If Δ=1+y1−y1−y1−y1+y1−y1−y1−y1+y=0, then value of y are
0, 3
2, – 1
– 1, 3
0, 2
Using C1→C1+C2+C3, we get
Δ=(3−y)1 1−y 1−y1 1+y 1−y1 1−y 1+y
Applying R3→R3−R2,R2→R2−R1 we get
Δ=(3−y)11−y1−y02y00−2y2y=(3−y)4y2∴ Δ=0⇒y=0 or y=3.