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JEE MAIN 2020
08-01-20 S2
Question
The system of linear equations
$\begin{aligned} & \lambda x+2 y+2 z=5
\\ & 2 \lambda x+3 y+5 z=8
\\ & 4 x+\lambda y+6 z=10 \text { has }\end{aligned}$
Select the correct option:
A
infinitely many solutions when $\lambda=2$
B
no solutio when $\lambda=8$
C
a unique solution when $\lambda=-8$
D
no solution when $\lambda=2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Δ = | 2 2 |    | 2 3 5 |    | 4 6 | ⇒ Δ = 4(18 − 5λ) − 2(12 − 20) + 2(12 − 12) ⇒ Δ = −2 − 6λ + 16 ⇒ Δ = (λ + 8)(2 − λ) ⇒ Δ = 0 for λ = 2 or λ = −8 ⇒ for λ = 2 Δₓ = | 5 2 2 |       | 8 3 5 |       | 10 2 6 | ⇒ Δₓ = 5(8 − 2) − 2(2 − 14) ⇒ Δₓ = 40 + 4 − 28 = 16 ≠ 0 ∴ System has no solution.
Question Tags
JEE Main
Mathematics
Medium
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