`a ,b ,c` are distinct real numbers not equal to one. If `a x+y+z=0,x+b y+z=0, and x+y+c z=0` have nontrivial solution, then the value of `1/(1-a)+1/(1-b)+(1)/(1-c)` is equal to a.`1` b. `-1` c.`z e ro` d. none of these

Options

A. `-1`
B. `1`
C. zero
D. none of these

Answer

B

Explanation

since the system has non-trivial solution we have
`|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|=0`
Applying `R_(1) toR_(1)-R_(2),R_(2) to R_(2) -R_(3) `we get
`Delta =|{:(a-1,,1-b,,0),(0,,b-1,,1-c),(1,,1,,c):}|=0`
`" or " c(1-a)(1-b)+(1-b)(1-c)-(1-c) (a-1)=0`
Dividing throughout by (1-a)(1-b)(1-c) we get
`(c ) /(1-c) +(1)/(1-a)+(1)/(1-b) =0`
`" or " -1 + (1)/(1-c)+(1)/(1-a)+(1)/(1-b)=0`
`" or " (1)/(1-c)+(1)/(1-a)+(1)/(1-b) =1`