Question:
The system of linear equations
$\lambda x+2 y+2 z=5$
$2 \lambda x+3 y+5 z=8$
$4 x+\lambda y+6 z=10$ has:
Correct Option: , 3
Solution:
$D=\left|\begin{array}{ccc}\lambda & 2 & 2 \\ 2 \lambda & 3 & 5 \\ 4 & \lambda & 6\end{array}\right|$
$D=\lambda^{2}+6 \lambda-16$
$D=(\lambda+8)(2-\lambda)$
For no solutions, $D=0$
$\Rightarrow \lambda=-8,2$
when $\lambda=2$
$D_{1}=\left|\begin{array}{ccc}5 & 2 & 2 \\ 8 & 3 & 5 \\ 10 & 2 & 6\end{array}\right|$
$=5[18-10]-2[48-50]+2(16-30]$
$=40+4-28 \neq 0$
There exist no solutions for $\lambda=2$
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