Let \(\Delta = \begin{vmatrix}1 & 1 & 1\\ a & b & c\\ a^{2} & b^{2} & c^{2}\end{vmatrix}\), where a, b, c are real numbers. Consider the following statements:
(A) \(\Delta = (a - b)(b - c)(c - a)\)
(B) If \(a = b\), then \(\Delta = 0\)
(C) For \(a = 1,\ b = 2,\ c = 4\), \(\Delta = -6\)
(D) Interchanging a and b leaves the value of \(\Delta\) unchanged for all a, b, c
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Apply the column operations \(C_2 \to C_2 - C_1\) and \(C_3 \to C_3 - C_1\):
\(\displaystyle \Delta = \begin{vmatrix}1 & 0 & 0\\ a & b - a & c - a\\ a^{2} & b^{2} - a^{2} & c^{2} - a^{2}\end{vmatrix}\)
Take \(b - a\) common from the second column and \(c - a\) from the third:
\(\displaystyle \Delta = (b - a)(c - a)\begin{vmatrix}1 & 0 & 0\\ a & 1 & 1\\ a^{2} & b + a & c + a\end{vmatrix}\)
Expanding along the first row:
\(\Delta = (b - a)(c - a)\left[(c + a) - (b + a)\right]\)
\(\Delta = (b - a)(c - a)(c - b)\)
\(\Delta = (a - b)(b - c)(c - a)\)
For a = 1, b = 2, c = 4:
\(\Delta = (1 - 2)(2 - 4)(4 - 1)\)
\(\Delta = (-1)(-2)(3) = 6\)
Interchanging a and b interchanges the first two columns, which changes the sign of \(\Delta\).
So \(\Delta = (a - b)(b - c)(c - a)\), it is 0 when a = b, it equals 6 for a = 1, b = 2, c = 4, and it changes sign when a and b are interchanged.
Checking the statements
(A) \(\Delta = (a - b)(b - c)(c - a)\) — Correct. \((b - a)(c - a)(c - b)\) has two factors with the sign reversed, so it equals \((a - b)(b - c)(c - a)\).
(B) \(\Delta = 0\) if a = b — Correct. The first two columns become identical.
(C) \(\Delta = -6\) for a = 1, b = 2, c = 4 — Incorrect. \((-1)(-2)(3) = 6\).
(D) Unchanged when a and b are interchanged — Incorrect. The first two columns swap, so for a = 1, b = 2, c = 4 the value changes from 6 to −6.