Q.The determinant $\begin{vmatrix} b^2 - ab & b - c & bc - ac \\ ab - a^2 & a - b & b^2 - ab \\ bc - ac & c - a & ab - a^2 \end{vmatrix}$ equals
(A) $abc(b - c)(c - a)(a - b)$
(B) $(b - c)(c - a)(a - b)$
(C) $(a + b + c)(b - c)(c - a)(a - b)$
(D) None of these

Concept — spot the column dependence. Factor $(b-a)$ from columns 1 and 3:

while $C_2=\begin{pmatrix}b-c\\a-b\\c-a\end{pmatrix}$.

Now

$C_1-C_3=(b-a)\begin{pmatrix}b-c\\a-b\\c-a\end{pmatrix}=(b-a)\,C_2,$ …

The idea

Every entry factors neatly, exposing a linear relation between the three columns. When columns are linearly dependent, the determinant is zero — no expansion needed.

Step 1 — Factor the columns

Column 1 entries are $b(b-a),\,a(b-a),\,c(b-a)$, so $C_1=(b-a)(b,\,a,\,c)^T$. Column 3 entries are $c(b-a),\,b(b-a),\,a(b-a)$, so $C_3=(b-a)(c,\,b,\,a)^T$. Column 2 is $C_2=(b-c,\,a-b,\,c-a)^T$.

Step 2 — Find the relation

Subtract:

So $C_1-C_3-(b-a)C_2=0$: the columns are linearly dependent.

Step 3 — Conclude …

Method: Detecting Linear Dependence Among Rows or Columns

A determinant is zero whenever its rows or columns satisfy a linear relationship — spotting this relationship evaluates the determinant instantly, without any term-by-term expansion.

Steps

Step 1: Factor each row/column to expose its structure

Look at each row or column and factor out any common term from its entries, rewriting each as a scalar multiple of a simpler vector.

Step 2: Test simple combinations for a relation

Check whether one column (or row) minus another equals a scalar multiple of a third — e.g. compute $C_1 - C_3$ and see if it matches $\lambda \cdot C_2$ for some expression $\lambda$.

Step 3: Conclude the determinant is identically zero

If a genuine linear relation among the rows/columns is found, the determinant equals zero for every value of the variables involved — no arithmetic expansion is required. …

Common Mistakes

Mistake 1: Attempting full term-by-term expansion instead of checking for column dependence

Why it's wrong: Directly expanding this $3\times3$ determinant, with entries like $b^2-ab$ and $bc-ac$, produces many terms and is highly error-prone; the entries are specifically constructed so that $C_1-C_3=(b-a)C_2$. Correct approach: before expanding, check whether one column can be written as a combination of the others — if so, the determinant is immediately zero with no arithmetic needed.

Mistake 2: Confusing "zero for a specific numeric check" with "identically zero for all $a,b,c$" …