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If the system of equations
$$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
$$
has a unique solution, then $\lambda$ is not equal to
Options:
$$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
$$
has a unique solution, then $\lambda$ is not equal to
Solution:
1210 Upvotes
Verified Answer
The correct answer is:
3
3
Given system of equations is
$$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
$$
It has unique solution.
$$
\begin{aligned}
& \therefore\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 2 & \lambda
\end{array}\right| \neq 0 \\
& \Rightarrow \quad 1(2 \lambda-6)-1(\lambda-3)+1(2-2) \neq 0 \\
& \Rightarrow \quad 2 \lambda-6-\lambda+3 \neq 0 \Rightarrow \lambda-3 \neq 0 \Rightarrow \lambda \neq 3
\end{aligned}
$$
$$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+3 z=10 \\
& x+2 y+\lambda z=0
\end{aligned}
$$
It has unique solution.
$$
\begin{aligned}
& \therefore\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 2 & \lambda
\end{array}\right| \neq 0 \\
& \Rightarrow \quad 1(2 \lambda-6)-1(\lambda-3)+1(2-2) \neq 0 \\
& \Rightarrow \quad 2 \lambda-6-\lambda+3 \neq 0 \Rightarrow \lambda-3 \neq 0 \Rightarrow \lambda \neq 3
\end{aligned}
$$
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