Question
Easy
If $\omega$ is the cube root of unity, then the inverse of $A=\begin{bmatrix}1&1&1\\ 1&\omega&\omega^{2}\\ 1&\omega^{2}&\omega\end{bmatrix}$ is :
1
$\frac{1}{3}\begin{bmatrix}1&1&1\\ 1&\omega^{2}&\omega\\ 1&\omega&\omega^{2}\end{bmatrix}$
2
$\frac{1}{4}\begin{bmatrix}1&1&1\\ 1&\omega&\omega^{2}\\ 1&\omega^{2}&\omega\end{bmatrix}$
3
$\frac{1}{2}\begin{bmatrix}1&1&1\\ 1&\omega^{2}&\omega\\ 1&\omega&\omega^{2}\end{ matrix}$
4
$\begin{bmatrix}1&1&1\\ 1&\omega^{2}&\omega\\ 1&\omega&\omega^{2}\end{bmatrix}$
Question Details
Time to Solve: 12
Exam: HTET
Level/Paper: Level 3
Chapter: Algebra
Topic: Determinants
Correct Answer
Option A
Explanation
To determine the inverse of the matrix \( A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{bmatrix} \), where \(\omega\) is a cube root of unity, we need to understand the properties of cube roots of unity and how they affect matrix operations. ### Properties of Cube Roots of Unity: 1. \(\omega^3 = 1\) 2. \(1 + \omega…Read More
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