If ω is a complex cube root of unity and A= ω 0 0 0 ω^2 0 0 0 1 then A^(-1)=…

Correct answer: (A) ω^2 0 0 0 ω 0 0 0 1. Given, ω is a complex cube root of unity ∴ ω^3=1 Given matrix A can be written as A=lA ⇒ ω 0 0 0 ω^2 0 0 0 1 = 1 0 0 0 1 0 0 0 1 A On apply R_1→ω^2R_1,R_2→ωR_2 , we get ⇒ ω^3 0 0 0 ω^3 0 0 0 1 = ω^2 0 0 0 ω 0 0 0 1 A ⇒ 1 0 0 0 1 0 0 0 1 = ω^2 0 0 0 ω 0 0 0 1 A A^(-1)= ω^2 0 0 0 ω 0 0 0 1

Q63 / 150 · MHT CET 2019 (Shift 1) · Mathematics

If $ω$ is a complex cube root of unity and $A = [\begin{matrix} ω & 0 & 0 \\ 0 & ω^{2} & 0 \\ 0 & 0 & 1 \end{matrix}]$ then $A^{- 1} = \dots$

Correct answer
A
$[\begin{matrix} ω^{2} & 0 & 0 \\ 0 & ω & 0 \\ 0 & 0 & 1 \end{matrix}]$
B
$[\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
C
$[\begin{matrix} 1 & 0 & 0 \\ 0 & ω^{2} & 0 \\ 0 & 0 & ω \end{matrix}]$
D
$[\begin{matrix} 0 & 0 & ω \\ 0 & ω^{2} & 0 \\ 0 & 0 & 0 \end{matrix}]$

Solution

Given,

ω

is a complex cube root of unity

∴ ω^{3} = 1

Given matrix A can be written as

A = l A

\Rightarrow [\begin{matrix} ω & 0 & 0 \\ 0 & ω^{2} & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A

On apply

R_{1} \to ω^{2} R_{1}, R_{2} \to ω R_{2}

, we get

\Rightarrow [\begin{matrix} ω^{3} & 0 & 0 \\ 0 & ω^{3} & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} ω^{2} & 0 & 0 \\ 0 & ω & 0 \\ 0 & 0 & 1 \end{matrix}] A

\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} ω^{2} & 0 & 0 \\ 0 & ω & 0 \\ 0 & 0 & 1 \end{matrix}] A

A^{- 1} = [\begin{matrix} ω^{2} & 0 & 0 \\ 0 & ω & 0 \\ 0 & 0 & 1 \end{matrix}]

Hence, the correct answer is option: 'A'.

If ω is a complex cube root of unity and A=ω000ω20001 then A-1=…

Solution

Relevant Questions

If $ω$ is a complex cube root of unity and $A = [\begin{matrix} ω & 0 & 0 \\ 0 & ω^{2} & 0 \\ 0 & 0 & 1 \end{matrix}]$ then $A^{- 1} = \dots$