If ω ≠ 1 is the complex cube root of unity and matrix H =[ω&0 0&ω], then H70 is equal to -

Question

If ω ≠ 1 is the complex cube root of unity and matrix H =[ω&0 0&ω], then H70 is equal to - (A) 0 (B) -H (C) H2 (D) H

Tardigrade Admin · Accepted Answer

H2 = [ω&0 0&ω][ω &0 0&ω ] = [ω2 &0 0&ω2 ] If Hk = [ωk &0 0&ωk ], then Hk+1 = [ωk+1 &0 0&ωk+1 ] So by mathematical induction, H70 = [ω 70 &0 0&ω 70 ] = [ω &0 0&ω ] = H

Q. If $ω \neq = 1$ is the complex cube root of unity and matrix $H = [ω 0 0 ω]$ , then $H^{70}$ is equal to -

0

-H

$H^{2}$

H

$H^{2} = [ω 0 0 ω] [ω 0 0 ω] = [ω^{2} 0 0 ω^{2}]$
If $H^{k} = [ω^{k} 0 0 ω^{k}]$ , then $H^{k + 1} = [ω^{k + 1} 0 0 ω^{k + 1}]$
So by mathematical induction,
$H^{70} = [ω^{70} 0 0 ω^{70}] = [ω 0 0 ω] = H$

Q. If ω=1 is the complex cube root of unity and matrix H=[ω0​0ω​], then H70 is equal to -

0

-H

H2

H

H2=[ω0​0ω​][ω0​0ω​]=[ω20​0ω2​] If Hk=[ωk0​0ωk​], then Hk+1=[ωk+10​0ωk+1​] So by mathematical induction, H70=[ω700​0ω70​]=[ω0​0ω​]=H

Q. If $ω \neq = 1$ is the complex cube root of unity and matrix $H = [ω 0 0 ω]$ , then $H^{70}$ is equal to -

$H^{2}$

$H^{2} = [ω 0 0 ω] [ω 0 0 ω] = [ω^{2} 0 0 ω^{2}]$
If $H^{k} = [ω^{k} 0 0 ω^{k}]$ , then $H^{k + 1} = [ω^{k + 1} 0 0 ω^{k + 1}]$
So by mathematical induction,
$H^{70} = [ω^{70} 0 0 ω^{70}] = [ω 0 0 ω] = H$