Let A= [2 -1 -1 1 0 -1 1 -1 0] and B=A - I. If ω=(√3 i-1/2), then the number of elements in the set n ∈ 1,2, ldots, 100: A n +(ω B ) n . = A + B is equal to

Question

Let A= [2 -1 -1 1 0 -1 1 -1 0] and B=A - I. If ω=(√3 i-1/2), then the number of elements in the set n ∈ 1,2, ldots, 100: A n +(ω B ) n . = A + B is equal to  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A= [ 2 -1 -1 1 0 -1 1 -1 0] ⇒ A2=A ⇒ An=A ∀ n ∈ 1,2, ldots, 100 Now, B = A - I =[ 1 -1 -1 1 -1 -1 1 -1 -1] B 2 =- B ⇒ B 3 =- B 2= B ⇒ B 5= B ⇒ B 99= B Also, ω3 k =1 So, n = common of 1,3,5, ldots, 99 and 3,6,9, ldots, 99 =17

Q. Let A=⎣⎡​211​−10−1​−1−10​⎦⎤​ and B=A - I. If ω=23​i−1​, then the number of elements in the set {n∈{1,2,…,100}:An+(ωB)n =A+B} is equal to ____

A=⎣⎡​211​−10−1​−1−10​⎦⎤​⇒A2=A⇒An=A ∀n∈{1,2,…,100} Now, B=A−I=⎣⎡​111​−1−1−1​−1−1−1​⎦⎤​ B2=−B ⇒B3=−B2=B ⇒B5=B ⇒B99=B Also, ω3k=1 So, n= common of {1,3,5,…,99} and {3,6,9,…,99}=17

Q. Let $A = ⎣ ⎡ 211 - 1 0 - 1 - 1 - 1 0 ⎦ ⎤$ and $B = A$ - I. If $ω = \frac{3 i - 1}{2}$ , then the number of elements in the set ${n \in {1, 2, \dots, 100} : A^{n} + (ω B)^{n}$ $= A + B}$ is equal to ____