Login

|b+c a-b a |

|c+a b-c b | = 3abc - a³ - b³ -c³

|a+b c-a c |

prove this by using the properties of determinants..

$Given | \begin{matrix} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{matrix} | C_{1} \to C_{1} + C_{3} C_{2} \to C_{2} - C_{3} = | \begin{matrix} a + b + c & - b & a \\ a + b + c & - c & b \\ a + b + c & - a & c \end{matrix} | = (-) | \begin{matrix} a + b + c & b & a \\ a + b + c & c & b \\ a + b + c & a & c \end{matrix} | = C_{1} \to C_{1} - (C_{2} + C_{3}) = (-) | \begin{matrix} c & b & a \\ a & c & b \\ b & a & c \end{matrix} | = | \begin{matrix} c & a & b \\ a & b & c \\ b & c & a \end{matrix} | = c (a b - c^{2}) - a (a^{2} - b c) + b (a c - b^{2}) = a b c - c^{3} - a^{3} + a b c + a b c - b^{3} = 3 a b c - a^{3} - b^{3} - c^{3}$

43

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?