Question

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

Answer 1

Correct Answer - A
Let `Delta = |{:(a-b-c, " "2a, " "2a), (2b, b-c-a, " "2b), (2c, 2c, c-a-b):}|`
Applying `R_(1) to R_(1) + R_(2) + R_(2) + R_(3)`, we get
`Delta = |{:(a+b+c, a+b+c, a+b+c), (" "2b, b-c-a, " "2b), (" "2c," " 2c, c-a-b):}|`
`= (a + b+ c)|{:(1, " "1, " "1), (2b, b-c-a, " "2b), (2c," " 2c, c-a-b):}|`
(taking common (a +b+c) from `R_(1)`)
Applying `C_(2) to C_(2) - C_(1) " and "C_(3) to C_(3)-C_(1)`, we get
`Delta= (a + b+ c)|{:(1, " "0, " "0), (2b, -(a+b+c), " "0), (2c," " 0, -(a +b+c)):}|`
Now, expanding along `R_(1)`, we get
`Delta = (a +b+c)1 {(a + b +c)^(2)-0}`
`= (a +b+c)^(3) = (a + b+ c)(x +a +b+c)^(2) " " ("given")`
`rArr (x +a +b +c)^(2) = (a + b+ c)^(2)`
`rArr x +a+ b+c = +- (a +b+c)`
`rArr x = -2(a+ b+c) " " [because x ne 0]`