If `[[axxb,bxxc, c xxa]]=lambda[[a, b, c]]^(2)`, then `lambda` is equal to

Options

A. 1
B. 2
C. 3
D. 0

Answer

A

Explanation

Given,` [ a xx b " " b xx c " " c xx a] = lambda [ "a b c"]^(2)`
`LHS = (a xx b ) * [ (b xx c) xx (c xx a ) ]`
` = (a xx b)*[{b*(cxxa)}c-{c*(cxxa)}b]`
`=(a xx b) *{["b c a"] c - 0} " " [ because c *(c xx a) = 0] `
`={(a xx b) *c} [ "b c a"]=["a b c"]["b c a"] `
`=[ "a b c"]["a b c"] " " [ because ["b c a"]=["a b c"]]`
`=["a b c"]^(2)=RHS`
`therefore lambda = 1`