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Solve! - Matrices and Determinants - JEE Main-5

If the system of linear equations 

x + y + z = 5

x + 2y + 2z = 6

x + 3y + \lambda z = \mu , (\lambda,\mu \epsilon R ) , has infinitely

many solutions, then the value of \lambda + \mu is :

  • Option 1)

    12

  • Option 2)

    9

  • Option 3)

    7

  • Option 4)

    10

 
Answers (1)
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x + 3y + \lambda z - \mu = p ( x + y + z - 5) + q ( x + 2y + 2z - 6 ) 

On comparing the coefficients 

p + q = 1   and    p + 2q = 3

=> ( p , q ) = ( -1 , 2 )

Hence, x + 3y + \lambda z - \mu = x + 3y + 3z - 7 

=> \lambda =3,\mu =7

=> \lambda + \mu =10

So, option (4) is correct.


Option 1)

12

Option 2)

9

Option 3)

7

Option 4)

10

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