For real numbers a, b and c, the value of \(\begin{vmatrix}1 & a & bc\\ 1 & b & ca\\ 1 & c & ab\end{vmatrix}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((a - b)(b - c)(c - a)\)
The first column is all ones, so the row operations \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\) give
\(\displaystyle \Delta = \begin{vmatrix}1 & a & bc\\ 0 & b - a & ca - bc\\ 0 & c - a & ab - bc\end{vmatrix}\)
\(ca - bc = c(a - b) = -c(b - a)\)
\(ab - bc = b(a - c) = -b(c - a)\)
Take the common factors \((b - a)\) from \(R_2\) and \((c - a)\) from \(R_3\):
\(\displaystyle \Delta = (b - a)(c - a)\begin{vmatrix}1 & a & bc\\ 0 & 1 & -c\\ 0 & 1 & -b\end{vmatrix}\)
Expanding along the first column:
\(\Delta = (b - a)(c - a)\left[(1)(-b) - (-c)(1)\right]\)
\(\Delta = (b - a)(c - a)(c - b)\)
Since \((b - a)(c - b) = (a - b)(b - c)\):
\(\Delta = (a - b)(b - c)(c - a)\)
Check with a = 1, b = 2, c = 3: the rows are (1, 1, 6), (1, 2, 3), (1, 3, 2), and \(1(4 - 9) - 1(2 - 3) + 6(3 - 2) = 2\), while \((-1)(-1)(2) = 2\).
Why the other options are incorrect
Option (2) \((b - a)(b - c)(c - a)\) — Incorrect. This is the negative of the value; for a = 1, b = 2, c = 3 it gives \(-2\).
Option (3) \((a + b + c)(a - b)(b - c)(c - a)\) — Incorrect. For a = 1, b = 2, c = 3 it gives 12, not 2.
Option (4) 0 — Incorrect. The value is 2 for a = 1, b = 2, c = 3; it is zero only when two of a, b, c are equal.