For unique solution ∣1+αααβ1+ββ112∣=0
Applying C1→C1+C2+C3, we get
∣α+β+2α+β+2α+β+2β1+ββ112∣=0
⇒(α+β+2)∣111β1+ββ112∣=0
⇒α+β+2=0
Clearly, point (2,4) satisfying the given condition.
JEE Main 2019 — Mathematics Algebra
An ordered pair (α,β) for which the system of linear equations
(1+α)x+βy+z=2
αx+(1+β)y+z=3
αx+βy+2z=2 has a unique solution, is :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
(−3,1)
(1,−3)
(2,4)
(−4,2)
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