The value of \( \begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix} \) is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( (a-b)(c-b)(a-c) \)
Subtracting the first row from the other two (\( R_2\to R_2-R_1 \), \( R_3\to R_3-R_1 \)):
\( \displaystyle \begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix}=\begin{vmatrix}1 & a & a^2\\0 & b-a & b^2-a^2\\0 & c-a & c^2-a^2\end{vmatrix} \)
Taking \( (b-a) \) out of row 2 and \( (c-a) \) out of row 3, and expanding along the first column:
\( \displaystyle =(b-a)(c-a)\begin{vmatrix}1 & b+a\\1 & c+a\end{vmatrix} \)
\( \displaystyle =(b-a)(c-a)(c-b) \)
Changing the sign of two factors leaves the product unchanged, so
\( \displaystyle (b-a)(c-a)(c-b)=(a-b)(c-b)(a-c) \)
Why the other options are incorrect
Option (2) \( (a-b)(b-c)(c-a)(a+b+c) \) — Incorrect. The extra factor \( (a+b+c) \) makes it of degree 4, but every term of this determinant has degree 0 + 1 + 2 = 3.
Option (3) \( abc(a-b)(b-c)(c-a) \) — Incorrect. The extra factor abc makes it of degree 6, but every term of this determinant has degree 3.
Option (4) \( (b-a)(c-b)(a-c) \) — Incorrect. Compared with the correct value, only one factor has its sign changed, so this is the negative of the determinant.