If \( f(x)=\begin{vmatrix}0 & x+a & x+b\\x-a & 0 & x+c\\x-b & x-c & 0\end{vmatrix} \) then the value of \( f(0) \) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
At x = 0 the determinant is
\( \displaystyle f(0)=\begin{vmatrix}0 & a & b\\-a & 0 & c\\-b & -c & 0\end{vmatrix} \)
Expanding along the first row,
\( \displaystyle f(0)=0\,(0+c^2)-a\,(0+bc)+b\,(ac-0) \)
\( \displaystyle =-abc+abc \)
\( \displaystyle =0 \)
Shortcut: The matrix at x = 0 is skew-symmetric (its transpose is its negative) and of odd order 3. For such a matrix
\( \displaystyle |A|=|A^T|=|-A| \)
\( \displaystyle |-A|=(-1)^3|A|=-|A| \)
so \( |A|=-|A| \), that is, \( |A|=0 \), without any expansion.