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Q. Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$,
$2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II
(A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions
(B) $\lambda \neq 8, \mu \in R$ 2. No solution
(C) $\lambda=8, \mu=15$ 3. Unique solution

AP EAMCETAP EAMCET 2019

Solution:

Given system of linear equations is
$x+2 y+3 z=6$
$x+3 y+5 z=9$
and $2 x+5 y+\lambda z=\mu$
Now, according to Cramer's rule,
$\Delta=\begin{vmatrix}1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & \lambda\end{vmatrix}$
$=1(3 \lambda-25)-2(\lambda-10)+3(5-6)$
$=\lambda-8$
$\Delta_{1}=\begin{vmatrix}6 & 2 & 3 \\ 9 & 3 & 5 \\ \mu & 5 & \lambda\end{vmatrix}$
$=6(3 \lambda-25)-(29 \lambda-5 \mu) +3(45-3 \mu)=\mu-15$
$\Delta_{2}=\begin{vmatrix}1 & 6 & 3 \\ 1 & 9 & 5 \\ 2 & \mu & \lambda\end{vmatrix}$
$=1(9 \lambda-5 \mu)-6(\lambda-10)+3(\mu-18)$
$= 3 \lambda-2 \mu+6$
and
$\Delta_{3}=\begin{vmatrix}1 & 2 & 6 \\ 1 & 3 & 9 \\ 2 & 5 & \mu\end{vmatrix}$
$=1(3 \mu-45-2(\mu-18)+6(5-6)$
$=\mu-15$
Now, if $\lambda=8$ and $\mu \neq 15$, then system of linear equations has no solution.
If $\lambda \neq 8$ and $\mu \in R$, then system of linear equations has unique solution.
And, if $\lambda=8$ and $\mu=15$, then system of linear equations has infinite number of solutions, because $\Delta_{2}=3 \lambda-2 \mu+6$ is also be zero.