Matrices and Determinants 4 Question 3

3. If the system of linear equations

$x+y+z=5$

$x+2 y+2 z=6$

$x+3 y+\lambda z=\mu,(\lambda, \mu \in R), \quad$ has infinitely many solutions, then the value of $\lambda+\mu$ is

(2019 Main, 10 April I)

(a) 7

(b) 12

(c) 10

(d) 9

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Answer:

Correct Answer: 3. (c)

Solution:

  1. Given system of linear equations

$ \begin{gathered} x+y+z=5 \\ x+2 y+2 z=6 \\ x+3 y+\lambda z=\mu \end{gathered} $

$(\lambda, \mu \in R)$

The above given system has infinitely many solutions, then the plane represented by these equations intersect each other at a line, means $(x+3 y+\lambda z-\mu)$

$=p(x+y+z-5)+q(x+2 y+2 z-6)$

$=(p+q) x+(p+2 q) y+(p+2 q) z-(5 p+6 q)$

On comparing, we get

$ \begin{array}{lc} & p+q=1, p+2 q=3, p+2 q=\lambda \\ \text { and } & 5 p+6 q=\mu \\ \text { So, } & (p, q)=(-1,2) \\ \Rightarrow & \lambda=3 \text { and } \mu=7 \\ \Rightarrow & \lambda+\mu=3+7=10 \end{array} $



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