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 a complex cube root of unity, then show that

$\mathrm{AB}+\mathrm{BA}+\mathrm{A}-2 \mathrm{~B}$ is a null matrix.

Answer

$\omega$ is the complex cube root of unity.

$\begin{aligned} & \omega^3=1 \\ & \omega^3-1=0 \\ & (\omega-1)\left(\omega^2+\omega+1\right)=0 \\ & \omega=1 \text { or } \omega^2+\omega+1=0\end{aligned}$

But, $\omega$ is a complex number.

$1+w+w^2=0 \ldots \ldots$ (i)

AB + BA + A – 2B

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

$+\left[\begin{array}{cc}1 & \omega \\ \omega^2 & 1\end{array}\right]-2\left[\begin{array}{cc}\omega^2 & 1 \\ 1 & \omega\end{array}\right]$

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

$+\left[\begin{array}{cc}1 & \omega \\ \omega^2 & 1\end{array}\right]-\left[\begin{array}{cc}2 \omega^2 & 2 \\ 2 & 2 \omega\end{array}\right]$

$\begin{aligned} & =\left[\begin{array}{ll}\omega^2+\omega+2 \omega^2+1-2 \omega^2 & 1+\omega^2+\omega^3+1+\omega-2 \\ \omega^4+1+1+\omega^3+\omega^2-2 & \omega^2+\omega+2 \omega+1-2 \omega\end{array}\right] \\ & =\left[\begin{array}{ll}\omega^2+\omega+1 & \omega^2+\omega+1 \\ \omega^2+\omega+1 & \omega^2+\omega+1\end{array}\right]\end{aligned}$

$\because\left[\because \omega^3=1 \therefore \omega^4=\omega\right]$

$ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $ $\ldots[$ From (i)]

which is a null matrix.

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