If the system of equations
$x+y +z= 6 $
$x+2y +3z= 10$
$3x+2y +\lambda z= \mu $
has more than two solutions, then $\mu - \lambda^2 =$
Answer & explanation
Correct answer: option 2
The correct answer is option (2) : 13
If the given system of equations has more than two solutions, then
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3\\ 3 & 2 & \lambda \end{vmatrix} = 0 $ and $\begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10\\ 3 & 2 & \mu \end{vmatrix}=0 $ [Using $D=D_3 =0 $]
$⇒\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 2\\ 3 & -1 & \lambda -3\end{vmatrix}=0 $ and $\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 4\\ 3 & -1 & \mu -18\end{vmatrix}=0 $
$\begin{bmatrix} Applying \, C_2→C_2-C_1, C_3 → C_3-C_1\\ and, C_2→C_2,C_3 → C_3-6C_1 \, \text{in first}\\ \text{and second determinants respectively}\end{bmatrix}$
$⇒\lambda -3 + 2= 0 $ and $\mu - 18 + 4 = 0 $
$⇒ \lambda = 1 $ and $\mu = 14 $
$∴ \mu - \lambda^2 = 14 - 1= 13 $