3 and 4 .Determinants and Matrices
medium

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

A

$5$ and $7$

B

$6$ and $8$

C

$4$ and $9$

D

$5$ and $8$

(JEE MAIN-2020)

Solution

For infinite many solutions

$D=D_{1}=D_{2}=D_{3}=0$

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

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

$\therefore \lambda=5$

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

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

$\mu=8$

Standard 12
Mathematics

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