Question

If $A=\left[\begin{array}{ccc}1& w& w^2\\ w& w^2& 1\\ w^2& 1& w\end{array}\right]$ and $B=\left[\begin{array}{ccc}w& w^2& 1\\ w^2& 1& w\\ 1& w& w^2\end{array}\right]$ where $w$ is a complex cube root of unity, then " $AB$ and $BA$ both are null matrices." that statement is _____?

• A. True
• B. False

Answer 1

The correct option is B False

Given

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

$\therefore AB=\left(\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right)\left(\begin{array}{ccc}\omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\end{array}\right)$

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

$=\left(\begin{array}{ccc}\omega (1+\omega +{\omega }^2)& \omega (1+\omega +{\omega }^2)& 1+{\omega }^2+{\omega }^4\\ {\omega }^2+{\omega }^4+1& \omega (1+{\omega }^{+}{\omega }^2)& \omega (1+\omega +{\omega }^2)\\ \omega (1+\omega +{\omega }^2)& \omega (+1+{\omega }^2+\omega )& \omega (1+\omega +{\omega }^2\end{array}\right)$ ( as $1+\omega +{\omega }^2=0$)

$=\left(\begin{array}{ccc}0& 0& 1+{\omega }^2+{\omega }^4\\ 1+{\omega }^2+{\omega }^4& 0& 0\\ 0& 1+{\omega }^2+{\omega }^4& 0\end{array}\right)$

$AB$ is not a null matrix.