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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
