Let x =[1 1 1] and A =[-1 2 3 0 1 6 0 0 -1]. For k ∈ N if X prime Ak X=33, then k is equal to:

Question

Let x =[1 1 1] and A =[-1 2 3 0 1 6 0 0 -1]. For k ∈ N if X prime Ak X=33, then k is equal to: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Allen (Dropped or 10) X =[1 1 1] ; A =[-1 2 3 0 1 6 0 0 -1] X T A K X =33 [1 1 1][-1 2 3 0 1 6 0 0 -1] k [1 1 1]=33 [1 1 1][-1 2 3 0 1 6 0 0 -1][1 1 1]=33 text As A2=[-1 2 3 0 1 6 0 0 -1][-1 2 3 0 1 6 0 0 -1]=[1 0 6 0 1 0 0 0 1] A4=[1 0 6 0 1 0 0 0 1][1 0 6 0 1 0 0 0 1]=[1 0 12 0 1 0 0 0 1] A8=[1 0 24 0 1 0 0 0 1] A10=[1 0 6 0 1 0 0 0 1][1 0 24 0 1 0 0 0 1]=[1 0 30 0 1 0 0 0 1] text for K arrow text Even A K =[1 0 3 K 0 1 0 0 0 1] X T A K X =33 (This is not correct) [1 1 1][1 0 3 K 0 1 0 0 0 1][1 1 1] =[1 1 3 K +1][1 1 1]=[3 K +3] ∴ 3 K +3=33 ∴ K =10 But it should be dropped as 33 is not matrix If K is odd X T A K X =33 X T AA K -1 X =33 [1 1 1][ -1 2 3 0 1 6 0 0 -1][1 0& 3 k -3 0 1 0 0 0 1][1 1 1]=33 [-1 3 8][3 k -2 1 1]=[33] [-3 k +13]=[33] k =20 / 3 (not possible)

Q. Let x=⎣⎡​111​⎦⎤​ and A=⎣⎡​−100​210​36−1​⎦⎤​. For k∈N if X′AkX=33, then k is equal to:

Q. Let $x = ⎣ ⎡ 111 ⎦ ⎤$ and $A = ⎣ ⎡ - 1 00 210 36 - 1 ⎦ ⎤$ . For $k \in N$ if $X^{'} A^{k} X = 33$ , then $k$ is equal to: