The correct option is B -1,2
∣∣
∣∣14201−2512x5x2∣∣
∣∣ = 0⇒ ∣∣
∣
∣∣06150−2−2x5(1−x2)12x5x2∣∣
∣
∣∣=0 (R1 → R1 − R2R2 → R2 − R3)⇒ 3.2.5.∣∣
∣
∣∣0110−(1+x)1−x21xx2∣∣
∣
∣∣=0⇒ (1+x)∣∣
∣∣0110−11−x1xx2∣∣
∣∣=0⇒ x+1=0 or x−2=0 ⇒ x=−1,2.
Trick: Obviously by inspection, x=-1,2 satisfy the equation.
At x=-1, ∣∣
∣∣14201−251−25∣∣
∣∣=0 as R2 ≡ R3
At x=2, ∣∣
∣∣14201−251420∣∣
∣∣=0 as R1 ≡ R3