TS EAMCET 9-Sep-2020 Shift 2 Solved Paper

For α=1, the system reduces to a homogeneous system which is always consistent. So, α≠1
For α≠1 we have
=|

α+2	α+2	α+2
1	α	1
1	1	α

|
( Applying R1→R1+R2+R3)
⇒‌‌D=(α+2)|

1	1	1
1	α	1
1	1	α

|
=(α+2)|

1	0	0
1	α−1	0
1	0	α−1

|
( Applying C2→C2−C1,C3→C3−C1)
⇒‌‌D=(α+2)(α−1)2
and =(α−1)|

1	0	0
1	α−1	0
1	0	α−1

|
( Applying C2→C2−C1,C3→C3−C1)
⇒‌‌D1=(α−1)3
Clearly D=0 for α=−2, but D1≠0
So, the system is inconsistent for α=−2