Solve this following

Question:

The values of $\lambda$ and $\mu$ for which the system of linear equations

$x+y+z=2$

$x+2 y+3 z=5$

$x+3 y+\lambda z=\mu$

has infinitely many solutions are, respectively

 

  1. 5 and 7

  2. 6 and 8

  3. 4 and 9

  4. 5 and 8


Correct Option:

Solution:

For infinite many solutions

$\mathrm{D}=\mathrm{D}_{1}=\mathrm{D}_{2}=\mathrm{D}_{3}=0$

Now $\mathrm{D}=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda\end{array}\right|=0$

1. $(2 \lambda-9)-1 .(\lambda-3)+1 .(3-2)=0$

$\therefore \lambda=5$

Now $D_{1}=\left|\begin{array}{lll}2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5\end{array}\right|=0$

$2(10-9)-1(25-3 \mu)+1(15-2 \mu)=0$

$\mu=8$

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