Question

The sum of distinct values of $\lambda$ for which the system of equations
$(\lambda -1)x+(3\lambda +1)y+2\lambda z=0$
$(\lambda -1)x+(4\lambda -2)y+(\lambda +3)z=0$
$2x+(3\lambda +1)y+3(\lambda -1)z=0$,
has non-zero solutions, is ________.

Answer 1

Answer: 3.00

$(\lambda -1)x+(3\lambda +1)y+2\lambda z=0$
$(\lambda -1)x+(4\lambda -2)y+(\lambda +3)z=0$
$2x+(3\lambda +1)y+(3\lambda -3)z=0$
$\left|\begin{array}{ccc}\lambda -1& 3\lambda +1& 2\lambda \\ \lambda -1& 4\lambda -2& \lambda +3\\ 2& 3\lambda +1& 3\lambda -3\end{array}\right|=0$
$R_1\to R_1-R_2\&R_2\to R_2-R_3$
$\left|\begin{array}{ccc}0& 3-\lambda & \lambda -3\\ \lambda -3& \lambda -3& -2\left(\lambda -3\right)\\ 2& 3\lambda +1& 3\lambda -3\end{array}\right|=0$
${\left(\lambda -3\right)}^2\left|\begin{array}{ccc}0& -1& 1\\ 1& 1& -2\\ 2& 3\lambda +1& 3\lambda -3\end{array}\right|=0$
$(\lambda -3{)}^2[(3\lambda +1)+(3\lambda -1)]=0$
$6\lambda (\lambda -3{)}^2=0\Rightarrow \lambda =0,3$
Sum $=3$