Home
Class 12
MATHS
If p q r!=0 and the system of equation ...

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 `1/p+b/q+c/r` is `-1` b. `0` c.`0""` d. `not-2`

A

-1

B

0

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
A

`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`
Promotional Banner

Similar Questions

Explore conceptually related problems

If pqr!=0 and the system of equation (p+a)x+by+cz=0ax+(q+b)y+cz=0ax+by+(r+c)z=0 has nontrivial solution,then value of (a)/(p)+(b)/(q)+(c)/(r) is -1 b.0 c.0d.neg-2

if the system of equation ax+y+z=0,x+by=z=0, and x+y+cz=0(a,b,c!=1) has a nontrival solution,then the value of (1)/(1-a)+(1)/(1-b)+(1)/(1-c)is:

If a!=b!=c!=1 and the system of equations ax+y+z=0,x+by+z=0 ,x+y+cz=0 have non trivial solutions then a+b+c-abc

Let a,b,c in R and the system of equations (1-a)x+y+z=0,x+(1-b)y+z=0,x+y+(1-c)z=0 has infinitely many solutions then the minimum value of 'abc' is

The system of linear equations x + y + z = 0 (2x)/(a) + (3y)/(b) + (4z)/(c ) = 0 (x)/(a) + (y)/(b) + (z)/(c ) = 0 has non trivia solution then