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 ________.

Answer 1

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+(3 \lambda-3) z=0$
$\begin{vmatrix}\lambda-1&3\lambda+1&2\lambda\\ \lambda-1&4\lambda-2&\lambda+3\\ 2&3\lambda+1&3\lambda-3\end{vmatrix}=0$
$R_{1}\rightarrow R_{1}-R_{2} \& R_{2} \rightarrow R_{2}-R_{3}$
$\begin{vmatrix}0&3-\lambda&\lambda-3\\ \lambda-3&\lambda-3&-2\left(\lambda-3\right)\\ 2&3\lambda+1&3\lambda-3\end{vmatrix}=0$
$\left(\lambda-3\right)^{2}\begin{vmatrix}0&-1&1\\ 1&1&-2\\ 2&3\lambda+1&3\lambda-3\end{vmatrix}=0$
$(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$
$6 \lambda(\lambda-3)^{2}=0 \Rightarrow \lambda=0,3$
Sum $=3$