Home
Class 12
MATHS
Consider the system of equations {:(2x...

Consider the system of equations
`{:(2x+lambday+6z=8),(x+2y+muz=5),(x+y+3z=4):}`
The system of equations has :
No solution if :

A

`lambda=2, mu=3`

B

`lambdane 2 , mu=3`

C

`lambda ne 2, mu ne 3 `

D

`lambda =2 , mu in R `

Text Solution

Verified by Experts

The correct Answer is:
B
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • DETERMINANTS

    VK JAISWAL ENGLISH|Exercise EXERCISE-4 : SUBJECTIVE TYPE PROBLEMS|12 Videos
  • DETERMINANTS

    VK JAISWAL ENGLISH|Exercise EXERCISE-2 : ONE OR MORE THAN ONE ANSWER IS / ARE CORRECT|6 Videos
  • CONTINUITY, DIFFERENTIABILITY AND DIFFERENTIATION

    VK JAISWAL ENGLISH|Exercise EXERCISE (SUBJECTIVE TYPE PROBLEMS)|22 Videos
  • DIFFERENTIAL EQUATIONS

    VK JAISWAL ENGLISH|Exercise EXERCISE (SUBJECTIVE TYPE PROBLEMS)|6 Videos

Similar Questions

Explore conceptually related problems

Consider the system of equations {:(2x+lambday+6z=8),(x+2y+muz=5),(x+y+3z=4):} The system of equations has : Exactly one solution if :

Consider the system of equations {:(2x+lambday+6z=8),(x+2y+muz=5),(x+y+3z=4):} The system of equations has : Infinitely many solutions if :

consider the system of equations : ltbr. 3x-y +4z=3 x+2y-3z =-2 6x+5y+lambdaz =-3 Prove that system of equation has at least one solution for all real values of lambda .also prove that infinite solutions of the system of equations satisfy (7x-4)/(-5)=(7y+9)/(13)=z

For what values of a and b, the system of equations 2x+a y+6z=8, x+2y+b z=5, x+y+3z=4 has: (i) a unique solution (ii) infinitely many solutions (iii) no solution

The system of equations -2x+y+z=a x-2y+z=b x+y-2z=c has

If the system of equations 2x+ay+6z=8, x+2y+z=5, 2x+ay+3z=4 has a unique solution then 'a' cannot be equal to :

For what values of p and q the system of equations 2x+py+6z=8, x+2y+qz=5, x+y+3z=4 has i no solution ii a unique solution iii in finitely many solutions.

For what values of p and q the system of equations 2x+py+6z=8, x+2y+qz=5, x+y+3z=4 has i no solution ii a unique solution iii in finitely many solutions.

For the system of equations: x+2y+3z=1 2x+y+3z=2 5x+5y+9z=4

Consider the system of equations x+y+z=6 x+2y+3z=10 x+2y+lambdaz =mu The system has no solution if (a) lambda ne 3 (b) lambda =3, mu =10 (c) lambda =3, mu ne 10 (d) none of these