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|[b+c,a,a] , [b,c+a,b] , [c,c,a+b]|=...

`|[b+c,a,a] , [b,c+a,b] , [c,c,a+b]|=`

A

`4abc`

B

`2(a+b+c)`

C

`(ab+bc+ca)`

D

none of these

Text Solution

Verified by Experts

Apply `R_(1) to R_(1) -(R_(2) + R_(3))`
Take(-2) common from `R_(1). "Apply" R_(2) to (R_(2) - R_(1)) " and "R_(3) to (R_(3) -R_(1))`
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