Solveeit Logo
Login

Question

Question: \[\left| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \right| =\]...

1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \right| =$$
A

a2+b2+c2a^{2} + b^{2} + c^{2}

B

(a+b)(b+c)(c+a)(a + b)(b + c)(c + a)

C

(a−b)(b−c)(c−a)(a - b)(b - c)(c - a)

D

None of these

Answer

(a−b)(b−c)(c−a)(a - b)(b - c)(c - a)

Explanation

Solution

∣1aa21bb21cc2∣=∣0a−ba2−b20b−cb2−c21cc2∣,\left| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \right| = \left| \begin{matrix} 0 & a - b & a^{2} - b^{2} \\ 0 & b - c & b^{2} - c^{2} \\ 1 & c & c^{2} \end{matrix} \right|, by R1→R1−R2R_{1} \rightarrow R_{1} - R_{2} $$R_{2} \rightarrow R_{2} - R_{3}

= (a−b)(b−c)∣01a+b01b+c1cc2∣(a - b)(b - c)\left| \begin{matrix} 0 & 1 & a + b \\ 0 & 1 & b + c \\ 1 & c & c^{2} \end{matrix} \right|

= (a−b)(b−c)∣00a−c01b+c1cc2∣(a - b)(b - c)\left| \begin{matrix} 0 & 0 & a - c \\ 0 & 1 & b + c \\ 1 & c & c^{2} \end{matrix} \right|, by R1→R1−R2R_{1} \rightarrow R_{1} - R_{2}

= (a−b)(b−c)(a−c)∣00101b+c1cc2∣(a - b)(b - c)(a - c)\left| \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & b + c \\ 1 & c & c^{2} \end{matrix} \right|

= (a−b)(b−c)(a−c).(−1)=(a−b)(b−c)(c−a)(a - b)(b - c)(a - c).( - 1) = (a - b)(b - c)(c - a).