The system of linear equations |
\[\lambda x+2y+2z=5\] |
\[2\lambda x+3y+5z=8\] |
\[4x+\lambda y+6z=10\] has |
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.You need to login to perform this action.
You will be redirected in
3 sec