If \( A=\begin{bmatrix}0 & b-a & c-a\\a-b & 0 & c-b\\a-c & b-c & 0\end{bmatrix} \) then the value of \( |A| \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Each entry below the diagonal is the negative of its mirror image above it:
\( \displaystyle a_{21}=a-b=-(b-a)=-a_{12} \)
\( \displaystyle a_{31}=a-c=-(c-a)=-a_{13} \)
\( \displaystyle a_{32}=b-c=-(c-b)=-a_{23} \)
All diagonal entries are 0, so \( A^T=-A \) and A is skew-symmetric of order 3.
\( \displaystyle |A|=|A^T| \)
\( \displaystyle =|-A| \)
\( \displaystyle =(-1)^3|A|=-|A| \)
\( \displaystyle 2|A|=0\ \Rightarrow\ |A|=0 \)
A skew-symmetric determinant of odd order is always 0.