Let A=100101010 satisfies An = An-2 + A2 -  I for n≥3. And trace of a square matrix X is equal to the sum of elements in its principal diagonal.Further consider a matrix ∪3×3 with its column as ∪1, ∪2, ∪3 such that A50∪1=12525,A50∪2=010,A50∪3=001Then answer the following questions:The values of |A50| equalsTrace of A50 equalsThe value of |U| equals

An−An−2=A2−I⇒A50=A48+A2−IFurther,A48=A46+A2−IA46=A44+A2−I⋮ ⋮ ⋮ ⋮A4=A2+A2−I A50=25A2−24I¯Here,A2=100101010100101010=100110101⇒ A50=25002525025025−24100010001             =10025102501∴ A50=1Also, tr⁡A50=1+1+1=3. Further,10025102501xyz=12525⇒xyz=∪1=100Similarly,U2=010 and ∪3=001⇒∪=1    0    00    1    00    0    1, i.e., |∪|=1An−An−2=A2−I⇒A50=A48+A2−IFurther,A48=A46+A2−IA46=A44+A2−I⋮ ⋮ ⋮ ⋮A4=A2+A2−I A50=25A2−24I¯Here,A2=100101010100101010=100110101⇒ A50=25002525025025−24100010001             =10025102501∴ A50=1Also, tr⁡A50=1+1+1=3. Further,10025102501xyz=12525⇒xyz=∪1=100Similarly,U2=010 and ∪3=001⇒∪=1    0    00    1    00    0    1, i.e., |∪|=1An−An−2=A2−I⇒A50=A48+A2−IFurther,A48=A46+A2−IA46=A44+A2−I⋮ ⋮ ⋮ ⋮A4=A2+A2−I A50=25A2−24I¯Here,A2=100101010100101010=100110101⇒ A50=25002525025025−24100010001             =10025102501∴ A50=1Also, tr⁡A50=1+1+1=3. Further,10025102501xyz=12525⇒xyz=∪1=100Similarly,U2=010 and ∪3=001⇒∪=1    0    00    1    00    0    1, i.e., |∪|=1