If `p q r!=0` and the system of equation `(p+a)x+b y+c z=0` `a x+(q+b)y+c z=0` `a c+b y+(r+c)z=0` has nontrivial solution, then value of `a/p+b/q+c/r` is `-1` b. `0` c.`0""` d. `not-2`

Options

A. -1
B. 0
C. 1
D. 2

Answer

A

Explanation

`Delta =|{:(p+a,,b,,c),(a,,q+b,,c),(a,,b,,r+c):}|=0`
Applying `R_(2) to R_(2) -R_(1) " and " R_(3) to R_(3) -R_(1) ` we get
`|{:(p+a,,b,,c),(-p,,q,,0),(-p,,0,,r):}|=0`
`" or " pqc+ [q(p+a)+bp]r=0`
Dividing by pqr , we obtain
`(a)/(p)+(b)/(q)+(c )/(r)=-1`