Prove that the determinants are equal

**sunitaparija** · March 20th 2009, 02:47 AM

b+c-a a a a b c
b c+a-b b = b c a
c c a+b-c c a b

**mr fantastic** · March 20th 2009, 03:24 AM

Originally Posted by sunitaparija

b+c-a a a a b c
b c+a-b b = b c a
c c a+b-c c a b

The brute force approach:

Calculate $\left| \begin{array}{ccc} b + c - a & a & a \\ b & c + a - b & b \\ c & c & a + b - c \end{array} \right|$ .

Calculate $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$ .

Alternatively, you could do row operations on $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$ .

**sunitaparija** · March 20th 2009, 03:27 AM

The Determinant sum has to be proved without expanding.

Thread: Prove that the determinants are equal

LinkBack

Thread Tools

Display

Prove that the determinants are equal

Similar Math Help Forum Discussions

Prove me that pi is no equal 2

[SOLVED] Prove the natural log can't equal 0

[SOLVED] How to prove this (about determinants)

Prove two angles are equal

prove H_T less than or equal to S_n

Search Tags