A is square matrix of order 3 × 3 such that a_ij=j-i , then which…

The correct answer is Option (2) → (B) and (C) only

Writing out \( a_{ij}=j-i \):

    \( \displaystyle A=\begin{bmatrix}0 & 1 & 2\\-1 & 0 & 1\\-2 & -1 & 0\end{bmatrix} \)

In general \( a_{ji}=i-j=-(j-i)=-a_{ij} \), so \( A^T=-A \). For a skew symmetric matrix of odd order:

    \( \displaystyle |A|=|A^T|=|-A| \)

    \( \displaystyle |-A|=(-1)^3|A|=-|A| \)

    \( \displaystyle |A|=-|A|\ \Rightarrow\ |A|=0 \)

So \( A^T=-A \) (for example \( a_{12}=1 \), \( a_{21}=-1 \)), |A| = 0, and the off-diagonal entries are not zero.

Checking the statements

(A) A is a symmetric matrix. — Incorrect. \( a_{12}=1 \) but \( a_{21}=-1 \).

(B) A is a skew symmetric matrix. — Correct. \( a_{ji}=-a_{ij} \) for all i, j, so \( A^T=-A \).

(C) |A| = 0. — Correct. A skew symmetric matrix of odd order has determinant 0.

(D) A is a diagonal matrix. — Incorrect. \( a_{12}=1\ne 0 \).