MCQ
The system of linear equations $\lambda x+2 y+2 z=5$ ; $2 \lambda x+3 y+5 z=8$ ; $4 x+\lambda y+6 z=10$ has
  • A
    infinitely many solutions when $\lambda=2$
  • B
    a unique solution when $\lambda=-8$
  • C
    no solution when $\lambda=8$
  • no solution when $\lambda=2$

Answer

Correct option: D.
no solution when $\lambda=2$
d
$D=\left|\begin{array}{ccc}{\lambda} & {3} & {2} \\ {2 \lambda} & {3} & {5} \\ {4} & {\lambda} & {6}\end{array}\right|=(\lambda+8)(2-\lambda)$

for $\lambda=2 ; \mathrm{D}_{1} \neq 0$

Hence, no solution for $\lambda=2$

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