If the system of linear equations x + y + z = 5 , x + 2 y + 2 z = 6 , x + 3 y + λ z = µ ,   (…

If the system of linear equations x+y+z=5, x+2y+2z=6,  x+3y+λz=µ, (λ, µ ∈R) , has infinitely many solutions, then the value of λ+µ is:

  1. 7
  2. 10
  3. 12
  4. 9

Solution

∆=11112213λ=λ-3
∆1=511622μ3λ=4λ-3
∆3=1511621μλ=λ-μ+4
∆4=11512613μ=μ-7
For infinite many solution ∆=∆1=∆2=∆3=0
⇒λ=3, & μ=7  from  ∆=0 & ∆4=0
&  λ=3, μ=7  Satisfies ∆3=0=∆1
∴  λ+μ=10

Asked in: JEE Main 2019 (10 Apr Shift 1)

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