question_answer

The system of linear equations

\[\lambda x+2y+2z=5\]

\[2\lambda x+3y+5z=8\]

\[4x+\lambda y+6z=10\] has

[JEE MAIN Held on 08-01-2020 Evening]

A) Infinitely many solutions when \[\lambda =2\]

B) No solution when \[\lambda =8\]

C) A unique solution when \[\lambda =-\,8\]

D) No solution when \[\lambda =2\]

Correct Answer: D

Solution :

\[=\text{ }\lambda \left( 18-5\lambda  \right)-2\left( 12\lambda -20 \right)+2(2{{\lambda }^{2}}-12)\]           \[\Rightarrow \,\,\,\,\,\Delta =-{{\lambda }^{2}}-6\lambda +16\] \[\Rightarrow \,\,\,\,\,\Delta =\left( \lambda +8 \right)\left( 2-\lambda  \right)\] \[\Rightarrow \,\,\,\,\,\Delta =0\] for \[\lambda =2\] or \[\lambda =-8\] \[\Rightarrow \,\,\,\,\,{{\Delta }_{x}}=5\left( 8 \right)-2\left( -2 \right)+2\left( -14 \right)\] \[\Rightarrow \,\,\,\,\,{{\Delta }_{x}}=40+4+-28=16\ne 0\] \[\therefore \] System has no solution.