The determinant of the matrix \(\left[\begin{array}{lll}1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2}\end{array}\right]\) is |
\(\left(a-b\right)\left(b-c\right)\left(a-c\right)\) \(\left(a-b\right)\left(c-b\right)\left(c-a\right)\) \(\left(a-b\right)\left(b-c\right)\left(c-a\right)\) \(\left(b-a\right)\left(b-c\right)\left(c-a\right)\) |
\(\left(a-b\right)\left(b-c\right)\left(c-a\right)\) |
The correct answer is Option (3) → \(\left(a-b\right)\left(b-c\right)\left(c-a\right)\) \(\left[\begin{array}{lll}1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2}\end{array}\right]\) $R_2→R_2-R_1$ $R_3→R_3-R_1$ \(\left[\begin{array}{lll}1 & a & a^{2}\\ 0 & b-a & (b-a)(b+a)\\ 0 & c-a & (c-a)(c+a)\end{array}\right]\) $⇒1\left((b-a)(c-a)(c+a)-(c-a)(b-a)(b+a)\right)$ $⇒(b-a)(c-a)[c+a-b-a]$ $⇒(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)$ |