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Question:

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:

10

7

9

12

Solution:

The correct option is B
10
x+3y+λz−μ=p(x+y+z−5)+q(x+2y+2z−6)
on comparing the coefficient;
p+q=1
and
p+2q=3
⇒(p,q)=(−1,2)
Hence
x+3y+λx−μ=x+3y+3z−7
⇒λ=3,μ=7