Question

Find $A^{-1}$, if $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]$, where $\omega$ is the cube root of unity

• A. $\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& {\omega }^2& \omega \\ 1& \omega & {\omega }^2\end{array}\right]$
• B. $\frac{1}{3}\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& 1& 1\\ 1& \omega & {\omega }^2\end{array}\right]$
• C. $\frac{1}{3}\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \\ 1& 1& 1\end{array}\right]$
• D. none of these

Answer 1

The correct option is A $\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& {\omega }^2& \omega \\ 1& \omega & {\omega }^2\end{array}\right]$

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& \omega & {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right]$
Here, $|A|=3({\omega }^2-\omega )$
Now, $adjA=C^T={\left[\begin{array}{ccc}{\omega }^2-\omega & {\omega }^2-\omega & {\omega }^2-\omega \\ {\omega }^2-\omega & \omega -1& 1-{\omega }^2\\ {\omega }^2-\omega & 1-{\omega }^2& \omega -1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{ccc}{\omega }^2-\omega & {\omega }^2-\omega & {\omega }^2-\omega \\ {\omega }^2-\omega & \omega -1& 1-{\omega }^2\\ {\omega }^2-\omega & 1-{\omega }^2& \omega -1\end{array}\right]$
$\Rightarrow A^{-1}=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& \frac{1}{\omega }& \frac{-(1+\omega )}{\omega }\\ 1& \frac{-(1+\omega )}{\omega }& \frac{1}{\omega }\end{array}\right]$
$\Rightarrow A^{-1}=\frac{1}{3}\left[\begin{array}{ccc}1& 1& 1\\ 1& {\omega }^2& \omega \\ 1& \omega & {\omega }^2\end{array}\right](\because {\omega }^3=1,1+\omega +{\omega }^2=0)$