JEE Mains 10-Apr-2023 Shift 1 Solved Paper

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=9 α=−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 .