# The sum of distinct values of

Question:

The sum of distinct values of $\lambda$ for which the system of equations

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$

has non-zero solutions, is

Solution:

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+(3 \lambda-3) z=0$

$\left|\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \& \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$

$\left|\begin{array}{ccc}0 & 3-\lambda & \lambda-3 \\ \lambda-3 & \lambda-3 & -2(\lambda-3) \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$(\lambda-3)^{2}\left|\begin{array}{ccc}0 & -1 & 1 \\ 1 & 1 & -2 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$

$6 \lambda(\lambda-3)^{2}=0 \Rightarrow \lambda=0,3$

Sum $=3$