An ordered pair (α,β) for which the system of linear equations (1+α)x+βy+z=2αx+(1+β)y+z=3αx+βy+2z=2 has a unique solution, is:
A
(1,−3)
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B
(−3,1)
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C
(−4,2)
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D
(2,4)
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Solution
The correct option is D(2,4) The system of linear equation are (1+α)x+βy+z=2αx+(1+β)y+z=3αx+βy+2z=2 for unique solution, Δ≠0 ⇒∣∣
∣∣1+αβ1α1+β1αβ2∣∣
∣∣≠0 Apply R1→R1−R2 ⇒∣∣
∣∣1−10α1+β1αβ2∣∣
∣∣≠0 ⇒1(2+2β−β)+1(2α−α)+0≠0⇒α+β+2≠0⇒α+β≠−2 Hence, (2,4) will satisfy this condition.