Solveeit Logo
Login

Question

Question: If \(\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} \right| = (a...

If ∣111abca3b3c3∣=(a−b)(b−c)(c−a)(a+b+c)\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} \right| = (a - b)(b - c)(c - a)(a + b + c) Where a, b, c

are all different, then the determinant

∣111(x−a)2(x−b)2(x−c)2(x−b)(x−c)(x−c)(x−a)(x−a)(x−b)∣\left| \begin{matrix} 1 & 1 & 1 \\ (x - a)^{2} & (x - b)^{2} & (x - c)^{2} \\ (x - b)(x - c) & (x - c)(x - a) & (x - a)(x - b) \end{matrix} \right|vanishes when

A

a+b+c=0a + b + c = 0

B

x=13(a+b+c)x = \frac{1}{3}(a + b + c)

C

x=12(a+b+c)x = \frac{1}{2}(a + b + c)

D

x=a+b+cx = a + b + c

Answer

x=13(a+b+c)x = \frac{1}{3}(a + b + c)

Explanation

Solution

∵\because ∣111abca3b3c3∣=(a−b)(b−c)(c−a)(a+b+c)\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} \right| = (a - b)(b - c)(c - a)(a + b + c) …… (i)

Now, ∣111(x−a)2(x−b)2(x−c)2(x−b)(x−c)(x−c)(x−a)(x−a)(x−b)∣=0\left| \begin{matrix} 1 & 1 & 1 \\ (x - a)^{2} & (x - b)^{2} & (x - c)^{2} \\ (x - b)(x - c) & (x - c)(x - a) & (x - a)(x - b) \end{matrix} \right| = 0

⇒ 1(x−a)(x−b)(x−c)∣(x−a)(x−b)(x−c)(x−a)3(x−b)3(x−c)3(x−a)(x−b)(x−c)(x−a)(x−b)(x−c)(x−a)(x−b)(x−c)∣=0\frac{1}{(x - a)(x - b)(x - c)}\left| \begin{matrix} (x - a) & (x - b) & (x - c) \\ (x - a)^{3} & (x - b)^{3} & (x - c)^{3} \\ (x - a)(x - b)(x - c) & (x - a)(x - b)(x - c) & (x - a)(x - b)(x - c) \end{matrix} \right| = 0Applying C1→C1(x−a),C2→C2(x−b),C3→C3(x−c)C_{1} \rightarrow C_{1}(x - a),C_{2} \rightarrow C_{2}(x - b),C_{3} \rightarrow C_{3}(x - c)

(x - a) & (x - b) & (x - c) \\ (x - a)^{3} & (x - b)^{3} & (x - c)^{3} \\ 1 & 1 & 1 \end{matrix} \right| = 0$$ $${\lbrack(x - a) - (x - b)\rbrack\lbrack(x - b) - (x - c)\rbrack\lbrack(x - c) - (x - a)\rbrack }{(x - a + x - b + x - c) = 0}$$ $(b - a)(c - b)(a - c)\lbrack 3x - (a + b + c)\rbrack = 0$ or $x = \frac{1}{3}(a + b + c)$ $\lbrack\because a \neq b \neq c\rbrack$