If α=−5 and β=8, then Δ=0 but at least one of Δ1,Δ2,Δ3 is non-zero, so the system is inconsistent. Hence option A is correct.
If α=−5 and β=9, then Δ=Δ1=Δ2=Δ3=0, so the system has infinitely many solutions. Hence option D is correct.
If α=−5, then Δ=0, so the system has a unique solution. Hence option C is correct.
For option B, the given values are α=−6 and β=9. Then
Δ=7(−6+5)=−7=0
so the system has a unique solution, not infinitely many solutions.
Therefore, the statement that is NOT correct is option C according to the solution. There is a discrepancy because the working shows option B is false, but the solution explicitly states The Correct Option is C. the correct option is C.
Elimination Check
Given:
2x−y+3z3x+2y−z4x+5y+αz=5=7=β
Find: Which option is not correct.
From the first two equations, eliminate variables to reduce the system. The extracted second approach in the solution states that the system reduces to
(α−2β+7)z=2α−8β+3
Hence:
unique solution if α−2β+7=0
no solution if α−2β+7=0 and 2α−8β+3=0
infinitely many solutions if both are zero.
Substitute α=−6 and β=9:
α−2β+7=−6−18+7=−17=0
So the system has a unique solution. This again shows that option B is false by the working.
However, the solution headline explicitly says The Correct Option is C. Therefore, the extracted answer must follow the solution conclusion, while noting the inconsistency between the heading and the algebraic working.
Common mistakes
Setting only Δ=0 and immediately concluding infinitely many solutions is incorrect. When Δ=0, you must also check the relevant sub-determinants or consistency conditions to distinguish between no solution and infinitely many solutions.
Assuming β does not matter once α=−5 is wrong. After Δ=0, the value of β determines whether the system is inconsistent or has infinitely many solutions.
Confusing the statement 'which is NOT correct' with 'which is correct' leads to selecting a true option. Always identify the false statement after analyzing each option.
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