Q.If $f(x) = \begin{vmatrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{vmatrix}$, then
(A) $f(a) = 0$
(B) $f(b) = 0$
(C) $f(0) = 0$
(D) $f(1) = 0$

Concept: Determinant Evaluation Using Identities — notice the skew-symmetric pattern on the diagonal (all zeros).

Step 1: Write $f(x)$ and observe that each row has a mix of $x \pm$ constants.

Step 2: Substitute $x = a$:

The first row has two zeros, so expanding along row 1:

This is zero only if $a=0$ or $a=b$ or $a=-c$ — not identically zero.

Step 3: Substitute $x = b$:

First row has zeros at columns 1 and 3, expand along row 1: …

Expand along the first row (the $(1,1)$ entry is $0$):

Multiplying out and collecting powers of $x$, the $x^2$ terms and the constant terms cancel:

$f(x) = 2x^3 + 2(ac - ab - bc)\,x = 2x\big(x^2 + ac - ab - bc\big).$ …

Method: Testing Specific Substitutions in a Determinant Function

When a question asks which value of $x$ makes a determinant function $f(x)$ vanish, it is usually far faster to substitute each candidate value directly and look for a structural pattern that forces zero, rather than expanding $f(x)$ symbolically in full.

Steps

Step 1: Substitute each candidate value one at a time

Plug each option's value of $x$ into the determinant separately, rewriting the matrix's entries with that specific substitution.

Step 2: Look for a structural pattern that forces the determinant to zero

Check for a row of zeros, two identical rows/columns, or — critically — a skew-symmetric pattern: a zero diagonal with off-diagonal entries in mirrored $+/-$ pairs, $a_{ij} = -a_{ji}$.

Step 3: Apply the odd-order skew-symmetric determinant rule

$\det(A) = 0 \text{ whenever } A^T = -A \text{ and } A \text{ has odd order}$ …

Common Mistakes

Mistake 1: Fully expanding $f(x)$ symbolically instead of substituting candidate values directly

Why it's wrong: Expanding $f(x)$ as a general cubic in $x$ with three extra parameters $a,b,c$ is long and highly error-prone; the question only asks which specific substitution gives zero, which is far faster to test directly. Correct approach: substitute each option's value of $x$ into the determinant and look for a structural reason (zero row, repeated row, or skew-symmetry) that forces it to vanish.

Mistake 2: Not recognizing that $f(0)$'s matrix is skew-symmetric of odd order …