Solveeit Logo
Login

Question

Question: \[\left| \begin{matrix} 1 & a & a^{2} - bc \\ 1 & b & b^{2} - ac \\ 1 & c & c^{2} - ab \end{matrix} ...

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

0

B

a3+b3+c3−3abca^{3} + b^{3} + c^{3} - 3abc

C

3abc3abc

D

(a+b+c)3(a + b + c)^{3}

Answer

0

Explanation

Solution

∣1aa2−bc1bb2−ac1cc2−ab∣=∣0a−b(a−b)(a+b+c)0b−c(b−c)(a+b+c)1cc2−ab∣\left| \begin{matrix} 1 & a & a^{2} - bc \\ 1 & b & b^{2} - ac \\ 1 & c & c^{2} - ab \end{matrix} \right| = \left| \begin{matrix} 0 & a - b & (a - b)(a + b + c) \\ 0 & b - c & (b - c)(a + b + c) \\ 1 & c & c^{2} - ab \end{matrix} \right|

by {R1→R1−R2R2→R2−R3 \left\{ \begin{aligned} & R_{1} \rightarrow R_{1} - R_{2} \\ & R_{2} \rightarrow R_{2} - R_{3} \end{aligned} \right.\

= (a−b)(b−c)∣01a+b+c01a+b+c1cc2−ab∣=0(a - b)(b - c)\left| \begin{matrix} 0 & 1 & a + b + c \\ 0 & 1 & a + b + c \\ 1 & c & c^{2} - ab \end{matrix} \right| = 0, {∵R1≡R2}\{\because R_{1} \equiv R_{2}\}.