If `A= [{:(a-b-c, 2a,2a), (2b, b-c-a, 2b),(2c, 2c, c-a-b):}]`
`= (a +b+c) (x +a +b+c)^(2), x ne 0 " and " a +b +c ne 0`, then x is equal to

Options

A. `2 (a+b+c)`
B. `-2 (a+b+c)`
C. `-(a+b+c)`
D. `abc`

Answer

B

Explanation

`Delta|{:(a-b-c, 2a, 2a),(2b, b-c-a, 2b),(2c, 2c, c-a-b):}|`
`R_(1)to R_(1)+ R_(2)+ R_(3)`
`Delta|{:(a+b+c, a+b+c, a+b+c),(2b, b-c-a, 2b), (2c, 2c, c-a-b):}|implies Delta (a+b+c)|{:(1,1,1),(2b, b-c-a, 2b),(2c, 2c, c-a-b):}|`
Now `C_(1) to C_(1) -C_(2)`
`therefore Delta = (a+b+c) |{:(0,1,1),(a+b+c, b-c-a, 2b),(0, 2c, c-a-b):}|implies Delta = (a+b+c) ^(2) |{:(0,1,1),(1, b-c-a, 2b),(0, 2c, c-a-b):}|`
Expanding along 1st column
`Delta = (a+b+c )^(3){-(c-a-b-2c)}`
`Delta = ( a+b+c) ^(3) = (a+b+c) (a+b+c)^(2)`
Either `x =0 or x =-2 (a+b+c)`