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If x in R,a(i),b(i),c(i) in R for i=1,2,...

If `x in R,a_(i),b_(i),c_(i) in R` for `i=1,2,3` and `|{:(a_(1)+b_(1)x,a_(1)x+b_(1),c_(1)),(a_(2)+b_(2)x,a_(2)x+b_(2),c_(2)),(a_(3)+b_(3)x,a_(3)x+b_(3),c_(3)):}|=0`, then which of the following may be true ?

A

`x=1`

B

`x=-1`

C

`|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`(a,b,c)` `|{:(a_(1)+b_(1)x,a_(1)x+b_(1),c_(1)),(a_(2)+b_(2)x,a_(2)x+b_(2),c_(2)),(a_(3)+b_(3)x,a_(3)x+b_(3),c_(3)):}|=0`
`impliesunderset(D_(1))(|{:(a_(1),a_(1)x+b_(1),c_(1)),(a_(2),a_(2)x+b_(2),c_(2)),(a_(3),a_(3)x+b_(3),c_(3)):}|)+underset(D_(2))(|{:(b_(1)x,a_(1)x+b_(1),c_(1)),(b_(2)x,a_(2)x+b_(2),c_(2)),(b_(3)x,a_(3)x+b_(3),c_(3)):}|)=0`
Applying `C_(2)toC_(2)-xC_(1)` in `D_(1)` and taking `x` common from `C_(1)` in `D_(2)`.
`implies|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|+x|{:(b_(1),a_(1)x+b_(1),c_(1)),(b_(2),a_(2)x+b_(2),c_(2)),(b_(3),a_(3)x+b_(3),c_(3)):}|=0`
Applying `C_(2)toC_(2)-C_(1)` and then taking `x` common from `C_(2)` in `D_(2)`
`implies|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|+x^(2)|{:(b_(1),a_(1),c_(1)),(b_(2),a_(2),c_(2)),(b_(3),a_(3),c_(3)):}|=0`
`implies(1-x^(2))|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0`
`impliesx=+-1`, or `|{:(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3)):}|=0`
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