The values of lambda and mu for which the sys | vedclass

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Question

For infinite many solutions

$D=D_{1}=D_{2}=D_{3}=0$

$\operatorname{Now} D =\left|\begin{array}{lll}1 & 1 & 1 \ 1 & 2 & 3 \ 1 &

Vedclass · Accepted Answer

$5$ and $8$

Vedclass · Answer

For infinite many solutions

$D=D_{1}=D_{2}=D_{3}=0$

$\operatorname{Now} D =\left|\begin{array}{lll}1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda\end{array}ight|=0$

$1 .(2 \lambda-9)-1 .(\lambda-3)+1 .(3-2)=0$

$	herefore \lambda=5$

$\operatorname{Now} D _{1}=\left|\begin{array}{lll}2 & 1 & 1 \ 5 & 2 & 3 \ \mu & 3 & 5\end{array}ight|=0$

$2(10-9)-1(25-3 \mu)+1(15-2 \mu)=0$

$\mu=8$

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

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