The system of linear equations
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:

1. (1) no solution when $\lambda=8$

2. (2) a unique solution when $\lambda=-8$

3. (3) no solution when $\lambda=2$

4. (4) infinitely many solutions when $\lambda=2$

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 $\lambda=2$