1) If △=[156017001] and △′=[10130346100] A) \triangle^{2}-3\triangle'=0 B) (\triangle+\triangle')^{2}-3(\triangle+\triangle')+2=0 C) (\triangle+\triangle')^{2}-3(\triangle+\triangle')+5=0 D) \triangle+3\triangle'+1 Answer: Option BExplanation:Given, △=[156017001] = 1(1−0)−5(0−0)+6(0−0) =1−0+0 △=1 If △=1 and △′=[10130346100] =1(0−18)−0(300−12)+1(18−0) =−18−0+18 △′=0 Now, (△+△′)2−3(△+△′)+2 =(1+0)2−3(1+0)+2 =-2+2=0