Question
The determinant $\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$ equals:
  1. abc(b - c)(c - a)(a - b)
  2. (b - c)(c - a)(a - b)
  3. (a + b + c)(b - c)(c - a)(a - b)
  4. None of these

Answer

  1. None of these

Solution:

$\begin{vmatrix}\text{b}^2-\text{ab}&\text{b}-\text{c}&\text{bc}-\text{ac}\\\text{ab}-\text{a}^2&\text{a}-\text{b}&\text{b}^2-\text{ab}\\\text{bc}-\text{ac}&\text{c}-\text{a}&\text{ab}-\text{a}^2\end{vmatrix}$

$=\begin{vmatrix}\text{b}(\text{b}-\text{a})&\text{b}-\text{c}&\text{c}(\text{b}-\text{a})\\\text{a}(\text{b}-\text{a})&\text{a}-\text{b}&\text{b}(\text{b}-\text{a})\\\text{c}(\text{b}-\text{a})&\text{c}-\text{a}&\text{a}(\text{b}-\text{a})\end{vmatrix}$

$=(\text{b}-\text{a})^2\begin{vmatrix}\text{b}&\text{b}-\text{c}&\text{c}\\\text{a}&\text{a}-\text{b}&\text{b}\\\text{c}&\text{c}-\text{a}&\text{a}\end{vmatrix}$ [Taking (b - a) common from C1 and C3]

$=(\text{b}-\text{a})^2\begin{vmatrix}0&\text{b}-\text{c}&\text{c}\\0&\text{a}-\text{b}&\text{b}\\0&\text{c}-\text{a}&\text{a}\end{vmatrix}$ [Applying C→ C1 - C2 - C3]

$=0$

Hence, the correct option is (d)

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