Consider the system of linear equations

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Question

$\begin{aligned} & \left|\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & \lambda^2 \ 1 & 3 & \lambda\end{array}ight|=0 \ &

Vedclass · Accepted Answer

System has unique solution if $\lambda 
eq 1$ and $\mu 
eq 13$

Vedclass · Answer

$\begin{aligned} & \left|\begin{array}{ccc}1 & 1 & 1 \ 1 & 2 & \lambda^2 \ 1 & 3 & \lambda\end{array}ight|=0 \ & \Rightarrow 2 \lambda^2-\lambda-1=0 \ & \lambda=1,-\frac{1}{2} \ & \left|\begin{array}{ccc}1 & 1 & 5 \ 2 & \lambda^2 & 9 \ 3 & \lambda & \mu\end{array}ight|=0 \Rightarrow \mu=13\end{aligned}$

Infinite solution $\lambda=1 \& \mu=13$

For unique $\operatorname{sol}^{\mathrm{n}} \lambda 
eq 1$

For no $\operatorname{sol}^{\mathrm{n}} \lambda=1 \& \mu 
eq 13$

If $\lambda 
eq 1$ and $\mu 
eq 13$

Considering the case when $\lambda=-\frac{1}{2}$ and $\mu 
eq 13$ this will generate no solution case

Consider the system of linear equations

$x+y+z=5, x+2 y+\lambda^2 z=9$

$x+3 y+\lambda z=\mu$, where $\lambda, \mu \in R$. Then, which of the following statement is NOT correct?

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Consider the system of linear equations $x+y+z=5, x+2 y+\lambda^2 z=9$ $x+3 y+\lambda z=\mu$, where $\lambda, \mu \in R$. Then, which of the following statement is NOT correct?