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 Obviously by inspection.
At x=−1,∣∣
∣∣14201−251−25∣∣
∣∣=0 as R2≡R3
At x=2,∣∣
∣∣14201−251420∣∣
∣∣=0 as R1≡R3.