If Δ(x)=1xx+12xx(x−1)x(x+1)3x(x−1)x(x−1)(x−2)xx2−1 then ∆(100) equals

Taking x common from C2, x + 1 from C3and x – 1 from R3, we getΔ(x)=x(x+1)(x−1)1112xx−1x3xx−2xApplying C1→C1−C3,C2→C2−C3 we get Δ(x)=x(x+1)(x−1)001x−1x2x2x=0                                  [ C1 and C2 are proportional]Thus  ,         ∆ (100) =0