Mathematics - Determinant Question with Solution | TestHub

$\begin{array}{rl}& (\lambda -1)x+(\lambda -4)y+\lambda z=5\\ & \lambda x+(\lambda -1)y+(\lambda -4)z=7\\ & (\lambda +1)x+(\lambda +2)y-(\lambda +2)z=9\end{array}$ For infinitely many solutions $\begin{array}{rl}& \mathrm{D}=\left|\begin{array}{ccc}\lambda -1& \lambda -4& \lambda \\ \lambda & \lambda -1& \lambda -4\\ \lambda +1& \lambda +2& -(\lambda +2)\end{array}\right|=0\\ & (\lambda -3)(2\lambda +1)=0\\ & {\mathrm{D}}_{\mathrm{x}}=\left|\begin{array}{ccc}5& \lambda -4& \lambda \\ 7& \lambda -1& \lambda -4\\ 9& \lambda +2& -(\lambda +2)\end{array}\right|=0\\ & 2(3-\lambda )(23-2\lambda )=0\\ & \lambda =3\\ & \therefore {\lambda }^2+\lambda =9+3=12\end{array}$