Let \( A=[a_{ij}]_{3\times 3} \) be a matrix such that \( a_{ij}=\displaystyle\frac{i-j}{i+j} \), then which of the following statements is/are TRUE?
(A) A is a symmetric matrix
(B) A is a skew symmetric matrix
(C) \( |A|=0 \)
(D) \( |A|\ne 0 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B) and (C) only
Interchanging i and j changes the sign of the numerator only:
\( \displaystyle a_{ji}=\frac{j-i}{j+i}=-\frac{i-j}{i+j} \)
\( \displaystyle a_{ji}=-a_{ij} \)
In particular \( a_{ii}=0 \). Written out,
\( \displaystyle A=\begin{bmatrix}0 & -\frac{1}{3} & -\frac{1}{2}\\\frac{1}{3} & 0 & -\frac{1}{5}\\\frac{1}{2} & \frac{1}{5} & 0\end{bmatrix} \)
So \( A^{T}=-A \). For a skew-symmetric matrix of odd order,
\( \displaystyle |A|=|A^{T}|=|-A| \)
\( \displaystyle =(-1)^3|A|=-|A| \)
\( \displaystyle 2|A|=0\ \Rightarrow\ |A|=0 \)
So A is skew-symmetric (not symmetric) and \( |A|=0 \).
Checking the statements
(A) A is a symmetric matrix — Incorrect. \( a_{12}=-\frac{1}{3} \) but \( a_{21}=\frac{1}{3} \).
(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 is skew-symmetric of order 3, so \( |A|=0 \).
(D) \( |A|\ne 0 \) — Incorrect. \( |A|=0 \).