Question

The sum of distinct values of $\lambda$ for which the systems 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

• A. 3
• B. 3.000
• C. 3.00
• D. 3.0

Answer 1

Answer: (A), (B), (C), (D)

$\left|\begin{array}{ccc}\lambda -1& 3\lambda +1& 2\lambda \\ \lambda -1& 4\lambda -2& \lambda +3\\ 2& 3\lambda +1& 3(\lambda -1)\end{array}\right|=0$
$R_2\to R_2-R_1$
$R_3\to R_3-R_1$
$\left|\begin{array}{ccc}\lambda -1& 3\lambda +1& 2\lambda \\ 0& \lambda -3& -\lambda +3\\ 3-\lambda & 0& \lambda -3\end{array}\right|=0$
$C_1\to C_1+C_3$
$\left|\begin{array}{ccc}3\lambda -1& 3\lambda +1& 2\lambda \\ -\lambda +3& \lambda -3& -\lambda +3\\ 0& 0& \lambda -3\end{array}\right|=0$
$(\lambda -3)\left[\right(3\lambda -1\left)\right(\lambda -3)-(3-\lambda \left)\right(3\lambda +1\left)\right]=0$
$(\lambda -3)[3{\lambda }^2-10\lambda +3-(8\lambda -3{\lambda }^2+3\left)\right]=0$
$(\lambda -3)(6{\lambda }^2-18\lambda )=0$
$\left(6\lambda \right)(\lambda -3{)}^2=0$
$\lambda =0,3$
sum of values of $\lambda =0+3=3$