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…
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.
\(\begin{array}{lll}A & B & C \\ 1 & 2 & 3 \end{array}\)
\(\begin{array}{lll}A & B & C \\ 3 & 1 & 2 \end{array}\)
\(\begin{array}{lll}A & B & C \\ 2 &3 &1 \end{array}\)
\(\begin{array}{lll}A & B & C \\ 3 &2 &1 \end{array}\)
Solution
Given system of linear equations is
\(\begin{aligned}
x+2 y+3 z & =6 \\
x+3 y+5 z & =9 \\
\text { and } \quad 2 x+5 y+\lambda z & =\mu
\end{aligned}\)
Now, according to Cramer's rule,
\(\begin{aligned}
& \Delta=\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 5 \\
2 & 5 & \lambda
\end{array}\right|=1(3 \lambda-25)-2(\lambda-10)+3(5-6) \\
& =\lambda-8 \\
& \Delta_1=\left|\begin{array}{lll}
6 & 2 & 3 \\
9 & 3 & 5 \\
\mu & 5 & \lambda
\end{array}\right|=6(3 \lambda-25)-2(9 \lambda-5 \mu) \\
& +3(45-3 \mu)=\mu-15 \\
& \Delta_2=\left|\begin{array}{lll}
1 & 6 & 3 \\
1 & 9 & 5 \\
2 & \mu & \lambda
\end{array}\right|=1(9 \lambda-5 \mu)-6(\lambda-10)+3(\mu-18) \\
& =3 \lambda-4 \mu+6
\end{aligned}\)
and
\(\Delta_3=\left|\begin{array}{lll}
1 & 2 & 6 \\
1 & 3 & 9 \\
2 & 5 & \mu
\end{array}\right|=1(3 \mu-45)-2(\mu-18)+6(5-6) = \mu - 15\)
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. Hence, option (2) is correct.
