Given ‌2x−y+3z=5 ‌3x+2y−z=7 ‌4x+5y+αz=β ‌∆=|
2
−1
3
3
2
−1
4
5
α
|=7α+35 ‌∆=7(α+5) For unique solution ∆≠0 α≠−5 For inconsistent & Infinite solution ‌∆=0 ‌α+5=0⇒α=−5 ‌∆1=|
5
−1
3
7
2
−1
β
5
−5
|=−5(β−9) ‌∆2=|
2
5
3
3
7
−1
4
β
−5
|=11(β−9) ‌∆3=|
2
−1
5
3
2
7
4
5
β
| ‌∆3=7(β−9) For Inconsistent system : - At least one ∆1,∆2&∆3 is not zero α=−5,β=8 option (A) True Infinite solution: ∆1=∆2=∆3=0 From here β−9=0⇒β=9α=−5& option (D) True β=9 Unique solution α≠−5,β=8→‌ option ‌(C)‌ True ‌ Option (B) False For Infinitely many solution α must be -5 .