Question

If $A=\left[\begin{array}{cc}1& \omega \\ {\omega }^2& 1\end{array}\right],B=\left[\begin{array}{cc}{\omega }^2& 1\\ 1& \omega \end{array}\right]$ where $\omega$ is complex cube root of unity then $AB+BA+A-2B$:

• A. Upper triangular matrix
• B. Lower triangular matrix
• C. Unit matrix
• D. Null matrix

Answer 1

The correct option is D Null matrix

Given,

$A=\left(\begin{array}{cc}1& \omega \\ {\omega }^2& 1\end{array}\right),B=\left(\begin{array}{cc}{\omega }^2& 1\\ 1& {\omega }^2\end{array}\right)$

$AB+BA+A-2B=\left(\begin{array}{cc}1& \omega \\ {\omega }^2& 1\end{array}\right)\left(\begin{array}{cc}{\omega }^2& 1\\ 1& {\omega }^2\end{array}\right)+\left(\begin{array}{cc}{\omega }^2& 1\\ 1& {\omega }^2\end{array}\right)\left(\begin{array}{cc}1& \omega \\ {\omega }^2& 1\end{array}\right)+\left(\begin{array}{cc}1& \omega \\ {\omega }^2& 1\end{array}\right)-2\left(\begin{array}{cc}{\omega }^2& 1\\ 1& {\omega }^2\end{array}\right)$

$=\left(\begin{array}{cc}{\omega }^2+\omega & 1+{\omega }^2\\ 1+{\omega }^4& {\omega }^2+\omega \end{array}\right)+\left(\begin{array}{cc}2{\omega }^2& 1+{\omega }^3\\ 1+{\omega }^3& 2\omega \end{array}\right)+\left(\begin{array}{cc}{\omega }^2& 1\\ 1& {\omega }^2\end{array}\right)-\left(\begin{array}{cc}2{\omega }^2& 2\\ 2& 2\omega \end{array}\right)$

$=\left(\begin{array}{cc}1+\omega +{\omega }^2& \omega +{\omega }^2+{\omega }^3\\ {\omega }^4+{\omega }^3+{\omega }^2& 1+\omega +{\omega }^2\end{array}\right)$

$=\left(\begin{array}{cc}1+\omega +{\omega }^2& \omega (1++\omega +{\omega }^2)\\ {\omega }^2(1++\omega +{\omega }^2)& 1+\omega +{\omega }^2\end{array}\right)$

$=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$ ( as $\omega$ is cube root of unity so $1+\omega +{\omega }^2=0$)

Hence $AB+BA+A-2B$ is null matrix.